Heat Flow Cold to Hot when both Conduction & Radiation Occurring?

PSI recently had a paper submitted where the author claimed that there are solutions to thermodynamics with heat flowing from cold to hot, as long as heat was flowing from hot to cold via a different mechanism at the same time.  Paper is here; author’s name has been removed.

Here we will present the review, just because it’s great reading for math and physics Übermenschen.

———-

For specific criticism and some reasons why your paper was not accepted.

PL:  “The First Law of Thermodynamics, conservation of energy for the radiator says: Rate of energy out/in by conduction at T to/from surroundings at Ts = rate of energy in/out by radiation from/to surroundings minus/plus the rate of energy accumulation/depletion within the radiating body.”

Firstly, that is one of the most convoluted, unreadable statements I’ve ever come across.  I’d ask for a re-write edit based on this sentence alone.  Thankfully you then wrote it as an equation:

“PL: Rate Out = Rate In – Rate of Accumulation

Qo(t) = Qi(t) – m Cp dT(t)/dt”

Since you’re using Q then these terms should mean heat, although you don’t call them heat in your text, and so again it needs editing and a re-submission based on that alone.  However, then something gets much worse.  The right-most term is indeed called heat and one will typically find it as dQ/dt = m Cp dT/dt or more simply dQ = m Cp dT.  But…this *is* the rate of heat energy in for an object, and it *is* the rate of heat energy out for an object, depending on the sign of dT, and, this *is* the rate of accumulation for an object.  Your equation is literally stating:

Rate of Accumulation = Rate of Accumulation – Rate of Accumulation

or

m Cp dT(t)/dt = m Cp dT(t)/dt – m Cp dT(t)/dt

which is senseless.

However, given that convoluted preliminary sentence, it seems that in this equation you’re actually trying to separate conduction from radiation, but that isn’t clear at all from the terms in the equation written since they simply denote heat out and in.  Your Qo(t) is actually the heat from conduction, and Qi(t) is the heat from radiation, given your convoluted intro sentence.  But given the way that you have inverted in writing in that convoluted sentence almost every single term in its relationship to what should be positive and negative, because you go from in/out to out/in several times, it becomes exceedingly difficult to track what you’re actually doing especially when you factor in that the individual terms drop any notational reference to the conduction and radiation that they’re supposed to correspond to.  It’s a hair away from being totally unreadable…strictly it is actually unreadable, and totally inconsistent.  And I’ll also note that that first equation which you start from has no reference stated for it from source material.

At this point, without going any further, your paper is being sent back for major revisions asking you to use proper English sentence structure, and proper & consistent mathematical and physics notation.  If the author didn’t know what that meant, and the paper then came back with the same English and mathematical grammatical syntax convolution, then the paper would immediately be recommended for outright rejection.  The author would need to go find somewhere else to publish.

Going further, for the sake here, not that I would have if this was a paper I was reviewing (it would have been sent back by now), you state:

“PL: Qo(t) = Qi(t) – m Cp dT(t)/dt

At steady state, T is constant, dT/dt = 0, out/in = in/out and Qo = Qi.  Let Qc be rate by conduction and Qr rate by radiation. Qo is Qc if Qc > 0. Qi is Qr if Qr > 0.  Qc = Qr”

If dT/dt = 0, this means that there is no heat flow, and so all Q’s should be equal to zero.  dT/dt = 0 defines all Q’s equal to zero.  Instead you are saying that the conductive heat input must be equal to the radiative heat loss, or in other words, that heat is entering and leaving at the same time.  There is some sort of underlying ambiguity which has been set up here but for now we will go with it given that heat is said to be leaving and entering at the same time via different mechanisms but at equal rates.

However, since you’ve stated that dT(t)/dt = 0, then since the object is in thermal equilibrium, Kirchhoff’s Law will be in effect; all of your subsequent discussion and where you arrive at heat flowing from cold to hot depends upon Kirchhoff’s Law not being in effect, but you started the analysis with the very conditions under which Kirchhoff’s Law is defined to be in effect…i.e. when dT(t) = 0, i.e. in thermal equilibrium.  So there’s a logical error here.  You’ve set up the conditions under which Kirchhoff’s Law is *defined* to be in effect, i.e. dT(t)/dt = 0, thermal equilibrium, but then you go on to dispense with it.

Your equation and text itself is inconsistent because your preliminary convoluted sentence is referring to the Q’s as those from convection and radiation, but then you write the equation as Q’s in and out.  So now I must correct this and sort it out:

m*Cp*dT refers to an object, and so any Q’s relating to this must also refer to same object, so that if dT is positive, then dQ is positive, i.e. if the object has risen in temperature then it has taken in heat – positive temperature change = positive heat input.  So getting rid of the heat in and out notation which makes no sense, and using heat from conduction and heat from radiation notation, then

dQc + dQr = m Cp dT

So now if dT = 0, then dQc = – dQr.  The heat input from conduction equals the heat output from radiation, or vice-versa.  And this must be a general result applicable to all situations.  If we very carefully read your text, it eventually becomes somewhat clear that you state the same thing.

So if we now look at your later equation (4) or just insert terms to dQc = -dQr…

dQc = k(Ts – T) –> that makes sense because the object is the reference; if there is positive heat +dQc into the object from conduction then we expect Ts to be greater than T, and k is always positive.

dQr = sigma * (α εs Ts4 – αs ε T4) –> that makes sense (in this writing) because if absorptivities and emissivities are unity, then positive heat dQr into the object from radiation is because Ts is greater than T.

So putting it together, and this is the same equation (4) as found in your text:

k(Ts – T) = -sigma * (α εs Ts4 – αs ε T4)

and given that your dQc = – dQr is a general result for the thermal equilibrium you’ve defined it thus applies to all situations of such, and so we can then look at an ideal case where absorptivities and emissivities are all unity:

k(Ts – T) = -sigma*(Ts4 – T4)

Thus, if Ts was less than T, then the left hand side would be a negative number, -x say.

-x = -sigma*(Ts4 – T4)

or

+x = +sigma*(Ts4 – T4)

The only way that the right hand side can be a positive number is if Ts is greater than T…however, this is in contradiction to the defined requirement that Ts was less than T.

Thus, there is a fundamental, general error, lurking somewhere in your creation here, and so it is no wonder that you could find heat flowing from cold to hot…once the initiating error is set up, subsequent errors can only follow.  And we have now proven that there is a general error embedded somewhere in your creation.  In other words it is mathematically impossible for dQc = – dQr.  This isn’t a possible expression in physics or the mathematics.  Certainly it can lead to heat flowing from cold to hot…because that is impossible…an impossibility can come from an impossible expression.

The impossibility must arise when we say that

dQc + dQr = m Cp dT

can have a condition where dT(t) = 0 without both dQc and dQr being zero.  In other words, the only condition where dT(t) = 0 is if both dQc & dQr equal zero.  If dQc and dQr were not zero, then it is not possible for dT(t) = 0.  dQc = – dQr only when they both equal zero.  Why would that be?

Is there a non (Ts – T) = 0 solution for

k(Ts – T) = -sigma*(Ts4 – T4)?

So

(Ts – T) = -(sigma/k) * (Ts4 – T4)

(sigma/k) is always positive and is just a scaling factor so let’s just remove it so that we have focus on the variable terms:

(Ts – T) = – (Ts4 – T4)

Well, again, this is actually just the same thing as above: if (Ts – T) > 0, and since all T > 0, then (Ts4 – T4) > 0, and so the negative sign is impossible.  The error is indeed in saying that

dQc + dQr = m Cp dT

can have a condition where dT(t) = 0 without both dQc and dQr being zero.  It’s mathematically impossible.  Probably related to the 2nd Law of Thermodynamics.

As I said at the beginning, there was something strange about saying that the heat loss from conduction must equal the heat input from radiation under thermal equilibrium as a general condition.  It’s much more sensible to say that, at thermal equilibrium, both of the heat exchanges from conduction and radiation must each be zero. Notwithstanding that thermal equilibrium was defined in the setup, but then the defined condition of thermal equilibrium, i.e. Kirchhoff’s Law, was dispensed with.

If we look at the full general equation:

k(Ts – T) + sigma(α εs Ts4 – αs ε T4) = m Cp dT

and given that we have proven that there are no mathematical solutions for dT = 0 without both terms on the left hand side also being zero, then that condition can only arise when both terms of the left hand side always have the same sign.  That is, either temperature is increasing and both terms contribute positively, or temperature is decreasing and both terms are negative. They both approach zero from the same side of positive or negative as dT approaches zero.  Why?

If k(Ts – T) was positive, then that means that heat was flowing from the warmer surrounding Ts to the cooler object T.  However, if the other term was negative due to the effects of absorptivity and emissivity, then that would mean that, for that radiative mechanism, heat was flowing from cold to hot.  That is a violation of the 2nd Law of Thermodynamics because the mathematical law of entropy increase does not allow two objects to spontaneously diverge in temperature by any mechanism, since that would mean that there existed a mechanism to spontaneously and passively decrease entropy.  The objects diverge in temperature and entropy decreases if heat flows from cold to hot because that would mean that the cold object loses thermal energy thus decreases in temperature, while the hot object gains that said energy and thus rises in temperature; this is a decrease in entropy.

I’ve said it elsewhere, but to repeat: the Laws of Thermodynamics are actually laws of mathematics, i.e. ontological mathematics or the mathematics of existence.  Such mathematics is a self-consistent and complete system of logic.  If you create an error of mathematics, then you will find the same error popping up other places.  That’s what happened here.  The logic of the physics was violated by setting up thermal equilibrium and then dispensing with the conditions of thermal equilibrium, i.e. Kirchhoff’s Law.  It was then found that the equations under those conditions lead to self-contradiction.  It was then found that such a contradiction arises due to a violation of the 2nd Law of Thermodynamics.  It’s all in the math, and in this case the math of the 2nd Law.

The reason why papers are simply sent back without further perusal when there are early indications that there are problems with the paper, is because it takes so much time to figure out where people eventually really go wrong.  And it is never of any use asking the author to agree with you.

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23 Responses to Heat Flow Cold to Hot when both Conduction & Radiation Occurring?

  1. Derek Alker says:

    To kill an error is as good a service as, and sometimes even better than, the establishing of a new truth or fact. Charles Darwin

    Seems appropriate, brilliant work Joe.

  2. This is how the interaction went after this review:

    PL: “I agree, that’s my equation”.

    JP: “Well, it says –x = +x.”
     
    PL: “Um, just get rid of the negative sign I had in my equation!”

    Now that’s pretty funny!

  3. Pingback: John Robson: If the elites are wrong about global warming, the damage will be enormous – Newsfeed – Hasslefree allsorts

  4. It’s a weird thing this 2nd Law because it seems straightforward, but here I am, a warm blooded creature lit by a hot light bulb on a planet lit by a hot sun and it seems to me that even though heat does *not* go from cold to hot, cold things do get hotter and things do get more organised. I am not saying this guy is right I am just saying the 2nd Law seems a bit strange in real life.
    Martin.

  5. Sunsettommy says:

    Here Is a comment I just got today,that bothered me:

    “>>
    Heat cannot be trapped. Heat cannot be stored, contained or reflected back. This is a law.
    <<
    I used to have a parabolic mirror (about 6" diameter) that would hold a cigarette at the focal point. When I focused sunlight on the end, it would light the cigarette. I wonder where the heat came from if it wasn't reflected?
    And heat can't be stored, but internal energy can. The First Law of Thermodynamics is often stated as dU = [delta]Q + [delta]W (where U is internal energy, Q is heat, W is work, and [delta] is supposed to be a lower case Greek delta, but HTML doesn't work on this site).
    Jim"

    He was replying to the previous comment from Richard 111,who wrote the first line.

  6. D. V. Schroeder, Thermal Physics, Addison Wesley Longman, 2000:

    “Much of thermodynamics deals with three closely related concepts: temperature, energy, and heat. Much of students’ difficulty with thermodynamics comes from confusing these three concepts with each other.”

    The commentator is making the same mistake – confusing energy with heat. It is solar energy being reflected, not heat. The strictness of language is fundamentally important in thermodynamics. It is am important distinction.

    The energy only manifests as heat when it is focused on the cigarette. It would not manifest as heat if it was focused on something already hotter than what the focused light energy could generate.

    Also:

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

    “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.”

    Thus, Richard111 is correct: heat is a transient phenomenon and cannot be trapped stored, or reflected. And to be sure, internal energy is just energy, not stored heat.

    The commentator is doing the usual thing of the confusion of terms, loose language, and resulting in either accidental or intended sophistry.

    The strictness of language in thermodynamics is paramount. It’s not f***ing Shakespeare!!! Although these jerk-faces all treat it that way! You CANNOT interchange and be loose with the terms temperature, energy, and heat!!!

  7. Sunsettommy says:

    He made a confused reply to me and my oh so obvious answer:

    Mine,

    “You wrote SUNLIGHT,which the mirror concentrate to a small point. He is talking about HEAT not being reflected,but IR can be reflected,just like Visible light.”

    his reply,

    “So sunlight isn’t a form of heat energy? What has been heating the Earth for 4.5 billion years if it wasn’t the Sun?”

    my reply,

    “When Visible light strikes a surface, the energy in it is converted to heat, since it encounters matter, that warms up.”

    http://www.globalwarmingskeptics.info/thread-235-post-12639.html#pid12639

  8. See…these idiots don’t abide definitions and are all confused. Sunlight energy is NOT heat unless it meets a situation where it can manifest as heat.

  9. Sunsettommy says:

    He says I am wrong in his reply,but then agrees with this posted, “G. J. V. Wylen, Thermodynamics, John Wiley & Sons, 1960:”

    I has replied,

    “When Visible light strikes a surface, the energy in it is converted to heat, since it encounters matter, that warms up.”

    His reply,

    “That is absolutely not the Thermodynamic definition of heat. Heat only appears at boundaries. It is a transfer of energy from one body to another due to a temperature difference across a system boundary. Heat may be transferred by conduction or radiation or both. (And sunlight contains more frequencies of EMR than just visible light.) Once heat enters a system, it is no longer heat. A bucket of hot water does not “contain” heat. But it does have a higher internal energy than a cold bucket of water with the same mass.”

    http://www.globalwarmingskeptics.info/thread-235-post-12641.html#pid12641

    He is all confused,since Visible light is NOT heat at all. Wasn’t disputing Thermodynamics at all. Not sure what he is getting on, since all I was pointing out is that Visible light is NOT heat at all.

    He is a mess when he writes like this,

    “So sunlight isn’t a form of heat energy? ”

    So confused

    gargle………

    Just posted this reply,

    “Solar radiation are composed of light waves of specific energy levels, NOT heat.”

  10. “Solar radiation are composed of light waves of specific energy levels, NOT heat.”

    Yes that is correct.

    This statement wasn’t correct though:

    “When Visible light strikes a surface, the energy in it is converted to heat, since it encounters matter, that warms up.”

    If the light is absorbed as heat then it is converted to thermal energy, i.e. an increase in temperature.

  11. But remember, light is not always absorbed as heat! It is only absorbed as heat, thus to increase temperature, if the object it strikes is of a lower surface energy flux density.

  12. Sunsettommy says:

    Thanks for the clarification.

    Here is his latest in reply to what I wrote:

    “Solar radiation are composed of light waves of specific energy levels, NOT heat.”

    His reply,

    “It’s called radiation heat transfer.
    The photon is the quanta of electromagnetic (EM) radiation and energy carrier for radiation heat transfer.
    Maxwell: “In Radiation, the hotter body loses heat, and the colder body receives heat by means of a process occurring in some intervening medium which does not itself thereby become hot.”
    So, yes, EMR is heat, by definition.”

    He seems to say my line is wrong,which you told me is correct. According to him it seems that energy AND heat are together in the Solar radiation traveling through interstellar space to Earth.

    Confusing!

    Then follows up with another reply:

    “Temperature is an intensive Thermodynamic property. Entropy is an extensive Thermodynamic property. Entropy and Temperature form a conjugate pair–their product is energy. Pressure (an intensive Thermo property) and volume (an extensive Thermo property) also form a conjugate pair–their product is also energy. If temperature is in Kelvin and entropy is in Joules/Kelvin, then their product is Joules.

    Energy comes in many forms: heat energy in calories or BTUs, work in joules, ft-lbs, newton-meters, watt-hours, ergs, dyne-centimeters, pascal-meters^3 and so on. Energy is an extensive Thermo property.

    Heat is the transfer of energy across a boundary due to a temperature difference. It is (heat) energy in motion–kinda what Thermodynamics means.

    The First Law deals with internal energy (a state variable), heat (a path variable), and work (another path variable). The signs of heat and work are not standard, unfortunately. I learned the Clausius standard, that is heat applied TO the system is positive and work done BY the system is positive. That gives us the equation dU = [delta]Q – [delta]W. Many are using the IUPAC standard which is heat applied TO the system is positive and work done ON the system is positive. That gives us the equation dU = [delta]Q + [delta]W. You need to keep your signs straight, but either equation is valid.

    Do I pass or am I confused?”

    He is making mountain out of a molehill, bla bla bla, is what I see.

  13. These people are just disgusting sophist goblins who like to hear themselves talk.

    Maxwell’s quote does NOT define all EMR as heat. Note that Maxwell refers to transfer from hot to cold, and thus only the EMR transferring in that direction acts as heat for the cooler object. Again, note the quote about confusion between heat, energy, and temperature.

    Light is energy. It, this energy, only acts as heat when transferring from hot to cold.

    Yes, your assessment is correct.

  14. Sunsettommy says:

    Lets see if I got it understood,

    Light photon is an energy carrier from a radiating source,that can become heat when it reach a boundary,which is a temporary event, going from high energy state to a low energy state,a one way transfer, that becomes equalized, which then cause heat to stop being manifested.

    He seems to say that Energy AND Heat occupy the same place in a light photon,that is where I get confused, as they are different things, being mentioned in the same place. How can Light have different states like Energy and Heat within itself at the same time,that is what he seems to be stuck on.

    You say it BECOMES heat, only when energy is being transferred to a cooler object?

  15. The *effect that light energy has will be as heat when the appropriate conditions are there…i.e. when going to a lower energy object. Otherwise the energy doesn’t act as heat…isn’t able to do that “work”.

  16. Heat is something that energy sometimes manifests as. Energy will act as heat when it is traveling down a gradient; it can’t act as heat when it travels up.

  17. Sunsettommy says:

    “Heat is something that energy sometimes manifests as. Energy will act as heat when it is traveling down a gradient; it can’t act as heat when it travels up.”

    Others see it:

    Bryan said
    June 23, 2013 at 3:57 am

    Ball4

    Read section 19.4 of link above.

    To sum up.

    All energy is measured in Joules.

    Not all Joules are equal in their ability to perform thermodynamic work in a given situation.

    No physics textbook will say heat is transferred spontaneously from a lower to a higher temperature object.

    In the case of a purely radiative transfer of energy between a higher temperature object and a lower temperature object.
    1. Photons are emitted and absorbed in both directions.
    2. Energy is transferred in both directions
    3. Heat transfer is a one way process always spontaneously from higher to lower temperature objects.

    https://noconsensus.wordpress.com/2013/06/05/psi-theory-destroyed/#comment-96152

  18. Sunsettommy says:

    It was still resisted by many anyway.

  19. squid2112 says:

    SunsetTommy,

    Just remember, slow cannot speed up fast. Lesser energy cannot increase the state of greater energy. These are fundamental and universal laws that all things in our universe must abide by. There are no exceptions! .. were it not for this very simple and fundamental behavior, our very universe could not exist! .. For all the reasons that Joseph has continually illustrated, from virtually every approachable direction, under no circumstances can a cooler object lead to increased temperature of a warmer object. period. .. It is irrefutable.

    I suggest you ask your friend to produce for you just a single empirical experiment that can prove otherwise. Just one. And no, not some stupid “thought” experiment, but an actual physical experiment that is proof that a cooler object can cause the heating of a warmer object.

    Also remember, placing two banana’s next to each other on your kitchen counter cannot cause either banana to warm. Neither can adding additional coffee to your cup make your coffee any hotter than it originally was. It is simply not possible in this universe!

    Finally, based upon the discussions you have been commenting about, it is quite clear to me that your friend is indeed painfully aware of these things and is simply trying to bullshit his way through your conversation by obfuscating and talking in circles. He obviously does not have control and command of the maths and principals behind Thermodynamics, and is simply trying to baffle you with the brilliance that he does not have.

  20. Can you guys believe what was actually going on with PL once I looked at his math? Isn’t it amazing what transpired? You should see the sh*t storm of emails of him pleading that heat can flow from cold to hot, that his equation can be correct if you just arbitrarily change terms, etc. It’s insane.

    This is not a serious, as in honest or well-intentioned or etc., move on PL’s part. This is not at all about science and legitimate debate. No one, no one in the world writes an equation which says that –x = +x, i.e. -2 = +2, admits that that’s what they’re equation says and is, and then just says, in some cockamamie argument, to get rid of the negative sign to make it work. This is NOT serious. These people prey on our naivety that we do not expect people to outright lie and deceive and sophize. We must all know by now and appreciate that they do. Cotton proved it. PL has proven it again.

    It is not serious to say what he said, therefore, there is some other purpose to it.

    To me it is just such a…an insane occurrence, with what happens. What happened is not serious. No one seriously does what was done. In no way do we ever see things like this in science. Where someone’s equation is, on mathematical certainty, proven to be wrong, and in being wrong, supports the 2nd Law. But then the person asks for the signs to be changed so that heat can flow from cold to hot and violate the 2nd Law. But then not just that, to then go on at length pretending the appearance of an argument which could confuse bystanders with sophistry. It’s quite literally insane on one level, but then so skillfully so! It becomes intelligently psychopathic.

    In any case, everyone continue their vigilance and wariness of what we’re facing and the vectors they may use to try to trip us up, etc.

    It’s almost like whatever spiritual force had been using Cotton for its sophistry and insanity, has now moved on to PL. They sound exactly the same if you were to read PL’s emails since he presented this,

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