The Sophistry of Backradiation

Another Tale of Two Versions

Sophistry finds its greatest expression and success in deceit when it can switch reference frames, goal posts, and contexts without switching the language.  We’ve seen this already in the two versions of “the greenhouse effect”, where we have two physically distinct mechanisms, one of which exists in reality and the other of which is a simulacrum of that reality upon which many other lies can and have been built, and where both use the same label.

It has now become apparent that the term “backradiation” is likewise being used by the pseudoscientific, pseudomeritocratic climate propaganda establishment in a sophistical manner to generate additional cognitive dissonance and sophistry.

 Form vs. Content

There are two ways to understand the term “backradiation”. The term is sophistical in and of itself, but the two versions have a difference which the sophists exploit to create cognitive dissonance.

The first most obvious way is in terms of its content, i.e. in the terms of what it is functionally required to do as part of the mechanism of the alarmist radiative greenhouse effect from which the term itself originates.  And that function is to cause heating, is to cause a surface source of thermal radiation to become even hotter still because its own radiant thermal energy is sent back to it (i.e., backradiation) after thermal absorption and re-emission at a cooler target.  More generally, the function is that the thermal radiation from a cool object will be absorbed by and thus cause heating (i.e. temperature increase) on a warmer object.  This is what is occurring in the diagram below from the University of Washington Department of Atmospheric Sciences.

 

greenhouse

 

The concept here is of course sophistry because radiation from a cool object does not heat up a warmer object – heat doesn’t flow from cold to hot.  It is sophistry because there is no such thing as “backdiffusion” of heat energy in thermodynamics, and diffusion works via the same fundamental electromagnetic force of physics that radiative transfer is mediated by.  If there is backdiffusion heating then there will be backradiation heating, and if there is not backdiffusion heating then there will not be backradiation heating.  There is no such thing as backdiffusion heating.

The second way to use the term “backradiation” is in terms of its form, i.e., in simple direct reference to the thermal radiation that a cooler object emits (with context in the vicinity of a warmer object).  All surfaces with non-zero emissivity and non-zero temperature will emit thermal radiation, and so whether we call a surface warm, or cold, etc., it does indeed emit thermal radiation.  This is sophistry because it is an invented term – there’s no need to call radiant energy from a cooler object “backradiation” just as there isn’t a need to call conductive energy  from a cooler object “backdiffusion”, because heat does not “backdiffuse” from a cooler object to a warmer one.  The label is of course invented in order to attach it to the idea that radiant thermal energy from a cool object must be absorbed by and thus increase the temperature of a warmer object.  Of course, there is no concept in thermodynamics that thermal energy will or even can diffuse from a cool object to a warmer one, just because the cool object contains thermal energy; in fact, this directly contradicts the set of Laws of Thermodynamics.

Energy does not physically, by physical conductive diffusion, transfer heat from a cool object to a warmer one.  However, the sophists have exploited various anthropocentric sense-perception-based propensities of the sensing-type majority population (and essentially all of science) to confuse them about what energy might do across a “non-physical” gap it might “jump” via radiation, as discussed in a previous post.  Of course, it doesn’t matter if the energy transfer is by physical diffusion or across a gap by radiation.  Either way, heat only transfers from hot to cold and so therefore the cold object can not increase the temperature of the warm object.  If it confuses them so as to how energy can “know” not to transfer heat radiatively from cold to hot across a gap, they should likewise be confused as to how energy can “know” not to diffuse heat conductively from cold to hot with objects in physical contact.  It is simply because they are sensing types that they accept the latter without question, but are so exploitable to confusion and sophistry with the former.

The Sophist Switch

The Slayers have always said that “backradiation doesn’t exist”.  We say that because “backdiffusion” doesn’t exist.  What it is supposed to do doesn’t exist, and what it is supposed to do is to cause a warmer object to become warmer still due to the thermal energy from a cool object.  This is only possible if heat flows from cold to hot, which it of course doesn’t. Hence backradiation doesn’t cause heating and likewise everyone already knows that backdiffusion doesn’t exist either.

At this point the sophist radiative greenhouse alarmist will switch the usage of the term “backradiation” to its form, and accuse us of denying that a cold object emits any radiation at all.  They make the sophist assumption that if you reject the notion that a cold object can heat a warmer object, then you must deny that backradiation exists in the form of radiant energy from a cooler object at all, because they implicitly assume that any radiation whatsoever can cause heating irrespective of its source temperature.  They either won’t admit to, or can’t understand, that while the thermal energy from a cool object can and does indeed exist, this energy is not capable of passing heat to a warmer object; this is true with either diffusive, or radiative, energy exchange.

If we look at either the radiative or diffusive heat flow equations (respectively)

Q’ = A*σ*(Thot4 – Tcool4)

Q’ = A*k*(Thot – Tcool)

we see that there are indeed two terms in the equation: a hot term and a cool term.  Only the greater portion of the energy can act as heat, and the greater portion is of course only available from the hotter object both in terms of quantity and more importantly in terms of quality, in terms of having higher frequency energy components that the cooler object does not contain.

Heat flow is of course all about Fourier Mathematics, and Fourier math is all about waves and the frequencies of those waves.  Higher temperature means higher frequency energy waves, and so these are higher frequencies which the lower temperatures do not contain.  Lower temperature can not increase the frequencies of a higher temperature, and thus increase that higher temperature’s temperature, because it doesn’t contain the higher frequencies required to add to the higher temperature.

difference of two planck curves

The Follow On

It is very funny sophist behaviour to watch, because after I write out the heat flow equation for them and explain that it has two terms and that only the difference can act as heat and raise the temperature of the cool object, they always then ask if I deny that the cool object emits any energy at all!

So let’s get this straight:  Do you see the radiative heat flow equation in the next line?

Q’ = A*σ*(Thot4 – Tcool4)

The equation has two terms – a hot term and a cool term.  Since the terms are there, then it means that both objects emit energy.  Therefore it would be impossible to write and use that equation while denying that one of the energy terms doesn’t exist.  The problem for the sophists, is that they are unwilling to admit that only Q’ is heat.

The equation defines that only the difference of energy is heat.  Each term is not itself heat.  Repeat:

Each term is not itself heat.

Only the difference of the energy is the heat and this heat flows from hot to cold and can only cause the cold object to raise in temperature.  You can only call A*σ*Thot4 or A*s*Tcool4 “heat” when they are by themselves with absolute zero background temperature, but even then it doesn’t make sense since empty space can’t be heated, i.e., can not attain a temperature, and so those terms are more properly still just total energy emission to space. Q’ is just the portion of A*σ*Thot4 which is acting as heat for the cooler object.

There is no two-way flow of heat which also has a “net” heat.  If there was a two-way flow of heat, then the cold object would raise the temperature of the warm object, and you just violated thermodynamics.  And think of that logic: if the cool object heats the warmer object, then as the warmer object heats the cooler object, the cooler object will cause the warmer object to warm up some more, requiring the cooler object to become warmer, resulting in the warmer object becoming warmer leading to the cooler object becoming warmer, which causes the warmer object to warm…ad infinitum.

This is a very basic logical sequence and the infinite recursion produced is there to debunk itself.

And even before that, a cold object can not send heat to a warmer object; heat does not diffuse from cold to hot and therefore heat does not radiate from cold to hot since in both cases the transfer of heat is mediated by electromagnetism.  What is so hard to understand about heat not flowing from cold to hot?!

There is two-way flow of energy; there is not two-way flow of heat.

Demo

It has now become clear that another thing which is confusing people is calling the thermal emission from an object “heat”.  This is wrong.  If we have for a single hot source

A*σ*Thot4

this term is actually its power, P = A*σ*Thot4, not heat!  This term is its energy emission, not its heat.  Heat is only the portion of the energy which can act to increase the temperature of another object, hence why heat is written as

Q’ = A*σ*(Thot4 – Tcool4)

You could only equate P with heat if it was emitting towards an object with absolute zero temperature, so that

Q’ = A*σ*(Thot4 – 04) = A*σ*Thot4 = P

Otherwise, heat is only the portion of P which is above and beyond A*σ*Tcool4.

Here is what goes wrong when you call any emission from any source “heat”.  Following on from the Gift of the Steel Greenhouse post, for an infinite plane which is perfectly insulated on its backside and is perfectly conductive, the differential equation for the temperature of a section of the plane of area A is:

1]                           mCp*dT/dt = ∑Q’i

where ‘m’ is the mass of the section, Cp is its thermal capacity, T is its temperature (and is actually T(t), a function of time, but I’ll just write T instead of T(t) for simplicity), and ∑Q’i is the sum of the “heat” flow into the section.  We start with the section having its own internal power source P0, and so

2]                           ∑Q’i = Q’0 = P0 – AσT4

Subbing that into equation 1 we get:

3]                             mCp*dT/dt = P0 – AσT4

Thermal equilibrium for the section is when the rate of change of temperature goes to zero, i.e., is when the temperature stops changing, and so the equilibrium temperature is:

4]                         mCp*dT/dt = 0 = P0 – AσT04

Solving for T0:

5]                              T0 = [P0/(Aσ)]1/4

which is a fixed constant value.

Now we bring in another plane with a small gap spacing between it and the original powered plane.  The new plane is not powered, not insulated, and it doesn’t have to be perfectly conductive.  As the powered plane raises in temperature it emits the requisite energy, and this energy heats the new plane.  And so, as the new plane is heated, it then emits its requisite energy and we must call this energy “heat”, and we must assume that it is absorbed by the original plane, and thus must add to the temperature forcing for the original plane.  The energy re-emitted by the new plane is simply equal to the energy that it receives which generates its temperature, that energy being AσT4 from the original plane. Thus, subbing the “new heat input” to the original plane with the new plane in place, we get:

6]                           ∑Q’i = P0 – AσT4 + AσT4

Subbing that into equation 1 we get:

7]                             mCp*dT/dt = P0 – 0

which has the solution

8]                              T(t) = [P0/(mCp)]*t

which goes to infinity as t goes to infinity.  That is, it grows without bound.

This is the runaway heating problem you get when you consider any emission from an object to be heat, and that it must be absorbed as heat by any other surface irrespective of the direction of heat flow.  The solution going off to infinity shows that it is wrong, not right!

Look at the diagrams below and try to visualize which one demonstrates the logical, thermodynamic, reality-based sequence of energy emission and heat flow, and which one doesn’t.

radiative

Sequence A

 

climate thermo

Sequence B

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144 Responses to The Sophistry of Backradiation

  1. blouis79 says:

    Sequence B is the thermosphere. I think the thermosphere is a measiurement error.

  2. blouis79 says:

    Or maybe the thermosphere is just another example of scientific madness. Wikipedia:Thermosphere “The highly diluted gas in this layer can reach 2,500 °C (4,530 °F) during the day. Even though the temperature is so high, one would not feel warm in the thermosphere, because it is so near vacuum that there is not enough contact with the few atoms of gas to transfer much heat. A normal thermometer would be significantly below 0 °C (32 °F), because the energy lost by thermal radiation would exceed the energy acquired from the atmospheric gas by direct contact.”

  3. The behaviour in the thermosphere isn’t the correct context or physics here, relevant to the diagrams etc.

  4. DurangoDan says:

    Joe, I admire your tenacity in refuting the atmospheric radiative greenhouse effect in so many ways. Once it is understood that cooler cannot heat warmer and that photons must have a higher frequency than the photons emitted by the receiving body in order to heat that receiving body, what is left? Unfortunately, scientifically proving the falsity of the greenhouse effect is just a baby step in convincing the masses of the complete fraud of anthropogenic climate change. I understand that it can be intellectually insulting to be a human, but at least we can appreciate George Carlin’s assessment of the human condition when he said, “the average person is a moron and half of them are dumber than that”. So true.

  5. geran says:

    I have some color deficiency in the old eyeballs, so these color exams are somewhat difficult. But, I think “A” is the real world, and “B” is the IPCC/GHE nonsense.

  6. Greg House says:

    Joe, is there a clear physical evidence that a colder object does emit IR specifically towards a warmer object? Like if we had 2 parallel planes, a cold one and a warm one. Two terms in an an equation can not be automatically considered the evidence for the existence of two processes going in the opposite directions, I thought you knew that. So is there any direct physical evidence, experimental maybe?

  7. markstoval says:

    This post clarifies a host of issues. It cuts to the very heart of the matter. You have explained most of this in other posts and at other times, but this one puts it all together in one place. You also wrote this piece in very understandable language which, I think, most educated men and women can understand; and no advanced degrees are needed to follow this line of reasoning.

    Now, can we get that physicist from Duke to read this post and understand it?

  8. And cheers Mark!

  9. Cheers DurangoDan. Well said.

  10. Yep that is correct Geran.

  11. Well Greg it is a good question, because we can ask about a direct analogy: Does a cooler object diffuse energy specifically towards a warmer object? There is an exchange or balancing of energy, sure, as in the equation, but does any energy actually diffuse from the cold object into the warm object? We wouldn’t really say that, would we.

    Hence my post about ontological mathematics and whether photons really see the gap or not. Radiative transfer is simply electromagnetism’s method for diffusing heat at what we experience as a distance…but the photons don’t really experience that distance from their own point of view.

  12. Greg House says:

    There is no balancing of energy as such in the equation, two terms are not the evidence for that, since there can be one process expressed by two terms. There is the equation and the question is if there is any evidence for cold radiating towards warm, specifically, if there is any physical evidence for it, like an experimental one.

  13. Well, if you do an experiment to measure the radiation from the cooler source, yes you can measure it. But that only means that the experiment was designed to measure it in situ to the experiment itself – it used a non-thermal electronic device with an appropriate photocathode or photodetector of some type that doesn’t function on thermal principles but on photoelectric principles. So does it mean that radiation travels from cold to hot? You can’t actually measure that, and you didn’t actually measure that with the spectrometer or IR detector or whatever. If you want to measure if radiation from a cold object heats up a warmer object, that certainly has been done and we certainly know that cold doesn’t heat up hot. Does it mean that radiation from cold didn’t travel to the hot at all? Or just that there was bosonic overlap of energy with no effect? Does the heat flow equation mean that radiation only travels from hot to cold, or that there is two-way travel of energy with only a balance which can act as heat?

    We do know that whichever way it goes, it doesn’t actually make a difference. You could interpret it either way, because both interpretations still result in heat only transferring from hot to cold and energy still being conserved.

    Hence, why I wrote the post about ontological mathematics and energy transfer from the point of view of the photons themselves, and electromagnetism and diffusion. What do they say?

    If energy does leave the cool object towards the warm object, it is immediately replaced by the warmer object (plus some), and so the portion of energy that left the cool object can be said to have not left at all since you can’t actually distinguish one unit of energy at a particular frequency from another.

    Then again, there should be a time delay in our non-speed of light reference frame if we have the two objects separated by great distance. Of course you need a giant space of vacuum and no other sources around. But if you had the perfect space for the experiment, then you could have two sources suddenly turned on. The cool one could be heated to some temperature and then power removed, and the hotter object could be sustained with a fixed input power. Measure the rate at which the cooler object cools, and when it begins warming from the warmer object; if the objects are separated by a light hour, then you could see when things start happening. Of course, you’d have to have high enough temperatures and large enough objects (i.e. large enough amounts of energy) to have an effect across that range of space.

    Anyway, back to the question: what are the actual energy exchanges?

    The Q’ equation is based on terms which have their own independent existence taken individually: Q’ = A*s*(Thot^4 – Tcool^4). The two terms are real terms that indicate real energy emission from each object taken individually. Their difference indicates a function of energy, called heat, or heating. So I say that they do exist. There is a two-way exchange of energy, with only the difference balance acting as heat. Still, does that exchange of energy indicate that energy from the cool object traveled to the hot object and had no effect because of bosonic overlap or resonance or something, or that the energy from the cool object wasn’t emitted at all? I would side with the former, because I do think that if there were a significant gap between the objects, the cool unpowered object would cool at a faster rate until the radiation from the warmer object reached it and began heating it. Say the cool object is a plane. Does it cool, before the radiation from the warm object reaches it, at the rate proportional to both sides of its surface area, or just the one side facing away from the hotter powered object?

    Well, that would only indicate that radiation had left it from both sides until the time where the radiation from the warm object arrived. At that point, is energy leaving the side facing the warmer object? Well, whatever radiation that side emits, it is replaced instantaneously and perfectly by the radiation from the warmer object.

    I say that it is emitted; there is a two-way exchange of energy, with only the difference balance acting as heat. The radiation from the cool object does not act as heat for the warmer object. Each term is not itself heat; only the greater portion of the energy from the hot term acts as heat.

    Radiation from cool does not raise the temperature of hot, as likewise diffusion from cool does not raise the temperature of hot.

  14. Greg House says:

    So I take it as there is no physical evidence for cold radiating towards hot, anyway you are not familiar with one. Let’s see if anyone else can present one. Not a “thought experiment”, of course, or suggestion how humble me COULD prove it experimentally if I wish. But as for now we have zero evidence.

    The second thing is the equation containing two terms, but we know that a single one-way process can be expressed in two term as well, so here again we do not have an evidence for “both ways”.

    Then I allow me to humbly suggest this “both ways” mantra to be dropped and get completely lost forever. It is not just inaccurate, but in my humble opinion confusing and counterproductive. Except for a small fans community I am sure such statements can only turn readers away. It would be sufficient to say one way or another that warming by back radiation is equivalent to infinite mutual heating or to production of energy out of nothing, without claiming that it exists.

  15. It is indeed the “both ways” thing that they use to sophize with, which was discussed in the OP here of course. They say that “if you admit there is two way energy exchange, then cold must then heat up hot since the cold radiation must be absorbed by the hotter object if you admit it traveled to the hotter object.”

    It’s sophistry and bullshit of course, and stupidity. Because each term is not itself heat in any case. Only the difference is heat, as per the equation. As to what happens to the energy from the cool object…nothing happens to it, and it certainly doesn’t raise the temperature of the warmer object anyway! Idiots.

  16. DurangoDan says:

    Sorry for not providing the citation at this time, but a paper I reviewed recently reported that heat transfer, following surface or atmospheric absorption of solar photons, within a system such as that comprising the earth surface and troposphere is almost completely by conductance which causes disparities in gaseous density which then generates convection. Radiative heat transfer such as infrared surface emittance or even atmospheric backradiation is virtually nonexistent in this system and so the question of whether photons go back and forth is not relevant. Perhaps this answers Greg’s question. No, backradiation doesn’t occur because the photons are not emitted because heat conductance prempts it. I suppose this leads to the conclusion that nature prefers conduction and only engages in radiation when a vacuum requires it. Sounds kind of strange but elegant.

  17. Arfur Bryant says:

    Joe, Thanks for this post.

    Following from Greg’s question about energy from cold objects…

    If we accept that all objects emit radiation, and this therefore includes, radiation between cold (C) and warm (W) objects, what actually happens to a photon emitted by C wen it arrives at W? My understanding is that an incoming photon can either be:
    a. transmitted
    b. deflected
    c. absorbed
    In the case of W to C, it would seem likely that the higher energy (higher frequency) photon is absorbed for thermal energy gain by C thus enlarging the electron orbit and increasing the temperature (average kinetic energy) of the molecules of object C.
    In the case of C to W, no such ‘absorption for energy gain’ can happen as the incoming photon frequency/energy is insufficient to raise the electron orbits beyond the maximum they already are. Therefore, the incoming photon from C is either transmitted or deflected (possibly depending on the physical nature of the surface of W). Or is it…?

    Is it possible that the photon from C is ‘absorbed for NO energy gain’ by W, or would that count as being deflected – i.e. being absorbed and instantly re-emitted at the same wavelength/frequency?

    Cheers,

    Arfur

  18. Arfur Bryant says:

    Sorry, didn’t realise a W and parentheses would end up being a WordPress logo! 🙂

  19. blouis79 says:

    What is missing in the two planes scenario is the “greenhouse gas” which does the work compared to a small gap of vacuum.

  20. johnmarshall says:

    Dr. Roy Spencer has reported that he has measured radiation, low frequency IR, from space. But he cannot determine whether this actually heats the warmer surface. So it is not evidence as we would like. My take is that every object above 0K emitts radiation but does not, cannot, heat warmer objects.
    Thanks Joe.

  21. DurangoDan says:

    Here’s a link to the source article that I paraphrased above perhaps not quite accurately: http://greenhouse.geologist-1011.net/ This is Tim Casey’s “The Shattered Greenhouse”. It’s well worth reading.

  22. Tim Casey is highly recommended.

  23. DurangoDan says:

    Here’s one of the relevant statements from “The Shattered Greenhouse”: “As you can see, Fourier admits that his work is constrained to the net movement of heat. In fact, nowhere does Fourier differentiate between radiative and, for example, “kinetic” heat transfer, because the means to tell the difference were not available when Fourier studied heat flow. What this tells us is that Fourier’s Law, and only Fourier’s Law, can describe the transfer of heat between bodies in thermal contact. Thus the distribution of heat between the atmosphere and the surface of the earth, with which it has thermal contact, cannot be correctly calculated using the radiative transfer equations derived from Boltzmann (1884) because the thermal contact of these bodies makes this a question of Fourier’s Law.” The paper explains that inclusion of back-radiation (the greenhouse effect) results in double counting heat transfer which is already included in conductive heat transfer. This has the effect of creating a second heat source in addition to the sun.

  24. Bryan says:

    Orthodox physics interpretation of photon two way flow plus suggested experiment.

    Two metal blocks A and B sit separated inside a vacuum filled adiabatic enclosure.
    Adiabatic means no heat can enter or leave the enclosure so no room for tricks

    Adiabatic enclosure consists of a perfect reflector face surrounded by a perfect insulator

    Initially both at the same temperature. The zeroth law of thermodynamics applies.
    Both emit and absorb equal amounts of radiation.

    Neither one is said to heat the other.

    One block (A) has a power supply which is now switched on causing the temperature of the block to rise.
    This in turn means that it will emit more radiation.

    A will now heat B causing its temperature to rise.
    B will in turn emit more radiation but this back radiation is caused by A.

    Now comes the clincher

    If B were not there at all the temperature A would be even higher.

    So B cannot be said in any meaning of the word as a cause of heating A

    Reply

  25. solvingtornadoes says:

    Greg House:
    “Joe, is there a clear physical evidence that a colder object does emit IR specifically towards a warmer object?”

    ST:
    Of course there is. The rate of cooling of the warmer object is less than it would be without its proximity to the cooler object. How was this ever not obvious?

  26. solvingtornadoes says:

    JP:
    “It’s sophistry and bullshit of course, and stupidity. Because each term is not itself heat in any case. Only the difference is heat, as per the equation. As to what happens to the energy from the cool object…nothing happens to it, and it certainly doesn’t raise the temperature of the warmer object anyway! Idiots.”

    ST:
    Joe, there are three different things at play here and unless you are very careful, specific, and deliberately explicit you will only open the door to more sophistry. There is energy, a noun. There is heat, a process. And there is increase or decrease in measured temperature, what is commonly called heating. If you are not perfectly clear as to which of these three you intend you will only open the door to confusion and more sophistry.

    If you pave the road to sophistry you can’t criticize those that travel it.

    I suggest that you reread what you wrote above and be extremely critical of where there might be any confusion as to which of these three meanings you intend in each instance and revise to remove any possibility of misinterpretation.

    Jim McGinn
    http://www.solvingtornadoes.com

  27. solvingtornadoes says:

    Greg House:
    “So I take it as there is no physical evidence for cold radiating towards hot, . . .”

    ST:
    Is there any evidence that it doesn’t?

    Greg House:
    or suggestion how humble me COULD prove it experimentally if I wish. But as for now we have zero evidence.

    ST:
    If you attempt to disprove it you will fail.

    Greg House:
    I allow me to humbly suggest this “both ways” mantra to be dropped

    ST:
    Based on what? You just confirmed you have no experimental evidence that contradicts it. Right?

  28. Bryan I think you meant that the radiation from B will cause A to get even hotter, if there is a radiative greenhouse effect, i.e. backradiation heating.

    Of course, however, the passive block B would simply be heated by the radiation from A, and the energy from A would then pass through B with conservation of energy and thermal equilibrium applying.

    In the OP here I already discussed this experiment using “infinite” planes, where all of the energy emitted from either plane will be absorbed by the other. In this case, if backradiation caused heating, then you get the runaway heating problem, instead of simply heat transfer through the passive plane with the equilibrium temperature being that of the power input per unit area to plane A.

  29. martin h says:

    To ST at 9:22 Actually, no, it is not obvious. Nothing I have ever encountered acts like that.

  30. solvingtornadoes says:

    Martin, have you never encountered a sweater, jacket, or coat?

  31. Arfur Bryant says:

    solvingtornadoes

    [“ST:
    Of course there is. The rate of cooling of the warmer object is less than it would be without its proximity to the cooler object. How was this ever not obvious?”]

    The rate of cooling of an object depends simply on the temperature difference between the warmer object and its surroundings (including the ‘target’ object). The mere presence of another cooler object can either reduce or increase the rate of cooling depending on the cooler object’s temperature. So no, it’s not obvious.

    And, while we’re being “specific”…

    ST[“Joe, there are three different things at play here and unless you are very careful, specific, and deliberately explicit you will only open the door to more sophistry. There is energy, a noun. There is heat, a process.”]

    Well, actually, heat is also a noun. 🙂

    However, I do agree with you that radiation is emitted from both the warm and cool objects (and in all directions). Heat, though, only flows one way. Any suggested reduction in rate of cooling is entirely dependant on the temperature difference, not the energy transfer. In the wider cAGW debate, I doubt the relevance of CO2 in this regard.

    Regards,

    Arfur

  32. markstoval says:

    I saw this exchange at 9:22 above:

    Greg House:
    “Joe, is there a clear physical evidence that a colder object does emit IR specifically towards a warmer object?”

    ST:
    Of course there is. The rate of cooling of the warmer object is less than it would be without its proximity to the cooler object. How was this ever not obvious?

    This made me think of a real world experiment. Take a stove and heat a pot of what to boiling temperature (100C) and then set it on the counter away from the heat and measure its rate of cooling. Now with the room temperature the same (a/c set at same temp should do it — 23C?) we do the same thing but set the pot of boiling water (100C) next to a pot of cooler water which is at say, 50 C, and then check the rate of cooling of the first pot (the pot of boiling water) again.

    As I understand it, S.T. claims the cooling rate of the first pot would be changed and Greg House says it would not change. Would this experiment settle the matter?

    ~ Mark

  33. DurangoDan says:

    Mark, Based on what Tim Casey has to say in “The Shattered Greenhouse”, thermal conductance will eclipse any radiant heat transfer and so you would not see any difference except what might be attributable to interference in convective air flow depending on the positioning of the pots. This experiment would fail unless conducted in a vacuum.

  34. solvingtornadoes says:

    Arfur Bryant says:
    The mere presence of another cooler object can either reduce or increase the rate of cooling depending on the cooler object’s temperature.

    ST:
    I agree, the cooler object would just have to be warmer than the surroundings.

    AB:
    Well, actually, heat is also a noun.

    ST:
    True. I guess I’m saying that heat (in contrast to what most people first assume) is a process, not a thing. Energy is a thing. Heat is the process by which energy changes its location.

    AB:
    However, I do agree with you that radiation is emitted from both the warm and cool objects (and in all directions). Heat, though, only flows one way.

    ST:
    Well, this is where things get real confusing. Being a process that is the result of energy changing its location, heat actually travels in all directions due to the fact that energy travels (changes location) in all directions also. Heating, being the measurable increase in temperature that results from the process of heat, only takes place in the cooler of the two objects. Heating is not a process. It is the result of a process. Heating sounds like a process, but it’s not. It’s the result of a process. Heat sounds like a thing, but it’s not. It’s a process.

    So, heat travels in all directions. Heating takes place in the cooler of the two objects and reduction in rate of cooling takes place in the warmer of the two objects.

    AB:
    Any suggested reduction in rate of cooling is entirely dependant on the temperature difference, not the energy transfer.

    ST:
    Same difference.

  35. solvingtornadoes says:

    MS:
    This made me think of a real world experiment. Take a stove and heat a pot of what to boiling temperature (100C) and then set it on the counter away from the heat and measure its rate of cooling. Now with the room temperature the same (a/c set at same temp should do it — 23C?) we do the same thing but set the pot of boiling water (100C) next to a pot of cooler water which is at say, 50 C, and then check the rate of cooling of the first pot (the pot of boiling water) again.

    As I understand it, S.T. claims the cooling rate of the first pot would be changed and Greg House says it would not change. Would this experiment settle the matter?

    ST:
    Yes, that would settle it.

    However, it might be difficult detecting temperature differences in pots of water. It might be easier to just use two metal plates.

  36. Truthseeker says:

    DurangoDan,

    If you conduct the experiment in a vacuum, both pots of water will “boil” before you apply any heat to them at alll …

  37. Pingback: Postma’s Obsession | Solving Tornadoes

  38. Bryan says:

    Joseph E Postma says
    “Bryan I think you meant that the radiation from B will cause A to get even hotter, if there is a radiative greenhouse effect, i.e. backradiation heating”

    No not quite.

    An adiabatic enclosure would have a reflective inner surface to stop radiant heat escaping.
    It would also have a insulated outer surface.
    A hollow polystyrene box with inner surface lined with polished aluminium would suffice for an experimental set up.
    The radiation from the powered object is emitted in all directions hitting container walls and being reflected .
    .The unpowered object would absorb some radiation and its temperature increase and it would in turn emit more radiation.
    The powered object would absorb some reflected radiation from the walls and some emitted radiation from B
    The powered object temperature would rise.
    This is no surprise because the heat cannot escape and the power is switched on.
    If the unpowered mass were not there then the temperature of the powered mass would be even higher.
    This is easy to see as without it being there E = CM(delta)T loss to the unpowered mass would not occur.
    So the colder object cannot be said to heat the warmer object

    NOW CONSIDER THE CLIMATE SCIENCE VERSION
    Instead of an adiabatic enclosure to study radiant heat transfer climate science, uses empty space as the surroundings (-273K) which means the maximum heat loss is from the powered source without the presence of the unpowered mass
    With the unpowered mass there , the heat loss would be less as it provides a partial insulating effect.
    This they say means the colder ‘warms’ the the hotter

    In basic thermodynamics theory the nature of heat transfer is studied first in an isolated system i.e. with the objects inside an adiabatic enclosure.
    This is why all reputable physics textbooks say heat flow is always spontaneously from a higher to a lower temperature

  39. What do you mean by “Postma’s Obsession”, ST?

  40. The two plane parallel walls is a much easier way to go, and is a good thought experiment that demonstrates the correct way to understand heat flow, and uses very simple math that anyone can follow. It is one of the most common example scenarios in any heat flow textbook. The second cooler (and in our case passive) wall does NOT cause the first wall to become hotter still; rather, heat either diffuses or radiates into the second wall and propagates through it, with the maximum temperature being that of what the original wall was on its own.

  41. Greg House says:

    markstoval says: “As I understand it, S.T. claims the cooling rate of the first pot would be changed and Greg House says it would not change.”
    ============================

    I did not say anything about cooling rate. Please do not lie.

  42. Arfur Bryant says:

    solvingtornadoes

    ST[“So, heat travels in all directions. Heating takes place in the cooler of the two objects and reduction in rate of cooling takes place in the warmer of the two objects.”]

    The first sentence is where we disagree. Heat does NOT travel in all directions. It only flows from hot to cold. If heat travelled in all directions, the addition of more cool objects would result in the warm object becoming even even warmer at some point. This does not happen. Anywhere. Radiation, however, does travel in all directions.

    For something to get warmer, its temperature must increase. If you add one cup of water at 10C to another cup of water at 10C you just get a (larger) cup of water at 10C. You have added energy but not made the end product warmer. For radiation (or conduction), only the transfer of (higher) energy from a hot object/surface/whatever to a cooler object/surface/whatever will result in the cooler object getting warmer. This is most definitely NOT what happens with so called ‘backradiation’. The radiation emitted by the cooler object is completely irrelevant to the warm object. It has no effect thermally because it does not have the ability to increase the thermal energy of the receiving molecules. An increase in temperature can only result from a raising of the internal (thermal) energy.

    The idea that backradiation from atmospheric CO2 can cause a reduction in cooling from the planet’s surface is misplaced.

    Regards,

    Arfur

  43. solvingtornadoes says:

    Joseph E Postma says:
    2015/05/28 at 9:59 AM
    What do you mean by “Postma’s Obsession”, ST?

    Originally, I didn’t even plan to publish that post. I saved it on my website when you didn’t first approve my post (it wouldn’t be the first time) and even sent a tweet to Mark Stoval in response to his question.

    No offense intended.

    Nevertheless, Joe, you have to admit, you are a bit obsessed with the back radiation issue. Personally, I think your talents are wasted on it. There are a lot other issues in regard atmospheric flow that are habitually (and with extreme prejudice) ignored by meteorologists.

  44. Well said Arfur, but I would make the careful distinction on the last sentence that backradiation from the atmosphere can not cause heating, i.e. heat transfer, to the surface, and raise its temperature.

  45. Well ST the backradiation and GHE thing is kind of the scam of the century.

  46. DurangoDan says:

    I just got back into this AGW thing recently having given up on the idea that people could be educated to understand that it is nothing but fraud. Anyhow, I agree with Joe that the GHE is the Achille’s heel of the scam and backradiation is the Achille’s heel of the GHE. I was just looking at the ScienceofDoom website and noted that they also use verbal deception in “proving” that backradiation doesn’t violate the 2nd LOT. Like the others, they try to make the case that photons from the cooler atmosphere must be absorbed by the surface and therefore must result in heating. They ignore the fact that if the cooler atmosphere is sending energetic photons to the surface, the warmer surface (being warmer) must be sending more energetic photons to the atmosphere and the net effect must be surface cooling. The whole scam is ridiculous but the people believe. Morons I tell you!

  47. Yep exactly Dan. I know the post at SoD you’re referring to and was thinking about doing a post on it. See if I have time.

  48. Arfur Bryant says:

    Joe,

    Thanks for the kudos. I take the inability of ‘backradiation from CO2’ to warm the surface as a given, However, the warmists and lukewarmists alike will then say “well, we didn’t actually mean it warms the surface, what we meant was it reduces the rate of cooling“.

    To me, the distinction is nugatory. The physics is the same. No reduction in the rate of cooling occurs because of ‘backradiation’ any more than no actual warming occurs because of it.

  49. Yes that is how they switch goal posts about it, for sure.

  50. DurangoDan says:

    ST might be on to something though and this relates to some of my previous comments referencing what Tim Casey has to say about heat transfer in systems in thermal contact. He says that conduction/convection is the primary mechanism for heat flows between the Earth’s surface and the atmosphere. Because we can see air, we tend to visualize radiative heat transfer as being the more important and this perception opens the door to the greenhouse gas focus. But if Tim Carey is correct and absorbtion/emittance plays only a very minor role in surface/troposphere heat flow, the GHE, if you choose to believe it (which I don’t), becomes far less significant. Anyhow, this might be worth exploring.

  51. DurangoDan says:

    Can’t see air, sorry.

  52. Arfur Bryant says:

    ST:[“Nevertheless, Joe, you have to admit, you are a bit obsessed with the back radiation issue.”]

    ST, you have to be kidding! 🙂

    The backradiation=cAGW argument is the fundamental, crucial error made by pro-cAGW supporters. It is precisely where Joe should be targeting his abilities as it is the area of physics which cannot be defended by them without resorting to sophistry!

  53. DurangoDan says:

    I can see how a warmist could point to reduced cooling on a humid cloudy night and say that the presence of so much greenhouse gas water vapor is radiating and slowing the rate of surface cooling. Of course this ignores the vastly reduced convection caused by the presence of the water vapor. Most would not understand this and the warmist would win again.

  54. Greg House says:

    DurangoDan says: “…The whole scam is ridiculous but the people believe. Morons I tell you!”
    =======================================

    It is not that simple. People who do not believe may also have some ridiculous reasons. Like “global warming pause” e.g. and some others. People make mistakes sometimes, including humble myself.

    The questions is how it could be explained to general public without scaring them with words like “thermodynamics” and “2nd law”. Another question would be how to avoid discussions on irrelevant points.

  55. solvingtornadoes says:

    Joseph E Postma says:
    2015/05/28 at 10:03 AM
    “The second cooler (and in our case passive) wall does NOT cause the first wall to become hotter still; rather, heat either diffuses or radiates into the second wall and propagates through it, with the maximum temperature being that of what the original wall was on its own.”

    ST:
    So . . . . now let’s address the issue that brought Mark Stoval to suggest the experiment . . . If you don’t know what I mean then, maybe, re-read his post and the posts that he was responding to.

  56. Arfur Bryant says:

    DurangoDan says:

    [“He (Tim Casey) says that conduction/convection is the primary mechanism for heat flows between the Earth’s surface and the atmosphere.”]

    General question:

    Is convection truly a form of heat transfer? I would argue that it is actually a form of heat transport, not transfer. Convection transports air from one level to another. Any heat transfer which takes place between molecules at the new level must surely be from either conduction or radiation.

    Having said that, I do agree that it plays a massive part…

  57. solvingtornadoes says:

    DurangoDan says:
    2015/05/28 at 2:50 PM
    “I can see how a warmist could point to reduced cooling on a humid cloudy night and say that the presence of so much greenhouse gas water vapor is radiating and slowing the rate of surface cooling.”

    ST:
    Well, yes, this helps us understand why deserts get so cold at night and why locations with high atmospheric moisture, let’s say San Francisco for example, stay so relatively warm at night.

    DD:
    “Of course this ignores the vastly reduced convection caused by the presence of the water vapor. Most would not understand this . . . ”

    ST:
    I don’t understand it. You are saying convection is vast? I think you need to establish this first. Then you have to describe how it is “vastly reduced,” by the presence of water vapor–which makes no sense to me. To me it seems you are asserting things here that have never been established empirically as best as I can tell.

    Maybe this is something you’ve not yet fully developed?

  58. Yes good point Arfur. Convection is really about heat diffusion into a fluid, with the fluid then moving on to wherever it is going. Diffusion of heat happens first, as the fluid is in contact, and then the fluid keeps moving on. Convection is about moving a heated parcel of fluid to somewhere else, with an exchange of fluid occurring at the position it is heated, and the heating occurs there via diffusive transfer…and/or by radiative transfer too in fact.

  59. DurangoDan says:

    ST. Okay I’m busted. Thought I might be bailed out by an astrophysicist but looks like I’m on my own. So here we go. When dry air is present, say a high pressure system, surface heat exchange takes place at the surface atmosphere boundary layer and the air almost instantaneously expands so it rises and cooler dry air takes its place and this cycle continues as the surface temperature tries to equilibrate with air temperature. When water is present in the surface layer and phase changes to water vapor the temperature doesn’t immediately change as the latent heat is stored in the molecular structure, but the density of moist air is less than dry air so it is buoyant compared to dry air. It rises and cools to the point where the dew point makes it condense forming a cloud. Once the cloud forms, the convection sort of gets jammed up and the convection process slows down. Hence heat transfer from the surface is slowed. I guess you could call this the actual GHE. Obviously CO2 does not contribute to this in any way nor does radiant heat transfer. Make sense?

  60. DurangoDan says:

    I forgot another important point. Our skin moisture evaporates faster in dry air, so we sense cooler air at night under dry conditions than under moist conditions at the same air temperatures so long as temperatures are less than skin temperatures. This causes the sense of relative cooling to be exaggerated. Even so the rate of cooling is faster in dry air than moist dry air per the previous explanation just not as much as we sense.

  61. Arfur Bryant says:

    DD and ST:

    Convection is reduced in moist air as follows:

    Convected dry air cools adiabatically at 3C/1000 ft
    Convected saturated (moist) air cools adiabatically at 1.5C/1000 ft in the lower atmosphere but this rate becomes closer to the DALR (dry adiabatic lapse rate) in the higher atmosphere (still troposphere) due to the paucity of water vapour at altitude. The difference between the DALR and SALR (1.5C/1000 ft) is because the water vapour in the convected parcel is condensing and releasing latent heat inside the parcel (adiabatic), thus reducing the cooling rate.

    When air is convected upwards, it cools at 3C/1000 ft until the air reaches its dew point, at which point the air becomes saturated and cloud forms. Further upward convection will result in cooling at 1.5C/1000 ft. If the convection trigger was thermal, this will reduce the convection rate. If the convection trigger was mechanical, it may not reduce the convection rate but the cooling rate will reduce.

    Apologies if you knew this already.

  62. johnmarshall says:

    All this discussion about heat change/loss is like knitting fog.
    Look at the basic heat equasion. Heat loss (from hot to cold) is proportional to the difference of the individual absolute temperatures to the 4th power.
    No argument.

    This means the greater the temperature difference the greater the heat loss of the hot article. A very steep expotential graph.
    One of the first experiments in 1st year physics at my school. Boil some water and measure cooling rate.

  63. Squid2112 says:

    ST, (and perhaps Arfur as well), please fellows, keep in mind you are conflating “heat” with something that can be moved around. It cannot! … “Heat” is not a “thing”, it is a “result”. The only “thing” that moves from place to place are the photons, those are not heat. They excite (or not) the molecules that they come into contact with. It is the molecules that are the heat (in essence). And the only way to excite a molecule further than it already is, is if it is contacted by a photon or other molecule of greater energy state. It absolutely must be by greater energy state, as no other mechanism can increase the energy state of the target molecule than an energy state greater than itself, otherwise, to has no increased energy effect whatsoever. This is precisely the reason why a cool object cannot, under any circumstances, heat a warmer object. It simply cannot be done and is a violation of nature physical laws that cannot be broken. This is why a Perpetuum Mobile is impossible.

  64. solvingtornadoes says:

    DD:
    ” . . . but the density of moist air is less than dry air so it is buoyant compared to dry air.”

    ST:
    The weight per volume of moist air is heavier, not lighter, than dry air. Therefore buoyancy is impossible and/or the exact opposite of what you indicate.

    http://solvingtornadoes.com/2014/08/29/the-fourth-phase-of-water/

    H.L. Mencken said, “For every complex problem there is an answer that is clear, simple…and wrong.”

  65. DurangoDan says:

    I googled “cloud cover slows nighttime cooling” and sure enough, almost all of the answers cited “downwelling infrared radiation from clouds. Most of the sites were meteorology based, many affiliated with universities. I think the falsifier in this explanation is that a car with the windows up does not get warmer on cloudy nights than the air around it. ST, please test this to confirm or refute. I’m sticking with my explanation which can be more concisely stated as moisture in air reduces the lapse rate and increases atmospheric stability thereby inhibiting convection.

  66. Arfur Bryant says:

    Squid 2112…

    No confusion here. This is what I said earlier in the thread:

    [“For radiation (or conduction), only the transfer of (higher) energy from a hot object/surface/whatever to a cooler object/surface/whatever will result in the cooler object getting warmer. This is most definitely NOT what happens with so called ‘backradiation’. The radiation emitted by the cooler object is completely irrelevant to the warm object. It has no effect thermally because it does not have the ability to increase the thermal energy of the receiving molecules. An increase in temperature can only result from a raising of the internal (thermal) energy.”]

    I agree with your post.

    Regards,

    Arfur

  67. solvingtornadoes says:

    Dan,

    I would suggest getting basic principles correct first. Most importantly, don’t do what everybody else does and dismiss contradictions. Don’t trust something just because it comes from a meteorologist or a university. Meteorological models of storm theory depend on the assumption that moist air is lighter than dry air. And, in fact, this notion is blatant nonsense.

    Embrace the contradiction. Contradictions are the key to opening the door to new insight.

    If moist air is heavier than dry air and convection is the only process that dictates atmospheric flow then one would have to conclude that all of the moist air in Earth’s atmosphere would hug the surface. The fact that it doesn’t proves that there must be some other factor or process involved. IOW, there MUST be something that meteorologists haven’t figured out yet.

    This post will give you a sense of where I am going with this:
    http://wp.me/p4JijN-by

  68. solvingtornadoes says:

    Arfur Bryant says:
    2015/05/29 at 1:24 AM
    “Convection is reduced in moist air as follows:”

    ST:
    Moist air is heavier than dry air. Therefore moist air has negative buoyancy.

    Check this out:
    http://wp.me/p4JijN-4y

  69. solvingtornadoes says:

    Squid2112:
    ST, (and perhaps Arfur as well), please fellows, keep in mind you are conflating “heat” with something that can be moved around. It cannot! … “Heat” is not a “thing”, it is a “result”. The only “thing” that moves from place to place are the photons, those are not heat.

    I agree with the spirit of what you are saying but I think there is a detail you need to get right. If I’m reading you correctly, you are suggesting there are two things, energy and heat. I am saying there are three, energy, heat, and heating.

    Energy is a particle. Heat is the process by which particles of energy change their location. Heating (and cooling) is the change in the measurable temperature that results from the process of heat.

    Only the last of these three is measurable, empirically detectable. So, all of our inferences are based on the last of these three, heating (not heat, which is immeasurable).

    If particles of energy travel in all directions at all times then heat must also (there is no getting around this, and no reason to need to get around it). Heating however (the increase in the measurable temperature of an entity) only goes from hot to cold.

    Squid2112:
    They excite (or not) the molecules that they come into contact with. It is the molecules that are the heat (in essence). And the only way to excite a molecule further than it already is, is if it is contacted by a photon or other molecule of greater energy state. It absolutely must be by greater energy state, as no other mechanism can increase the energy state of the target molecule than an energy state greater than itself, otherwise, to has no increased energy effect whatsoever.

    ST:
    You are conflating heat and heating.

    Squid2112:
    This is precisely the reason why a cool object cannot, under any circumstances, heat a warmer object. It simply cannot be done and is a violation of nature physical laws that cannot be broken.

    ST:
    I disagree. A cooler object can and does heat a warmer object. However, it cannot increase the measurable temperature of the warmer object. I can, and does, reduce the rate of cooling of the hotter object.

  70. johnmarshall says:

    #squid2112
    Exactly.
    Heat [JP Edit: Not heat, but temperature; heat is energy transfer, temperature is energy content] is an intrinsic property of matter and is due to the molecular kinetic energy. Molecular vibrations are proportional to the absolute temperature. The frequency of the vibrations is the absolute temperature in essence and the reason why temperatures below absolute zero are impossible. A negative frequency is not possible as any vibration is a positive temperature.

  71. markstoval says:

    Greg House says:
    2015/05/28 at 10:31 AM

    markstoval says: “As I understand it, S.T. claims the cooling rate of the first pot would be changed and Greg House says it would not change.”
    ============================

    I did not say anything about cooling rate. Please do not lie.

    Well, that seems to be the first time I have been called a liar for misunderstanding someone’s position on a matter of science. I notice you did not correct my words and say what you really meant since you say I am wrong. Interesting.

  72. markstoval says:

    Another try at a cheap experiment

    Since some say that my two pots of water at different temperatures close together in a room at normal room temperature would not prove or disprove anything to do with “back radiation”, I am going to try one more and see if it flies.

    Put a metal object on a bench at normal room temperature. Use ultraviolet heating by some UV lamp to heat up the metal object via radiation. Use a very sensitive thermometer to measure the temperature before the lamp is turned on and then after the lamp is turned on to show the amount of heating.

    Once we have heated the metal object via ultraviolet radiation, we try to heat it some more via cooler longwave radiation. I don’t know if a long wave lamp is an off the self device or not. Surely someone has tried to produce a longwave source. After all, we have spent mega billions on this CO2 fantasy by now. Notice that I did not set up the experiment in a vacuum. That is because the earth is not a vacuum and it is said that longwave radiation will heat the earth’s surface in the presence of the atmosphere. So, seems like the experiment should be in earth-like conditions.

    My field has always been math, and setting up experiments is not my best talent. I just think there should be some simple experiment that proves or disproves “heating by back radiation” to the common man. And, since the longwave radiation coming back (if it really does) from the atmosphere will not warm the warmer surface then I think using a longwave source should be sufficient to show it does or does not warm the warmer object.

    Thoughts anyone?

  73. Arfur Bryant says:

    solvingtornadoes:

    ST[‘ST:
    I disagree. A cooler object can and does heat a warmer object. However, it cannot increase the measurable temperature of the warmer object. I can, and does, reduce the rate of cooling of the hotter object.”]

    You really should rethink this statement.

    A cooler object can neither heat nor increase the ‘measurable temperature’ of a warmer object.

    As I said upthread, the only way a cooling rate can only be affected is by changing the temperature difference between a hot object and its surroundings. If you have a room at 20C and introduce a hot (hotter than 20C) baked potato into the room, the potato will cool at ‘Cooling rate X’ as it finds equilibrium with the room. If you now introduce another potato which is at room temperature the cooling rate will not change unless the cooler baked potato gets warmer than room temperature which it may or may not do depending on its proximity to the warmer object.. If, instead, you introduce a frozen potato (0C) to the room then the cooling rate will increase.

    So your three sentences (after you disagree) are all wrong.

    Finally,

    ST[“Moist air is heavier than dry air.”]

    Please provide a reference for this assertion.

    Regards,

    Arfur

  74. Jan-Ove Pedersen says:

    Your suggestion on a simple experiment is interesting.
    This experiment is not simple but it does show that back radiation does not exist. You may already read it.

    http://principia-scientific.org/publications/New_Concise_Experiment_on_Backradiation.pdf

  75. johnmarshall says:

    JP Edit
    I do not think so JP. Heat is the property supplied by the vibrational (kinetic) energy which provides the heat at that energy level. Temperature is a measure of the heat/energy at that vibrational level. Heat is a form of energy that cannot be transferred radiationally because photons are massless so cannot kinetically vibrate. Photon vibration is that supplied by the matter radiating at that vibrational temperature. The higher the temperature the higher the vibrational frequency. The photons, in turn, add energy as heat in the matter that adsorbes them.

  76. johnmarshall says:

    #markstoval.
    I wish that “long wave radiation” could be called “low frequency” radiation. It would get over worries about the energy levels available to do the heating. Remember Plancks Law, The low frequency radiation cannot increase the frequency of a higher level vibration so cannot heat the subject.

  77. Hi John,

    I am using this definition of heat:

    “Heat is defined as the form of energy that is transferred across a boundary by virtue of a temperature difference or temperature gradient. Implied in this definition is the very important fact that a body never contains heat, but that heat is identified as heat only as it crosses the boundary. Thus, heat is a transient phenomenon. If we consider the hot block of copper as a system and the cold water in the beaker as another system, we recognize that originally neither system contains any heat (they do contain energy, of course.) When the copper is placed in the water and the two are in thermal communication, heat is transferred from the copper to the water, until equilibrium of temperature is established. At that point we no longer have heat transfer, since there is no temperature difference. Neither of the systems contains any heat at the conclusion of the process. It also follows that heat is identified at the boundaries of the system, for heat is defined as energy being transferred across the system boundary.”

    G. J. V. Wylen, Thermodynamics, John Wiley & Sons, 1960.

    So I think that when you say “Temperature is a measure of the heat/energy at that vibrational level”, that “heat” should be replaced by “thermal energy”. Heat is defined by the equation we all know which indicates it is a flow between objects, not what is contained inside the objects. What is contained inside the objects which gives us the concept of “temperature” is “thermal energy”.

    I know what you mean when you write it, and you what I mean, and we know what each other mean, etc., but if we agree to that textbook definition of heat and the equation for heat, then I think a difference in the terms used is there.

  78. solvingtornadoes says:

    JP Edit:
    . . . heat is energy transfer, temperature is energy content

    ST:
    I don’t disagree, but I think it is important to be cognizant of the fact that temperature is measured based on the amount of energy/heat that flows into the thermometer. So, it does not give us a direct or literal measurement of the energy content of the entity being measured. Instead it gives us an indication of the rate that energy is flowing out of the entity. The amount of energy in the entity, what you refer to as “energy content,” is inferred from this measurement.

  79. solvingtornadoes says:

    Arfur Bryant says:
    ST[‘ST:
    I disagree. A cooler object can and does heat a warmer object. However, it cannot increase the measurable temperature of the warmer object. I can, and does, reduce the rate of cooling of the hotter object.”]

    You really should rethink this statement.

    A cooler object can neither heat nor increase the ‘measurable temperature’ of a warmer object.

    ST:
    I went to considerable effort to draw a distinction between heat and heating. If you ignore that then there isn’t much I can do to help you.

    Can you explain to us why it is you are so, seemingly, stubborn about not accepting this distinction that I suggested?

  80. solvingtornadoes says:

    ST:
    “Moist air is heavier than dry air.”

    AB:
    Please provide a reference for this assertion.

    ST:
    Please provide a reference that disputes it.

  81. solvingtornadoes says:

    markstoval says:
    “Another try at a cheap experiment”

    ST:
    The experiment has already been done. In fact, it’s been done thousands if not millions of times.

    The results are hardly debatable.

    It would seem that the dispute has nothing to do with empirical reality and everything to do with the semantics we use to describe empirical reality.

    One of the strange things about scientific discourse is that although people will sometimes concede a point based on empirical evidence they never or almost never make concessions with respect to their semantics lacking in conciseness, being ambiguous or even deceptive. Instead what they do, most often, is fall back into a consensus of those to whom with which they share the same semantic assumptions.

    I think this is why scientific paradigms tend to persist for hundreds of years until somebody comes along and untangles the semantics and exposes the underlying stubbornness, as I am doing with meteorology.

  82. Heat was defined above fellows.

  83. squid2112 says:

    Squid2112:
    This is precisely the reason why a cool object cannot, under any circumstances, heat a warmer object. It simply cannot be done and is a violation of nature physical laws that cannot be broken.

    ST:
    I disagree. A cooler object can and does heat a warmer object. However, it cannot increase the measurable temperature of the warmer object. I can, and does, reduce the rate of cooling of the hotter object.

    Squid2112:

    A cooler object can and does heat a warmer object.

    I completely disagree. A molecule of lower energy (vibrational state, kinetic energy) cannot increase the energy of a molecule that is already more excited (higher vibrational state, kinetic energy). If this were the case, then heat could “pile”, which as Joseph has shown, is also not possible. I would refer back to my original bicycle wheel analogy, as it is a nearly perfect analogy to this situation. And, therefore, a cooler object, in fact, cannot and does not heat a warmer object. It simply could not be otherwise.

  84. squid2112 says:

    markstoval says:
    2015/05/30 at 5:45 AM
    Another try at a cheap experiment

    I like your experiment. I would only add that, one should also be able to use such an experiment to measure the “reduced cooling” produced by the “cooler longwave radiation”, and also use to disprove that “slowing cooling” will create heat within a system, for all of those people that seem to believe that adding a blanket creates heat.

  85. squid2112 says:

    johnmarshall says:
    2015/05/31 at 5:09 AM

    The photons, in turn, add energy as heat in the matter that adsorbes them.

    I am not certain that I am accurately understanding this phrase, but I think you may mean to add a qualifier here, and that is the the photons can only add additional energy if, and only if, the target matter is at a lesser kinetic energy state. Once again, within the principle that heat cannot “pile” (I believe to be a sound principle), and conforming to the principle that a molecule of lesser energy cannot additionally excite a molecule of greater or equal energy, which again, I believe to be a demonstrably sound principle as Joseph has already shown.

    I am harping on this because I believe it to be so utterly important to the entire conversation and the entire subject of the so-called “greenhouse effect”. This, in itself, is the very hinge pin of the entire premise for which the GHE is built upon, and without violation of this principle, you cannot have a “greenhouse effect”. In other words, if a lesser or equal energy molecule cannot further excite another molecule, then the so-called “greenhouse effect” is simply not possible. Case closed.

  86. Arfur Bryant says:

    solvingtornadoes,

    No need to get stuffy about this, ST. We’re all just trying to get the facts straight, I’m sure.

    I completely understand your distinction between ‘heat and heating’ but you are still wrong when you state: [“A cooler object can and does heat a warmer object.”]

    According to Joe’s linked definition by Wylen, at no point does the water heat the copper block. There is no heat transfer from cold to hot, so your assertion is incorrect. Radiation is not heat. This is not me being stubborn, it is me understanding that you cannot just make stuff up and get all ‘stamping feet’ because someone calls you out on your assertions.

    You stated: [“Heat is the process by which particles of energy change their location.”] You assume, therefore, that transfer of energy is heat.

    No. Wrong. If you were right, an ice cube would be able to ‘heat’ a baked potato. It doesn’t.

    This assumption leads you to believe that, somehow, the backradiation “reduces the cooling rate”. Again, wrong. No transfer of energy from a cool object can raise the thermal energy level (hotness, or temperature) of a warmer object. I repeat (and you ignore) that the cooling rate can only be affected by a change in temperature difference between the hot object and its surroundings.

    Because radiation IS NOT heat!

    ***

    Next topic:

    ST:
    “Moist air is heavier than dry air.”
    AB:
    Please provide a reference for this assertion.
    ST:
    Please provide a reference that disputes it.

    Not the most considerate way to answer a request but, ok, here are a few:

    http://en.wikipedia.org/wiki/Water_vapor

    http://www.scienceforums.net/topic/37651-why-is-denser-air-heavier-than-less-dense-air/
    (Mr Skeptic)

    http://www.theweatherprediction.com/habyhints/260/

    http://science.howstuffworks.com/nature/climate-weather/atmospheric/is-humid-air-heavier-than-dry-air.htm

    http://www.conservationphysics.org/atmcalc/atmoclc2.pdf

    http://www.kean.edu/~csmart/Hydrology/Lectures/Lecture%2003%20moisture%20v.ppt

    From the wiki link:
    “Using Avogadro’s Law and the ideal gas law, water vapor and air will have a molar volume of 22.414 L/mol at STP. A molar mass of air and water vapor occupy the same volume of 22.414 litres. The density (mass/volume) of water vapor is 0.804 g/L, which is significantly less than that of dry air at 1.27 g/L at STP. This means water vapor is lighter than air.”

    I would still like to see a reference from you which states the opposite – as per my polite request.

    Please note that air which has water droplets in it is heavier than dry air, but that’s not what you said…

    Kind regards,

    Arfur

  87. solvingtornadoes says:

    Arfur Bryant says:
    Convection is reduced in moist air as follows:
    Convected dry air cools adiabatically
    Convected saturated (moist) air cools adiabatically
    When air is convected upwards,
    Further upward convection will
    If the convection trigger was
    If the convection trigger was mechanical, it may not reduce the convection rate

    Convection is a myth. Convection plays little or no role in our atmosphere. If it did then clouds would come falling out of the sky since they are, indisputably, heavier than dry air.

    Moisture in our atmosphere is never in the gaseous phase. This notion is nothing but an urban legend. IOW, there is no steam (monomolecular H2O) in Earth’s atmosphere. Consequently, using Avogadro’s law, moist air is always heavier than dry air. Electrostatic forces are what allows miniature clumps/droplets of H2O to remain suspended amongst N2 and O2 molecules. But this only explains how they can get up to about 1000 meters. Storms, powered by the jet streams, do the rest of the work to bring heavier-than-air moisture up high in the sky.

    http://wp.me/p4JijN-by

    http://wp.me/p4JijN-45v

  88. johnmarshall says:

    OK Joe, you stick to that 55year old definition.
    To my mind the heat is all the way through the object because all the objects molecules are vibrating with that thermal energy. It is felt at the surface if you touch it because your nerves tell you if it is too hot by burning you but the heat that you finger removes to produce the burn is replace by heat from the interior so heat is inside as well. Temperature is a relative quantity, in this case relative to Absolute Zero thermodynamically but in Celcius relative to the freezing point and boiling point of water. Heat transfer is simply and induced vibration from a hot object to a cooler one by contact( direct hot molecule hitting a cooler one at the contact point), or convection and by radiation through induced photons from the surface of the hotter object.

  89. solvingtornadoes says:

    Durango Dan:
    “Once it is understood that cooler cannot heat warmer and that photons must have a higher frequency than the photons emitted by the receiving body in order to heat that receiving body, . . .”.

    ST:
    Are you sure about this, Dan? I have never heard of this before. What is the empirical basis for this assertion?

  90. Well John it is the proper definition. Look man we’re saying the same thing, I just think you should use the term “thermal energy” rather than “heat” in certain spots, since heat is a process, a flow, whereas the energy inside which gives the temperature is just the thermal energy content.

  91. solvingtornadoes says:

    Arfur Bryant:
    Not the most considerate way to answer a request but, ok,

    ST:
    You ignored my request. You failed to dispute anything I stated. You need to delineate your assumptions first. Explain to us why you assume moist air contains steam. Without that your argument is nonsense.

    I’ve had the same conversation over and over again. It always goes the same way. You point me to a bunch of sheep that make the same mistake you make and you think you’ve presented an argument. You haven’t presented an argument. First tell us why you believe that moist air contains steam. The fact that other people make the same assumption is not an intelligent argument. Without that assumption your argument is, obviously, nonsense. If you can’t explain why you think this assumption is reasonable (and I know you can’t) then the discussion is over (so the discussion is over): http://wp.me/p4JijN-5A

    Arfur, answer this one question: Do you or do you not have any direct evidence of gaseous H2O at temperatures below its boiling point? If you do I will send you a check for 10,000 dollars.

    Humid air is heavier than dry air, not lighter.

    Arfur Bryant:
    “Using Avogadro’s Law and the ideal gas law, water vapor and air will have a molar volume of 22.414 L/mol at STP.

    ST:
    Water vapor is only an ideal gas above it’s boiling point. So this explanation is nonsense.

    Arfur Bryant:
    Please note that air which has water droplets in it is heavier than dry air, but that’s not what you said…

    ST:
    Please note that all air that is at temperatures below the boiling point of H2O exists in droplets, not individual molecules of H2O.

    Prove me wrong. Science isn’t about what you believe or how many people carry the same belief. It is about reproducible experimental evidence. And you have none.

  92. Greg House says:

    Sorry to interrupt you guys, but I have one question: WTF are you talking about?

  93. johnmarshall says:

    OK Joe. If you wish.
    Similar volumes of different substances, water and air for instance, can have the same temperature but vastly different heat contents.

    #squid2112
    All objects emit photons vibrating at a rate that is proportional to its absolute temperature. Cooler objects receiving this radiation adsorbe these photons which induces an increase in vibrational frequency in those molecules. This is the heat transfer and rise in temperature.

  94. DurangoDan says:

    ST: The frequency of a photon is induced by the temperature of the emitting body. Saying that the frequency of photons from the emitting body must be higher than the photons emitted by the receiving body in order to increase the temperature of the receiving body is the same as saying heat flows only from warmer to cooler. That’s all.

    Regarding moist air being buoyant in dry air, Arfur’s explanation fits perfectly well with empirical observation. I know that you don’t like the idea of “cold steam”, but unless Arfur’s explanation somehow damages our life experience (as belief in the GHE clearly does) I don’t see why we need your alternative explanation. If you can show a relative benefit in adopting your explanation, let’s hear it.

    One interesting aspect of water vapor being lighter than air is the measurement of barometric pressure as an indicator of the strength of hurricanes. My explanation of the reduced barometric pressure is that water vapor has displaced the oxygen and nitrogen molecules and since water vapor (MW: 18) is lighter than dry air (MW: 28), the pressure at the bottom of that cyclonic region of moist air is therefore reduced. The lower the pressure, the more water vapor in the air and the higher the potential energy release when the water vapor gives up its latent heat. Beautiful, even though we lost our beach house to Ivan in 2004.

  95. Arfur Bryant says:

    solvingtornadoes,

    ST[“Convection is a myth. Convection plays little or no role in our atmosphere. If it did then clouds would come falling out of the sky since they are, indisputably, heavier than dry air.”]

    I fully support your right to believe whatever you wish, ST. Just don’t present it as fact without substantiation. Using your own material does not count as substantiation or independent reference!

    Back to the subject at hand (backradiation), which you now want to ignore:

    You said that a cold object can heat a warmer object. I notice that you feel no longer inclined to support your statement. Both I and Durango Dan have repeatedly told you why you are wrong.

    You have also stated that moist air is heavier than dry air. I asked for references, you told me to provide references to show otherwise. I provided said references to support my view. You have still not provided any references other than your own material to support your assertion. This is not the way to behave in what is meant to be a rational debate.

    I don’t particularly care if you want to espouse a plasma theory. It is of no interest to me; I was more interested in your defence of the ‘backradiation = warming’ theory. However, it now seems obvious to me that you have confused moist air with clouds and steam. Moist air has no water droplets in it. Water droplets (of whatever size) are an addition to moist air. Moist air is lighter than dry air but you want to consider clouds as being ‘moist air’. For someone who entered this thread berating Joe for not being ‘specific’, you appear not to have taken your own advice.

    So be it. I wish you well in your belief.

    With kind regards,

    Arfur

  96. Will Janoschkas says:

    Arfur Bryant says: 2015/05/27 at 1:27 AM
    “If we accept that all objects emit radiation, and this therefore includes, radiation between cold (C) and warm ( W ) objects, what actually happens to a photon emitted by C wen it arrives at W?”

    Why should anyone ever accept that? Such has never been demonstrated! The amount of EMR flux detached from a radiating surface is always limited by any opposing radiance or electromagnetic field strength at that frequency. No EMR flux shall emit in a direction of higher radiance at that frequency! Would you accept that heat flux in a conductor between two temperatures requires flux proportional to each absolute temperature in opposing directions, or is the only flux proportional only to the difference in thermal potential between the two temperatures!

  97. Rosco says:

    I haven’t read all of this narrative by Joe as yet but wanted to jot down my thoughts as I see my favourite bullshit diagram – the University of Washington “Greenhouse Effect” diagram !

    It is simple to dismiss the concept of backradiation versus the solar radiation and demonstrate that University “Climate Scientists” simply have no idea about what they teach as “science”.

    I’ve seen all the “experts” – including Willis, Anthony Watts, Robert Brown and even the more moderate Roy Spencer – manipulating the Stefan-Boltzmann equation into all sorts of fantastic results including justifying that most stupid of scenarios – The Steel Greenhouse.

    (They are so smug they think anyone except them cannot do algebra – I got Distinctions in my engineering degree Maths exams.)

    All based on the simple arithmetic manipulations of the SB equation and the 2 way flow of “heat” or energy. Of course the thing most often overlooked is that if a cold object is emitting radiation to a hotter object the hotter object is emitting significantly more across the whole spectrum so the energy from the cold object cannot possibly cause any heating of the hotter object – as proven centuries ago by Pictet.

    But the answer to the stupidity of “climate science” lies in the first diagram above showing how the 239.7 solar radiation combines with 239.7 back radiation to cause heating equivalent to 479.4 W/sqm at ~303 K

    As such the University is teaching that the solar radiation at 239.7 W/sqm is EQUIVALENT in power to back radiation at 239.7 W/sqm and that is BULLSHIT !!

    All of the experts simply equate the power in W/sqm as if the discrete fluxes are equal and surely everyone knows they are not ! Fluxes with the same power can originate from entirely different circumstances and this needs to be accounted for.

    This is the problem of using the SB equation – it doesn’t define a radiation flux completely. Only Planck’s equation gives a real picture of the relationship between radiation fluxes and temperature – at least that is the current standard of the science – it could be mistaken also.

    Thus in the diagram used above the 239.7 W/sqm solar radiation – even if a real value – cannot be equated to the back radiation at 239.7 W/sqm simply because the power values are the same.

    The solar radiation is orders of magnitude more powerful than the back radiation !!

    It is easy to prove this with a magnifying glass – sunlight focussed by a magnifying glass starts fires and back radiation doesn’t under any circumstances including focussed by IR transparent materials or metal reflective parabolic mirrors no matter how large !

    To any person thinking straight that simply demolishes the very foundation of “climate science” – they do not fully define the types of flux they manipulate by algebra.

    Failing to do this is simply stupid as even children know you can start fires with a magnifying glass and sunlight.

    I tried to show this with my experiment a few years ago but was greeted by stupid comments such as some who claimed I showed cold heating hot. Of course I didn’t show that at all because the source of the energy I used to heat the thermometer came from objects at ~3000 K – the spotlight filaments – but the fluxes were attenuated by inverse square.

    I still believe I falsified many of the claims of climate science with this simple experiment – they equate values that are simply, and obviously, not equal and that is stupid beyond belief.

  98. SkepticGoneWild says:

    HI Joseph,

    Remember this Curt Wilson guy who performed the infamous light bulb experiment for Anthony Watts? Well, he’s over at Roy Spencer’s site yammering on and on about backradiation, insulation warming you up, the “Knowledge of Photons”, and so on. Apparently he, “studied mechanical engineering, undergraduate and graduate, at MIT and Stanford, with significant coursework in thermodynamics and heat transfer” and periodically UCLA asks him to teach engineering courses for undergraduate and graduate students. He’s quite the arrogant one.

    Roy Spencer even says, “Curt knows what he’s talking about. Listen and learn”.

    Curt’s arguments pretty much line up with the sophistry you have presented in your article above, and from some of your previous posts.

    Some, though I have not heard before. For example, Curt states:

    “A blackbody by definition absorbs all radiation incident upon it…”

    So by virtue of the inherent definition of a blackbody, he claims that a cooler blackbody’s EM radiation will be completely absorbed by the warmer blackbody, causing it to warm further.

    And he gives the typical example of a person experiencing hypothermia being wrapped in a space blanket to raise their core body temperature. He seems to be a Science of Doom clone, since he make multiple references to that site.

    Just thought you might be interested:

    http://www.drroyspencer.com/2015/05/trmm-satellite-coming-home-next-month/#comment-192280

  99. solvingtornadoes says:

    Arfur:
    No need to get stuffy about this, ST. We’re all just trying to get the facts straight, I’m sure.

    I completely understand your distinction between ‘heat and heating’ but you are still wrong when you state: [“A cooler object can and does heat a warmer object.”]

    Jim McGinn of Solving Tornadoes:
    The devil is in the details of which words each one of us, somewhat arbitrarily, choose to represent different concepts. In the context of the definitions that I provided, it is perfectly reasonable to state that, “A cooler object can and does heat a warmer object.” I think my thinking on this issue is consistent within itself and that is the best any of us can hope for.

    Maybe part of the confusion is my use of the word “heat” to describe a process and the word, “heating” to describe the results of a process. I can see how some might suggest that for purposes of conceptual clarity that this could be reversed. We can go around in circles forever debating which word is the right word. The larger point I’m trying to make here is that when people are dogmatic about semantics progress stops and it turns into a shouting match.

    Whatever the case, when a packet of energy comes off one entity and is received by another entity the energetic state of the second entity is increased as a result. The temperature and/or relative temperature of the two entities makes no difference. If you think the laws of thermodynamics indicate otherwise I would only suggest that you be cognizant of the fact that the laws of thermodynamics do not contradict this interpretation in that these laws deal with the net transfer of energy.

  100. markstoval says:

    @ Joe P.

    Please read this by Monckton at wuwt.

    Has NOAA / NCDC’s Tom Karl repealed the Laws of Thermodynamics?

    http://wattsupwiththat.com/2015/06/05/has-noaa-ncdcs-tom-karl-repealed-the-laws-of-thermodynamics/

    Now the fellow he is attacking is dead wrong on several counts, but C. M. attacking someone for violating the laws of thermodynamics seemed to be especially ironic to me for some reason. Please note his responses in the thread to those who don’t see his idea of the violation of the laws. I think that he has lost his way on this issue. I wonder if you would care to comment at some point on this one.

    Regards, Mark

  101. I don’t know either Greg. I think ST says that water vapour can only exist as steam and thus only if T >= 100C. And so the “water vapour” content in the air is actually composed of tiny liquid water droplets, not actual vaporous H2O.

    From my own research on the lapse rate question, one requires the latent heat density out of the condensation of vaporous H2O to liquid H2O, not liquid to ice H2O, in order to push the lapse rate down from 9.8 K/km to 6.5 K/km. You also need the same condensation of water vapour to explain why gardens and other surfaces are kept warm over night when dew forms – if dew was simply tiny water droplets forming bigger droplets and then precipitating out as dew, then you wouldn’t get any latent heat release and hence wouldn’t get any slowed cooling.

    So in my opinion ST’s stuff isn’t supported by the evidence, and I usually don’t let him post his stuff here. Why here anyway ST? Bring people to your own site on the question. Don’t make it part of this site. Interesting too that you recently said that I shouldn’t waste my time anymore on the greenhouse problem…when it is THE ONLY problem that matters right now!

    And ST, you don’t need to use tinyurl’s to your site. I moderate every comment here.

  102. solvingtornadoes says:

    Greg House says:
    2015/06/03 at 9:17 PM
    “Sorry to interrupt you guys, but I have one question: WTF are you talking about?”

    ST:
    Read upthread.

  103. Great stuff Rosco.

  104. It would be good if Joe would correct these misunderstandings about the basics of thermal radiation, but perhaps I’ve just missed those posts:

    DurangoDan wrote: “Once it is understood that cooler cannot heat warmer and that photons must have a higher frequency than the photons emitted by the receiving body in order to heat that receiving body, what is left?”

    All objects emit across the spectrum, but the PEAK wavelength in the curve will reflect the body’s temperature insofar as it operates in a fashion similar to an idealized black body. The W/m^2 flux is higher from a warmer body than from a cooler body, and the momentum contained in a single photon is dictated by its wavelength (as opposed to its mass ( zero ) and velocity ( c ) ). But it is not the case that (a) all photons from an object are at some magic wavelength dictated by its temperature (the photons are in a broad spectrum of wavelengths — look at a graph of the sun and earth, for example) or (b) the maximum energy flux is at the highest frequency (again, look at the curves for the sun and earth).

    Arfur Bryant wrote: “The radiation emitted by the cooler object is completely irrelevant to the warm object.”

    That cannot be correct. Since the heat flow from hot to cold is the net of radiation between the two, the heat LOSS RATE of the warmer object is slowed by the radiation from the cooler object. If that were NOT the case, Q’ would not be reduced by a cooler object.

    That is precisely why there is an effect by so-called — and here demonized — “backradiation”: it can’t HEAT the surface, but as the heat equations show, it does slow the rate of energy loss from the surface. Not a trivial, dismissive thing.

  105. Oh lord SkepticGoneWild…these people just don’t stop. I wonder if empirical refutation of their own theory would even stop them at this point? They would likely ignore it and pretend it means the opposite just like they do with every theoretical debunk of their belief.

  106. Sorry just don’t have the time Mark. Don’t worry, I have the final solution for them completed. After that, it will simply be a matter of war.

  107. Greg House says:

    Joseph E Postma says: “I don’t know either Greg.”
    =================================

    I know now: it is all about more energetic photonic vibrators.

  108. solvingtornadoes says:

    [JP Edit: You’re no longer welcome here Doug, I mean Jim. I’ll leave this one here though for everyone else to laugh at! LOL!!]

    Joe:
    I don’t know either Greg. I think ST says that water vapour can only exist as steam and thus only if T >= 100C. And so the “water vapour” content in the air is actually composed of tiny liquid water droplets, not actual vaporous H2O.

    Jim McGinn of Solving Tornadoes:
    Right. There is no steam in Earth’s atmosphere. It’s physically impossible. The water in moist air is comprised of little droplets/clusters, often too small to be seen. Consequently meteorologist’s claims that moist air is lighter than dry air are about as reliable as Al Gore’s claims about CO2-caused warming.

    Without the belief that moist air contains steam meteorology has no choice but reject the models they depend on to convince the public they know what they are talking about. They will be forced to find new models.

    The boiling point of water is determined by pressure and temperature. There is a wealth of laboratory evidence that proves/demonstrates this. There is zero laboratory evidence indicating the existence of gaseous H2O at temperatures and pressures below its boiling point. Zero.

    The fact that meteorologists chose to believe nonsense doesn’t mean that the rest of us are obligated to follow blindly.

    Joe:
    From my own research on the lapse rate question, one requires the latent heat density out of the condensation of vaporous H2O to liquid H2O, not liquid to ice H2O, in order to push the lapse rate down from 9.8 K/km to 6.5 K/km.

    Jim McGinn of Solving Tornadoes:

    The concept of “latent heat” is a nonsense concept. Empirically it’s meaningless (it’s never been measured or even detected). It’s sophistry.

    Joe:
    So in my opinion ST’s stuff isn’t supported by the evidence

    Jim McGinn of Solving Tornadoes:
    Yet you can’t clearly explain your opinion. You are confused. And when people become confused they tend to start pretending that they understand what they don’t understand. This is the reason so many people believe in AGW. They become confused and start pretending.

    Joe:
    I usually don’t let him post his stuff here. Why here anyway ST?

    Jim McGinn of Solving Tornadoes:
    Because in order to verify a scientific breakthrough you need to get through to people that recognize the importance empiricism over consensus. Is that not you?

    Joe:
    Bring people to your own site on the question. Don’t make it part of this site.

    Jim McGinn of Solving Tornadoes:
    You should be flattered that I am presenting you the opportunity to confirm a scientific breakthrough.

    Joe:
    Interesting too that you recently said that I shouldn’t waste my time anymore on the greenhouse problem…when it is THE ONLY problem that matters right now!

    Jim McGinn of Solving Tornadoes:
    You have an engineer’s understanding of thermodynamics. You do the math accurately. But you misinform people when it comes to properly conceptualizing what is actually taking place. And since you are so good at the math people tend to assume that you must also understand it conceptually. And, well, you just don’t (as has been demonstrated vividly recently in this blog.) I don’t know the math that well, but I can conceptualize the phenomena and I know how to avoid the semantic confusion that is so common in scientific discourse. You are out of your element in that respect.

    BTW, who was it back in September of 2013 that informed the various participants in the “back-radiation” discussion about the fact that the laws of thermodynamics have to do with net flow of energy and not just the flow? It wasn’t you, Joe. It was me. (And, frankly, it still seems you don’t get it.)

    The worst thing you can do in science is claim you understand something that you actually do not understand. Because after a while you start believing your own lies. Then you become the thing you are fighting against.

  109. Arfur Bryant says:

    Joe,

    What we have here is people asserting different things (myself included) which are being corrected by another assertion but not with any corroborating evidence…

    Will Janoschkas says:
    2015/06/04 at 6:00 PM
    Arfur Bryant says: 2015/05/27 at 1:27 AM
“If we accept that all objects emit radiation, and this therefore includes, radiation between cold (C) and warm ( W ) objects, what actually happens to a photon emitted by C wen it arrives at W?”
    Why should anyone ever accept that?

    Will, do you not accept that all objects above 0K emit radiation?

    My question is: Does that radiation reach a warmer object and, if it does, does it increase the thermal energy of the receiving (warmer) object?

    **

    solvingtornadoes says:
    2015/06/05 at 11:48 AM
    Whatever the case, when a packet of energy comes off one entity and is received by another entity the energetic state of the second entity is increased as a result.

    ST: Please provide any objective evidence that radiation (‘energy packets’) from a cooler entity will enhance the thermal energy of receiving warmer entity.

    I have never seen such evidence…

    **

    Tom OregonCity says:
    2015/06/05 at 7:11 PM
    Arfur Bryant wrote: “The radiation emitted by the cooler object is completely irrelevant to the warm object.”
    That cannot be correct. Since the heat flow from hot to cold is the net of radiation between the two, the heat LOSS RATE of the warmer object is slowed by the radiation from the cooler object. If that were NOT the case, Q’ would not be reduced by a cooler object.
    That is precisely why there is an effect by so-called — and here demonized — “backradiation”: it can’t HEAT the surface, but as the heat equations show, it does slow the rate of energy loss from the surface. Not a trivial, dismissive thing.

    Tom, when I said “completely irrelevant” is was talking about raising the thermal energy level of the warmer object.

    Q: Why do you say the Heat Flow equation is the net of radiation?

    To my mind, the heat flow is just about heat flow, not radiation. For example: a ‘radiation equation’ of 5 less 3 = 2 is not the same as a ‘heat flow equation’ of 2 less 0 = 2. Where do you get the ‘net’ from? The radiation from the cooler object cannot add thermally to the warmer object. If your view were to be correct, simply adding more cooler objects (of identical temperature to each other) would eventually raise the temperature of the warmer object. The heat loss rate only depends on the temperature difference…

    *** ***

    Joe, do you see what I mean? This is the fundamental cognitive issue at the heart of the ‘cAGW by backradiation from CO2’ theory!

    I, for one, would like it to be sorted! 🙂

    Regards to all,

    Arfur

  110. I have an exercise for everyone. Start with this diagram:

    and add 4 more layers of glass (in place of an atmosphere, but the backradiation result is the same) and calculate the temperature which the bottom surface should have, given a noon-time input flux of 1100 W/m^2. Try it for 10 layers. 20 layers.

    Discuss why such devices which would lead to the useful result the calculation indicates do not exist.

  111. Also reference Willis’ steel greenhouse if you need help figuring out the math & “physics” required when you add more layers:

  112. squid2112 says:

    ST:

    Whatever the case, when a packet of energy comes off one entity and is received by another entity the energetic state of the second entity is increased as a result.

    ST, this is not true. The only way the energetic state of the second entity is increased is if, and only if, the energetic state of the source is greater than the energetic state of the target. Please refer to my bicycle wheel analogy, as that describes perfectly what is happening. And, because of this, therefore, a cooler object cannot make a warmer object warmer still. No way, no how.

    And please, remember, while “energy” is subject to the “net” flow, “heat” is not. “Heat” is the result. If the net “energy” flow is to the cooler, the result can only be that the cooler object is further heated, the warmer object cannot be further heated. It is as simple as that.

  113. Arfur: Why is this complex?

    You wrote: “My question is: Does that radiation reach a warmer object and, if it does, does it increase the thermal energy of the receiving (warmer) object?”

    Of course it does; by what physical process could it be prevented from doing so? But that doesn’t mean the warmer object’s temperature increases, because the warmer object is emitting much more than it receives, and the T^4 differential means that happens even if the cooler object completely surrounds the warmer one (simple geometry). Because that warmer object is emitting more than it is receiving from all cooler objects around it, those cooler objects cannot make it warmer: T^4, again (it is losing more energy than it gets from the cooler objects).

    That increase in “thermal energy” does not manifest itself as an increase in temperature, so using the description “thermal energy” could be confusing. That increase in energy from radiation absorbed from a cooler object instead simply offsets some fraction of the radiation being emitted from the warmer object, which in turn does not permit the warmer object from cooling as fast as it would otherwise, or, if the warmer object is also being heated by an object warmer still (the sun, for example), than the presence of the offset energy received from the cooler objects allows more energy from the yet hotter object to WARM the warmer (middle temperature) object a bit faster, since that HOT source energy can’t be shed as quickly, either.

    On the other side of the argument, in order to prove that a cooler object cannot emit energy which is absorbed by a warmer object, you would need to demonstrate that the cooler object “knows” how to avoid doing that (thermal emissions are omnidirectional), or that the warmer object “knows” where the energy is coming from, and refuses to accept photons from cooler objects (remember, objects emit across a broad spectrum of wavelengths, not just at some specific wavelength, even though some wavelengths may be more abundant because of emissivity or similarity of the object to an idealized black body).

    Net does indeed mean the difference between emissions in both directions. Or, conversely, would you suggest that once two objects are in thermal equilibrium with each other, they simply stop emitting toward each other, in spite of the fact that all objects above absolute zero emit thermal energy?

    This discussion should focus first on the first principles of radiative energy, and how objects emit and absorb. Diving head-first into the swamp of warmists vs skeptics isn’t wise if the fundamental physics of the issue is not solid. It’s not dangerous, or wrong, to acknowledge radiation between all objects. It’s fundamental. The abuse by the warmists is elsewhere, in the feedbacks and model insufficiency and such. Before we get to the FLUX involved, the basic process has to be understood.

  114. Squid2112 & ST wrote:

    ST: “Whatever the case, when a packet of energy comes off one entity and is received by another entity the energetic state of the second entity is increased as a result.”

    Squid2112: “ST, this is not true. The only way the energetic state of the second entity is increased is if, and only if, the energetic state of the source is greater than the energetic state of the target.”

    Each “packet” of energy is emitted according to the emissivity and temperature of the emitting entity. Each “packet” of energy received by an entity is absorbed according to its emissivity curve, regardless of its temperature.

    Whether the absorbing entity goes up or down in temperature is a function of the total FLUX absorbed minus the total FLUX emitted. If it is receiving more energy from all sources than it is emitting, its temperature will rise. If it is receiving less energy from all sources than it is emitting, its temperature will fall.

    No magic, no hidden process, no political agenda. No entity understands the temperature of objects surrounding it, it simply absorbs according to its energy curve. The earth, for example, receives and absorbs energy from the sun which is at a LOWER “black body temperature” — that is, at very long wavelengths, such as those which would be emitted from an object colder than the earth (the overlap isn’t much, but it’s there, nevertheless). It is not the energy of a “packet” that warms the earth, it is the total FLUX, of all “packets”, at all wavelengths, emitted from the sun to the earth that cause the earth to be warmed by the sun — and even those “packets” that are from the lowest energy portion of the emitting curve of the sun participate in heating the earth. Look at the graphs for solar emissions and earth emissions, and you will see solar emissions which are received by earth which overlap earth’s emitting curve.

    Consider this: if two objects are emitting, one slightly cooler than the other, their emission curves will largely overlap (look, for example, at the bell-shaped curve of earth’s emission profile). That means that lots of photons being sent from the warmer object toward the colder one are at an energy level LOWER than the peak emissions of the cooler object. And yet, I doubt anyone here would suggest that those photons are rejected by the colder object. Likewise, there are photons being emitted from the colder object that are at an energy level HIGHER than the peak emission wavelength of the warmer object. Why suggest those are not likewise absorbed, and by what process?

    It’s all about total flux, not the wavelength of individual “packets”.

  115. Only the warmer object has more energy at all frequencies, and more energy at higher frequencies, and energy at higher frequencies that the cooler object will not have at all.

  116. Joe wrote: “Only the warmer object has more energy at all frequencies, and more energy at higher frequencies, and energy at higher frequencies that the cooler object will not have at all.”

    Well, perhaps we are talking across each other. The warmer object emits more energy in total, but the energy isn’t “at a frequency” until it is indeed emitted, I’m sure we agree. For the individual emission — a photon, as one description — the energy contained is a function of wavelength, not a function of the total energy possessed by the emitting object. Is that not why discussion of “packets” can be confusing, when it’s the total flux that can describe why temperature rises or falls by the receiver?

    The sun, for example, emits energy at wavelengths LONGER than the higher energy emissions of the earth. That energy is absorbed by the earth, too.

    What I was trying to point out is that it’s not the wavelength of individual emissions that dictates whether the emitter warms the absorber, it’s the total flux of energy emitted and absorbed by an object that dictates whether an object is warming or cooling, regardless of where incoming energy originates.

    Are we in agreement on that?

  117. I think so for the most part, but with the understanding that with thermal emission the flux contribution itself shifts to higher and higher frequencies as temperature increases; those higher frequencies are characteristic of the higher temperature.

    Another case though is that of a “green 9 Watt laser” that I saw in a physics laboratory once. The frequency here is uniform and the energy itself of course nothing to do with thermal emission and representative of blackbodies, and the frequency is of course visible; however, this laser can burn a hole through 1/4″ steel without trying. And so of course, the 9 Watts of emission is actually concentrated in less than one square millimeter laser beam, resulting in a flux density one million times the power. I guess one could also think of the temperature required for something to glow “green hot”, and that would be quite hot indeed, probably hot enough to melt steel! Wien’s Law says that for a peak emission at 520 nm (“green”), the temperature is ~5500K! Notwithstanding that lasers can concentrate flux to a greater degree than thermal emission would. Still, what would a 9 Watt red laser do with the same beam size? Same effect on matter, or less, than the green beam?

  118. Arfur Bryant says:

    Tom,

    [“Arfur: Why is this complex?
    You wrote: “My question is: Does that radiation reach a warmer object and, if it does, does it increase the thermal energy of the receiving (warmer) object?”
    Of course it does; by what physical process could it be prevented from doing so?”]

    Ok, this is a reasonable debate. You have raised a number of issues and I will try to respond to each issue…

    By what physical process could it be prevented from doing so? Well, by not being absorbed, of course! The radiation can be trasnmitted or reflected. Being reflected can also be considered as being absorbed for no energy gain and instantly re-emitted.
    So there is no “of course” about it!

    TOC[“That increase in “thermal energy” does not manifest itself as an increase in temperature, so using the description “thermal energy” could be confusing. That increase in energy from radiation absorbed from a cooler object instead simply offsets some fraction of the radiation being emitted from the warmer object, which in turn does not permit the warmer object from cooling as fast as it would otherwise…”]

    Tom, when radiation from a hotter source is absorbed, the electron orbit of the receiving surface is raised because the energy (higher frequency) of the incoming radiation is sufficient to allow the electrons to jump the energy gap to the new (wider) orbit.
    Radiation from a cooler object does not have the ability to jump this gap, so the receiving surface gains no thermal energy. There is no distinction here between ‘gaining thermal energy’ and ‘increasing temperature’. Yes, there may be other wavelengths being emitted but any increase in temperature of an absorbing object depends on the shortest wavelength, emitted at higher frequency.

    Talk of a cooler object ‘knowing’ it cannot emit energy which is absorbed by a warmer object is specious. It is an irrelevant argument. Any radiation from a cooler object simply is not absorbed for energy gain. That’s it.

    [“TOC:Net does indeed mean the difference between emissions in both directions. Or, conversely, would you suggest that once two objects are in thermal equilibrium with each other, they simply stop emitting toward each other, in spite of the fact that all objects above absolute zero emit thermal energy?”]

    If you are correct about net radiation, then you must explain this (as I said before):
    If you then add more cooler objects, there sum of radiation (according to you, absorbed by the warmer object in order to reduce the heat loss) from the increasing number of cooler object MUST, at some point, reach a level where the warmer object’s temperature increases – because, according to you, the ‘net’ of incoming radiation is greater than the outgoing. The reason this doesn’t happen is because heat flow is only about the difference in temperature of the emitter and absorber, NOT the net of radiation!

    TOC[“Or, conversely, would you suggest that once two objects are in thermal equilibrium with each other, they simply stop emitting toward each other, in spite of the fact that all objects above absolute zero emit thermal energy?”]

    Don’t twist my words, I never said that objects stop emitting when they reach equilibrium. When two object reach equilibrium they are still emitting but, since there is no heat flow, neither object can affect the other’s temperature. No absorption for thermal energy gain is happening.

    TOC[“The abuse by the warmists is elsewhere, in the feedbacks and model insufficiency and such. Before we get to the FLUX involved, the basic process has to be understood.”]

    I agree wholeheartedly that the basic process has to be understood but I disagree that the warmists’ abuse is elsewhere. They have got this absorption thing wrong and they have built a house of cards upon that error, which they now need to continue building on no solid foundation. The error of feedbacks and model inputs are just some of the tools used to try to hide the erroneous basic assumption.

    As for flux, the total flux is always from hot to cold. That’s kind of the point.

    Regards but I may not be able to respond for a day or two…

    Arfur

  119. DurangoDan says:

    My college physics has about 40 years of rust on it but I think we have been making a mistake by focusing on radiative heat transfer. I believe that Tim Casey as described in The Shattered Greenhouse has it right when he states that matter in thermal contact equilibrates temperature by kinetic energy exchange. The whole backradiatiion discussion is irrelevant in this context. I think Squids ? bicycle wheel analogy has it right. In other words photons are not the primary driver of Earths surface and atmosphere energy exchange process. Maybe this is the real fatal flaw in the radiative greenhouse effect. Virtually all of the radiative transfer takes place in the near vacuum of the upper atmosphere. The surface and near surface conduct heat only from warmer to cooler. It’s kinetic energy not photons. Hopefully that is clear. The atmosphere is but a refrigeration cycle heat engine with water as the phase change medium. Radiative GHG gases can only expedite the cooling effect in the region of the upper atmosphere. Water rules the climate and it doesn’t pay attention to CO2 or methane.

  120. RF says:

    I’ll second what DurangoDan says. To my mind the radiation-“backradiation” issue is something of a red herring. We most often talk about radiation when bodies are isolated by a vacuum, or in some instances when very high temperatures are involved. When there is thermal contact, most of the energy transfer is kinetic. Moreover, Fourier’s Law, which governs heat transfer for bodies in thermal contact, by default accounts for all forms of heat transfer, whether kinetic or radiative. I think that Tim Casey’s “The Shattered Greenhouse” should be shared far and wide. Highly recommended reading.

    Having said all that, No, cool objects do not make warm objects warmer still. That’s pure nonsense, and I enjoy reading JP’s posts pointing out just how silly and dishonest the “backradiation” sophistry is. (I also like squid2112’s bicycle wheel analogy; deceptively simple yet quite telling nonetheless.)

  121. Yes good point about Fourier’s Law RF.

  122. “If there was a two-way flow of heat, then the cold object would raise the temperature of the warm object, and you just violated thermodynamics.”

    This is incorrect. If the warmer object emits 10 joules/sec. and receives 5 joules/sec. (from the colder object) it is cooling at a rate of 5 joules/sec.. If the colder object is receiving 10 joules/sec. and emitting 5 joules/sec., it is warming at a rate of 5 joules/sec. Bottom line: the warmer object cools and the colder object warms. The colder object reduces the heat loss from the warmer object, so when outside energy is added (from the sun) there is more heat entering the system than leaving. That is what causes the warmer object to warm further. The back radiation merely sends back some to the energy that would otherwise escape–so the system is warmer than it would be if the radiation escaped.

    Psun – AoeT(hot)^4 < P(sun) – Aoe(T(hot)^4 – T(cold)^4)

    The above inequality shows how a system warms when the colder object is added.

    [JP: The scenario is with the warmer source being the power source, and so with the cooler passive object sending 5 W to the warmer object, the climate alarmist GHE says that this will heat the warmer object power source some more…which is incorrect.

    Also see this post for the debunk of the idea that the cooler object retards heat loss thus requiring the warmer object to warm up some more. It’s false physics.]

  123. “A*σ*Thot4

    this term is actually its power, P = A*σ*Thot4, not heat! This term is its energy emission, not its heat. Heat is only the portion of the energy which can act to increase the temperature of another object, hence why heat is written as

    Q’ = A*σ*(Thot4 – Tcool4)”

    P is both Q’ and A*o*Thot4. In both cases you have a measure in joules per second. Q’ isn’t heat, but the rate of heat. Heat is Q = mc(T). The relationship between heat and temperature is linear. The relationship between power and temperature is exponential.

  124. See this definition of heat:

    “Heat is defined as the form of energy that is transferred across a boundary by virtue of a temperature difference or temperature gradient. Implied in this definition is the very important fact that a body never contains heat, but that heat is identified as heat only as it crosses the boundary. Thus, heat is a transient phenomenon. If we consider the hot block of copper as a system and the cold water in the beaker as another system, we recognize that originally neither system contains any heat (they do contain energy, of course.) When the copper is placed in the water and the two are in thermal communication, heat is transferred from the copper to the water, until equilibrium of temperature is established. At that point we no longer have heat transfer, since there is no temperature difference. Neither of the systems contains any heat at the conclusion of the process. It also follows that heat is identified at the boundaries of the system, for heat is defined as energy being transferred across the system boundary.”

    G. J. V. Wylen, Thermodynamics, John Wiley & Sons, 1960.

    “Q = mcT” is the thermal energy content, although loosely & colloquially, but incorrectly, people call it heat sometimes.

    But under these strict definitions, then only Q’ is heat, and P is only heat if the background is zero Kelvin.

  125. Greg House says:

    Joseph E Postma says: “See this definition of heat: “[…] a body never contains heat […] When the copper is placed in the water and the two are in thermal communication, heat is transferred from the copper to the water, until equilibrium of temperature is established. At that point […] Neither of the systems contains any heat at the conclusion of the process.”
    ============================

    Really? So, something that a body did not contain has been transfered from it. Very scientific. Poor students who have to learn that.

  126. A body does not contain running, yet it does it and it transfers energy.

  127. Greg House says:

    But you can not say “running has been transfered from the body”, can you? The same goes for that heat definition above. Just stick to the text of that definition please, no need to give analogies.

  128. Stick to it then and stop confusing yourself.

  129. Greg House says:

    Who is confused? As an amateur psychologist I can say that such a definition can only cause some brain damage to students.

  130. ““Heat is defined as the form of energy that is transferred across a boundary by virtue of a temperature difference or temperature gradient.””

    It’s not that difficult. It indeed transferred what it contained – energy, but in the form of heat.

  131. According to my physics text (College Physics 10th Edition), the P in P = eAoT^4 and P = eAo(Ta^4 – Tb^4) is not heat in either case, but radiation in watts. How much an object radiates depends its environment. The second term in the second equation above is the temperature of the environment surrounding the black or gray body. The P in that equation is the net energy radiated. Quoting from the text: “Radiant energy is distinct from heat, though both correspond to energy in transit. Heat is heat; electromagnetic radiation is electromagnetic radiation–don’t confuse the two.” The heat your text seems to be describing is conduction, not radiation.

    I would never say that cold heats hot, but I would think that at a quantum level there is a radiation exchange or exchange of photons that results in an expected value of a warmer object heating the colder object.

    [JP: I agree with this part and I think it is saying the same thing as I am intending: “Radiant energy is distinct from heat, though both correspond to energy in transit. Heat is heat; electromagnetic radiation is electromagnetic radiation–don’t confuse the two.”]

  132. “Q = mcT” is the thermal energy content, although loosely & colloquially, but incorrectly, people call it heat sometimes.”

    From College Physics 10th Edition: “Heat (Q) is thermal energy in transit …” It seems logical to assume P is also thermal energy in transit or heat, but my textbook says no. Radiation is distinct from heat. The math also implies they are distinct: Q=mcT; P=eoAT^4. Heat has a linear relationship with temperature; radiation has an exponential relationship with temperature. Q is measured in joules; whereas, P is measured in joules per second.

    [JP: Yes agreed with this. Q is thermal energy content. Q’ (sometimes P, but not always) is heat flow.]

  133. DurangoDan says:

    When molecular kinetic energy, commonly known as heat (and measured as temperature), is transferred from the Earth’s surface to the atmosphere by a process of molecular contact commonly known as conduction of heat, the atmosphere which is a gaseous fluid responds by expanding at the immediate surface / atmosphere interface. This bit of atmosphere now being less dense than the air above it, rises and allows cooler denser air to contact the surface initiating a very efficient heat transfer process commonly known as convection. This convection feeds back on itself carrying the air and heat higher and higher above the surface. In dealing with conductive heating it is easy to visualize by the bicycle wheel analogy why heat only flows from warmer to cooler. If warmists had to explain the greenhouse effect in the context of conductive heat transfer I don’t see how they could do it effectively. This is why the graphics most commonly used to display the greenhouse effect show radiative heat from the surface to the atmosphere and back again. Since radiative heating is not the predominant process by which the Earth’s surface sheds heat to the atmosphere, the very use of radiative heating in this diagram is fraudulent. Radiative heat transfer by photons makes for a much easier magic trick when it comes to creating the illusion that cold can heat hotter. Going along with the radiative mechanism facilitates the deception.

    [JP: Well said.]

  134. Rosco says:

    Greg House wrote

    “I know now: it is all about more energetic photonic vibrators.”

    Sounds like a sex toy to me.

  135. Squid2112 says:

    Hahahaha … well said Rosco! …. filing this one away …

  136. Shawn Marshall says:

    Joe
    What is wrong with this experiment?
    1) Construct 2 identical 1m^2 earth boxes 6?in. Deep; fill with dessicated sand
    2) construct a darkroom inside an unconditioned building to contain the 2 earth boxes and to shut out all light and other radiation; also a wall is needed between the boxes to block possible IR from CO2 interfering with the other box
    3) construct 2 helical coils of flexible water tubing fixed at ?3in from the bottom of the boxes
    4) connect input ends of tubes to a hot water tank with identical length tubing
    5) connect the output end of tubing to a set of valves, one to return hot water to source, the other to fill identical catch basins to measure the volume of water
    6) install 2 ‘clouds’ constructed of non-infrared blocking glass; 1m^2 x ? 6in.
    7) Locate the clouds at a height of 2? meters above the sand boxes to avoid restriction of convection; the dark room should be well ventilated at the top of the walls by large apertures covered with shrouds
    8) saturate the sand boxes with identical quantities of distilled water ?20gal
    9) place thermocouples in the center of each box on top of sand
    10) Circulate hot water through each box until a steady temperature of 90 F is achieved
    11) fill one cloud with CO2; the other with a natural air mix of Oxygen and nitrogen
    12) stop the circulating water and open the valves to the catch basins
    13) maintain the temperature in each box at 90 F by flowing hot water whenever the temperature declines by .5 F, all the while charting the temperatures for both boxes
    14) Maintain experiment for x? Hours
    15) premise – if the CO2 can warm the box beneath it by back radiation then that box will require less hot water to maintain temperature

  137. @Shawn,

    The CO2 box would have to get hotter than the 90F input water, not just cycle a little differently.

  138. What I find amusing about the Uni of Washington Diagram is that it shows 239.7W/m2 going in and 239.7W/m2 going out. Balance=zero. All the other figures are just added on out of thin air! The extra 239.7W/m2 shown going back to the surface. The extra 239.7W/m2 added to together to give the surface radiation in the equation underneath the diagram. The arbitrary calculation applied to the result of adding 239.7W/m2 to itself to achieve the 303K result. All of it completely invented to mask the fact that 239.7 goes in and 239.7 goes out. If I have $239.70 in my savings account to start with and each day I both spend and receive another $239.70, my savings won’t grow. The same way, the Earth keeps it’s average temperature.

  139. Shawn Marshall says:

    I was unclear about my GHE experiment above – the heating water is above 90F and flow is regulated to keep each box at 90F. Thus if back heating from CO2 exists, that box will require less hot water by volume. The effect is calculable given GHE claims or it does not exist. This basic experiment would prove or disprove GHE once and for all would you agree? I would like to challenge an undergrad thermo class to run the test or redesign it to something more indicative if the expected effect is too small.

  140. Abe says:

    Back radiation remains a story about how an insulating gas in suspension around the energy sensors of a planet

    blocked 20% of the energy from the fire of a star,

    causing every energy sensor on earth,

    to indicate more energy arriving,

    than was arriving, when there was

    more energy arriving.

    This Backerdistical Magick is followed by the news it’s believers feel that

    if we put up more insulating gas in suspension around the planet,

    such that 25% of energy in gets blocked/diffracted away,

    every heat sensor on the planet will show still MORE energy arriving,

    than there was when… yeah, that’s right –

    when MORE was arriving.

  141. Shawn I think that doing Fourier’s and de-Saussure’s experiment is the most direct and obvious way to go about it. I wrote about it in the last post on the blog but I have another post describing it in more detail to come at some point.

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