What is Heat?

A science and physics lesson on what heat is and how our knowledge of heat debunks the pseudoscientific field of climate science.

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47 Responses to What is Heat?

  1. Brilliant as usual. 💯

  2. Marshall Rosenthal says:

    >

  3. CD Marshall says:

    Just linked this to a clown, perfect timing for his comment and excellent performance btw. Need more MIT level lectures , which I had just watched one on superposition prior to your video popping up.

  4. On a layman’s level the lattice shown in the presentation can be visualised as objects in motion and the collective rate of motion of all the atoms within it as a depiction of the structure’s temperature. Heat, then, becomes the difference in the rates of motion within two objects. A slow object thus cannot make a faster object go faster. This also transfers to the phenomenon of light which we can visualise as the various rates of motion of the atoms/electrons from the object the light came from and that only if they contain higher (faster, more violent, intense, etc, etc) states of motion than the object they connect with can they cause the atoms within it to increase to a more intense state themselves.

  5. Joseph E Postma says:

    Yah after I finished I realized I forgot some of those sorts of comments. Oh well.

    Maybe I’ll do a follow up.

  6. CD Marshall says:

    Someone was trying to use that argument at PSI saying a smaller object moving faster runs into a larger object it is transferring its energy to the largest object…Or something to that point of cold can transfer energy to a hotter object or something like that can’t find the original comment anymore.

    So I said if a smaller object is moving faster and collides into the larger object it has more kinetic energy, thus it is still a transfer of higher to lower not a hot object receiving “heat” from a colder object.

    …And still energy can transfer both ways, they were obfuscating the correct terminology for heat on purpose as per the usual climate clown advocate.

  7. CD Marshall says:

    I really need a very good explanation for Hydrostatic Balance of the Earth. Yes i get the just of it, but I really want a better explanation than the one I’ve been getting. They seem to be “fumbling” with a coherent physics explanation.
    The Hydrostatic Balance keeps the atmosphere from collapsing and/or dissipating out to space thus it is a very precarious balance, but it is not thermal equilibrium or accredited to the ghge.

  8. Kev-In-ZA says:

    Fantastic practical instruction on Internal Energy, Heat and Work. Nicely done Joe. Even though I have studies advanced Heat Transfer and Thermodynamics as part of engineering degree, I thoroughly enjoyed the detailed instruction on how all the DOFs in the object and all the resulting vibrational mods and energy levels end up making up the Planck Function of the emission. Beautiful mathematical function emerging out of random vibration of gazillions of electrically charged particles moving within associated electro-magnetic fields. I still need to spend time understanding the QM and emission theory for my own edification.

    Not sure if you are going to do a follow on to this talking about the primary heat transfer modes between objects of matter, and also how they are somewhat two sides of the same heat-transfer coin (action though close contact vs action at a distance)

  9. arfurbryant says:

    Over fifty years ago a British comedy duo made this short song which says it all. I urge you to listen all the way through. Fabulous and exactly what Joe is talking about…!
    https://www.google.com/search?q=thermodynamics+music+work+is+heat&oq=thermodynamics+music+work+is+heat&aqs=chrome..69i57j33i22i29i30l4.23983j1j7&client=ms-android-huawei&sourceid=chrome-mobile&ie=UTF-8#
    Enjoy!

  10. arfurbryant says:

    A better link?

  11. Zelator says:

    Lol Excellent Arfur

  12. Herb Rose says:

    Hi Joe,
    The comment CD was referring to was mine. When a moving object is struck from behind by an object with more velocity (traveling in the same direction) the velocity (energy) of the moving object will increase even if the striking object has less kinetic energy (less mass). Since temperature is a measurement of kinetic energy, not energy, a cooler object can add heat to a hotter object.
    In the atmosphere O2 and N2 are absorbing energy (uv) from the sun. This is what causes O2 molecules to split forming oxygen atoms, N2O, and ozone at different altitudes in the atmosphere. Because this energy is being absorbed the uv light decreases as it comes to the surface of the Earth but O2 and N2 (because of their bond length) will continue to absorb energy whenever uv light strikes them. Since there is no lattice structure between gas molecules they do not distribute energy (except during collision and by radiation) to a neighboring molecules as is don in solids and liquids. The few molecules in the colder atmosphere contain more energy (have greater velocity) but less mass than the molecule lattice forming the surface of the Earth and are able to transfer energy to it.

  13. Kev-In-ZA says:

    @arfurbryant, what a brilliant fine. I think the essence of the 2LoT got lost when they generalized to entropy can never decrease, and that completely lost the practical meaning that anyone can understand.

  14. But as I said in the video Herb, temperature is not energy nor is it merely a “measure” of kinetic energy. Temperature is energy flux density as can be seen via the Steffan Boltzman Law F = sT^4, or the internal thermal energy equation dU = mCpdT => dT = dU/m/Cp which shows that temperature is about energy normalized to mass and its efficiency at holding internal energy.

    And so in fact, the smaller mass moving faster has higher kinetic energy density, thus it speeds up the heavier slower mass.

    Nice try, but as always, these arguments will only ever be used by me for physics education…not reality inversion.

  15. boomie789 says:

    Made a meme

  16. Barry says:

    Thanks for this Joseph it has really helped me as a layperson to visualize how this thing works and why cold doesn’t make hot hotter. I totally believe in established thermal dynamic principles but have trouble trying to explain the basics to others. Having this simple diagram makes it so easy to explain without having your grasp of it. Keep up the great work.

  17. Zelator says:

    Lol Boomie that reminds me of Roddy Piper in “They Live” PUT THE GLASSES ON !!!! See the Truth!!

  18. arfurbryant says:

    The reason for posting the Flanders and Swann video is to question, if two public school educated musicians can so clearly understand the Laws of Thermodynamics seventy years ago, why can’t modern-day ‘climate scientists’ do the same today?

  19. Philip Mulholland says:

    Herb Rose,

    You write

    When a moving object is struck from behind by an object with more velocity (traveling in the same direction) the velocity (energy) of the moving object will increase

    You are mixing terms in a glorious kaleidoscope of nonsense; it would greatly help your understanding of science if you applied dimensional analysis.
    Velocity has dimensions LT^-1 and is measured in metres per second whereas Energy has dimensions ML^2T^-2 and is measured in Joules.

    even if the striking object has less kinetic energy (less mass).

    Glorious nonsense: Kinetic Energy has dimensions ML^2T^-2 whereas Mass has dimensions of well just M.

    Since temperature is a measurement of kinetic energy, not energy,

    You are now saying that kinetic energy with dimensions ML^2T^-2 is not energy with dimensions ML^2T^-2. You have excelled yourself.

    a cooler object can add heat to a hotter object.

    A cold object cannot power a hot object, but I guess that concept completely eludes you.

  20. Philip Mulholland says:

    In reply to arfurbryant
    Galaxy Song – Monty Python’s The Meaning of Life

  21. arfurbryant says:

    @Philip
    Brilliant. Thanks for reminding me!😉

  22. Philip Mulholland says:

    Following on from Joe’s comment:
    “The internal thermal energy equation dU = mCpdT => dT = dU/m/Cp which shows that temperature is about energy normalized to mass and its efficiency at holding internal energy.”
    I have been trying to find out more about Enthalpy.
    This website has some very useful explanation pages:
    https://www.linseis.com/en/wiki-en/
    https://www.linseis.com/en/properties/enthalpy/
    https://www.linseis.com/en/wiki-en/what-does-thermal-conductivity-mean/

  23. CD Marshall says:

    I looked up “The first Law of Thermodynamics in terms of enthalpy” at MIT I got this…Didn’t really help.

    dU = dQ – dW (for any process, neglecting DKE and DPE)

    dU = dQ – pdV (for any quasi-static process, no DKE or DPE)

    H = U + pV therefore dH = dU + pdV + Vdp

    so dH = dQ – dW + pdV + Vdp (any process)

    or

    dH = dQ + Vdp (for any quasi-static process)

  24. CD Marshall says:
  25. Kev-In-ZA says:

    @Philip Mulholland: I’m a little rusty on my thermodynamics, but Enthalpy (H) in practical terms is Internal Energy (U) plus the Pressure x Volume (pV). Enthalpy is a slightly more complete measure of the the “energy” state of some matter. Since you can get Work out of Pressure acting over Distance (think steam engine, pressurized tyre blow-out, etc), and when you have high temperature gas as the working fluid in a process cycle, you can generate pressure to do work. Hence, Enthalpy is a key practical thermodynamic property for some process analysis.

    I’m not sure, but I don’t think it is of much use for climate physics beyond the altitude temperatures-pressure lapse rate, but you can get that without invoking Enthalpy.

    Hope that helps a bit.

  26. Michael says:

    It feels to me, as a layman who seeks to learn more on this subject, that the concept of heat transfer being put forth by Mr. Postma makes perfect sense in the context of our every day lives. The idea that a cooler object can add heat to a warmer object is asinine. It seems that alarmists want to simply complicate what should be a simple explanation in order to mislead. Or maybe they are that dense. My physics education ends at high school level and this, at a base level, makes sense even to me.

  27. Michael – You are exactly correct: it is asinine, they are complicating it all beyond comprehension and reason, it is all very misleading, but I don’t know if it all because they are malicious or stupid – I believe it is both.

    Thanks for the support! It does all make sense. Science does make sense. What these people are doing makes no sense at all though.

  28. LOL@Klimate Katastrophe Kooks says:

    Starting at 15:00… not all energy is heat, only the differential in energy density (remember that temperature is a measure of energy density, equal to the fourth root of energy density divided by the radiation constant) can be considered heat.

    Now remember also that energy only flows if there is an energy density gradient. This includes energy in the form of photons.

    For instance, a bound electron in an atom or molecule is always attempting to reach a lower energy state. It is the energy density of the quantum vacuum which sets the bound electron ground state orbital radius because once the bound electron reaches an orbital radius which:

    1) has an integer number of de Broglie waves in the orbital, and…

    2) is near enough to ambient energy density such that release of a photon and subsequent quantum jump would put the bound electron at an energy state lower than ambient…

    … it cannot emit because energy cannot spontaneously flow up an energy density gradient.

    The same goes for excited bound electrons, except that ambient energy density is locally higher than the quantum vacuum ground-state energy density, exciting the electron to a higher orbital radius. Once that local excitation energy is removed and local ambient energy density falls, the bound electron can emit a photon and fall to a lower orbital radius with the same conditions as 1) and 2) above.

    So 2LoT holds, macroscopically and quantumly.

    [1] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.11.790
    [2] https://web.archive.org/web/20190713220130/https://arxiv.org/ftp/quant-ph/papers/0106/0106097.pdf
    [3] https://web.archive.org/web/20190713225420/https://www.researchgate.net/publication/13330878_Ground_state_of_hydrogen_as_a_zero-point-fluctuation-determined_state
    “We show here that, within the stochastic electrodynamic formulation and at the level of Bohr theory, the ground state of the hydrogen atom can be precisely defined as resulting from a dynamic equilibrium between radiation emitted due to acceleration of the electron in its ground-state orbit and radiation absorbed from zero-point fluctuations of the background vacuum electromagnetic field, thereby resolving the issue of radiative collapse of the Bohr atom.”

    [4] https://web.archive.org/web/20180719194558/https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150006842.pdf

    “The energy level of the electron is a function of its potential energy and kinetic energy. Does this mean that the energy of the quantum vacuum integral needs to be added to the treatment of the captured electron as another potential function, or is the energy of the quantum vacuum somehow responsible for establishing the energy level of the ‘orbiting’ electron? The only view to take that adheres to the observations would be the latter perspective, as the former perspective would make predictions that do not agree with observation.”

    Idealized blackbody objects assume emission to 0 K and ε=1. At > 0K they emit as though to a 0 K ambient, and their radiant exitance is not proportional to energy density gradient. They maximally emit and absorb. Of course, they also don’t exist… they’re idealizations. They use this form of the S-B equation: q = σ T^4

    Real-world graybody objects assume emission to > 0 K and ε<1. They emit when their temperature is > 0 K above their ambient, and their radiant exitance is proportional to the energy density gradient between object and ambient. They use this form of the S-B equation: q = ε σ (T_h^4 – T_c^4) A_h

    https://i.imgur.com/QErszYW.gif

    The climastrologists misuse the S-B equation… they use it by calculating as though all objects are emitting to 0 K, then they subtract cooler object energy flow from warmer object energy flow. They essentially treat all objects as though they’re idealized blackbody objects, just with ε<1 (sometimes… the Kiehl-Trenberth Energy Balance graphic treated the planet exactly as an idealized blackbody object with emission to 0 K and ε=1).

    Except doing it their way inflates radiant exitance of all objects (because they’re calculating for emission to 0 K via q = ε σ T^4).

    That’s not how the S-B equation is supposed to be used, it’s supposed to be used to subtract cooler object energy density from warmer object energy density:

    T = 4^√(e / (4σ / c))

    T^4 = (e / (4σ / c))

    q = ε σ (T_h^4 – T_c^4) A_h

    q = ε σ (e_h / (4σ / c) – e_c / (4σ / c)) A_h

    Energy only flows with an energy density gradient; free energy is defined as the capacity to do work; if there is no free energy, no energy can flow because no work can be done.

    Or, to put it as succinctly as possible, heat is an energy flux. That’s it, that’s all.

    The layman use of ‘heat’ is not the scientific definition of ‘heat’.

    An aside:
    Thermodynamics is completely analogous to electrical theory. I put a thermodynamics problem into a circuit simulator. Most people can think in terms of current flow and voltages more easily.
    https://tinyurl.com/yzo8hak9

  29. LOL@Klimate Katastrophe Kooks says:

    We can look at the definitions of emissivity, reflectivity, transmissivity and absorptivity to determine that the above is correct, and that energy does not flow unless there is an energy density gradient.

    α = absorptivity
    ρ = reflectivity
    τ = transmissivity
    ε = emissivity

    α + ρ + τ = 100% where τ = 0% for opaque objects (ie: the radiation incident upon an opaque object is either absorbed or reflected). ε + ρ = 100% for opaque objects (ie: radiation measured from a surface is a mix of reflected and emitted, because there’s no way to easily distinguish between the two), thus at thermodynamic equilibrium, when no object can emit because there exists no energy density gradient and thus energy cannot flow (remember 2LoT), per the definition of emissivity (“the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature.“), ρ must = 100% alone… energy is reflected from the object at thermodynamic equilibrium, not absorbed by it; thus energy from cooler objects certainly isn’t absorbed by a warmer object under any circumstances. And if ρ = 100% and τ = 0%, then α = 0%.

    Thus, at thermodynamic equilibrium between two objects, a standing wave is set up, the photons in the intervening space being those emitted by the objects as they (and the intervening space) came into thermodynamic equilibrium, those photons perfectly reflected from each surface to create that standing wave.

    Should one object change temperature, that standing wave becomes a traveling wave, with the group velocity proportional to the energy density differential between the objects.

    Now, rather than objects, let’s visualize a cavity at thermodynamic equilibrium. If the walls absorbed and emitted even at thermodynamic equilibrium as the climastrologists claim, that would double the energy density in the intervening space. And since temperature is a measure of energy density (equal to the fourth root of energy density divided by the radiation constant), that would mean the intervening space would not be in thermodynamic equilibrium, it would be hotter than the walls and would thus transfer energy to the cavity walls, heating them up…. then the walls would emit at thermodynamic equilibrium back into the intervening space, raising its energy density and thus temperature even more… round and round it goes, where it stops nobody knows.

    So the climastrologist take on thermodynamics demonstrably violates Stefan’s Law, and thus violates 1LoT, 2LoT, and creates a perpetuum mobile, to boot.

    That’s what physicist Dr. Charles R. Anderson, PhD shows here:
    https://objectivistindividualist.blogspot.com/2018/08/the-nested-black-body-shells-model-and.html

  30. LOL@Klimate Katastrophe Kooks says:

    Hey, Joe.

    I think it’d be pretty neat if you did your own analogization of thermodynamics and electrical theory (as I do in my prior post), to include a live circuit simulator embedded right in the article.

    You can embed the falstad.com circuit simulator into the web page via an iframe.

    Look at the source code here to see how he did it:
    http://falstad.com/circuit/

    As I said in my prior post, most people can think in terms of voltages and current flows more easily than radiant exitance and energy density.

    So when a climate loon tells them that energy can flow at thermodynamic equilibrium, they can analogize that to two 1.5 volt batteries connected ‘+ to +’ and ‘- to -‘ and they can easily see that there will be no current flow, just as for two objects at the same temperature, there will be no energy flow.

    So when a climate loon tells them that a cooler object can warm a warmer object, they can analogize that to a 1.5 volt battery connected ‘+ to +’ and ‘- to -‘ to a 12 volt battery, and they easily see that the 1.5 V battery won’t do any work upon the 12 V battery, just as a 300 K object will do no work upon a 400 K object.

  31. CD Marshall says:

    “All shall kiss the climate ring”
    Even hyperphysics…
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/grnhse.html#c1

  32. Joseph E Postma says:

    Nice find CD. Doing a new OP for that one 🙂

  33. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “For example two boxes of ideal gas with different specific heats have energy density u=cv*T . If they have the same energy density u but different cv then they have different temperature T.”

    That formula deals with the change of temperature of an ideal gas during a constant volume process.

    Δe = c_v * ΔT where ΔT is the change of temperature of the gas during the process,and c is the specific heat capacity. The subscript “v” is a reminder that this is a constant volume process.

    Temperature is a measure of energy density. If energy density changes, temperature changes:

    T = 4^√(e / (4σ / c))

    T^4 = (e / (4σ / c))

    q = ε σ (T_h^4 – T_c^4) A_h

    q = ε σ (e_h / (4σ / c) – e_c / (4σ / c)) A_h

  34. Yes temperature is energy density normalized to entropy which is what the c_v does. You can read that right in the equation there. An equal change in temperature is a requisite change in energy normalized to c_v. Only with equal substances can you say energy density difference is the same as a temperature difference.

  35. CD Marshall says:

    Technically U does not have a T if I’m following this correctly in regards to the atmosphere. A parcel of air can have LTE and actually have varied internal energy molecular states. In fact they would have to with a billion +/- collisions a second you would think “true” LTE is ideally not possible except for a “proxy” mathematical equation.

    Or something like that?

  36. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “So temperature is not a measure of energy density!”

    Well, let’s noodle our way through this…

    The Equipartition Theorem dictates that temperature is a measure of energy density (explained below).

    ΔT = Δe / c_v

    Specific heat capacity has a ‘stair-step’ effect because of the Equipartition Theorem. Below certain energy levels, certain DOF (Degrees of Freedom) are ‘frozen out’ and thus energy cannot be equipartitioned into those rotational and vibrational mode quantum states. In this manner, it is quite akin to latent heat capacity (which deals with phase change).

    Here’s hydrogen:

    The three steps signify the energy levels that must be reached to equipartition energy into the rotational and vibrational mode quantum states. Translational mode corresponds to three degrees of freedom, rotational mode to another two, and vibrational mode to yet another two.

    It was this finding which helped to solidify quantum theory by properly explaining specific heat capacity.

    In dealing with solely translational mode energy and neglecting vibrational mode and rotational mode energy for the moment, the Equipartition Theorem states that molecules in thermal equilibrium have the same average energy associated with each of three independent degrees of freedom, equal to:

    3/2 kT per molecule, where k = Boltzmann’s Constant
    3/2 RT per mole, where R = gas constant

    At common atmospheric temperatures:
    Specific gas constant = 28.26916348 J mol-1 K-1 for CO2
    Specific gas constant = 20.8 J mol-1 K-1 for O2
    Specific gas constant = 20.78614962 J mol-1 K-1 for N2
    Specific gas constant = 12.58362 J mol-1 K-1 for Ar

    ==========

    ASIDE: The above is why a higher atmospheric concentration of CO2 will convectively transit more energy from surface to emission height, the tropospheric thermalization by CO2 only having the net effect of increasing CAPE (Convective Available Potential Energy) which increases convection. Thus CO2 is a net atmospheric radiation coolant at all altitudes except for negligible warming at the tropopause, where it absorbs a greater proportion of cloud-reflected solar insolation and radiation from cloud condensation.

    NASA JPL:

    Clough & Iacono:

    https:/co2islife.files.wordpress.com/2017/01/spectralcoolingrates_zps27867ef4.png

    ==========

    Thus the Equipartition Theorem equation:

    KE_avg = 3/2 kT

    … serves well in the definition of translational mode kinetic energy (which we sense as temperature).

    It does not do as well at defining the specific heat of polyatomic gases, simply because it does not take into account the increase of internal molecular energy via vibrational mode and rotational mode excitation. Energy imparted to the molecule via either photon absorption or collisional energetic exchange can excite those vibrational mode or rotational mode quantum states, increasing the total molecular energy E_tot, but not affecting temperature at all.

    While that stored (specific heat capacity) energy doesn’t affect the current temperature of the gas (because temperature is a measure of translational mode kinetic energy only), if that gas were to lose energy, the Equipartition Theorem would dictate that some of the stored (specific heat capacity) energy would be equipartitioned back into translational mode kinetic energy upon collision with other molecules… thus the gas changes temperature more slowly when its specific heat capacity is higher.

    Thus, for temperature changes small enough that we don’t breach those energy levels at which equipartition of energy into those rotational and vibrational modes occurs or is ‘frozen out’, specific heat capacity can be considered a constant.

    ΔT = Δe / c_v

    For identical parcels of gas, the same change in energy density gives the same change in temperature, and vice versa.

    And if specific heat capacity is considered a constant, then even the formula above shows that temperature is a measure of energy density.

    U = T^4 4σ/c
    e = T^4 4σ/c
    T = 4^√(e/a)
    e = aT^4

    All of those formulae are functionally identical.

  37. LOL@Klimate Katastrophe Kooks says:

    “net atmospheric radiation coolant”
    – should be –
    “net atmospheric radiative coolant”

  38. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “energy density is U/V”

    U = T^4 4σ/c
    The above formula is the Stefan-Boltzmann relation between energy density and temperature.

    Energy density is not U/V, U already has units of J m-3.

    U = E / V

    ∴ T^4 4σ/c = E/V
    ∴ E = T^4 4σ/c * V

    If volume changes, of course temperature is going to change, for the same E.

  39. LOL@Klimate Katastrophe Kooks says:

    Also remember that temperature in this context is a measure of translational mode kinetic energy only… energy will be equipartitioned into (from translational mode) / out of (to translational mode) E_rot, E_vib and E_elec when energy is added to / removed from the system (depending upon which quantum states are not ‘frozen out’), slowing the rate of temperature change for higher specific heat capacity molecules.

    There is also rotational and vibrational temperature, used to express the rotational mode and vibrational mode partition function. It also has units of temperature (K).

    θ_rot = (hc B/hc)/k_B

    θ_vib = hνc/k_b

    Where:
    B/hc = rotational constant
    h = Planck constant
    c = speed of light
    k_B = Boltzmann constant
    ν = characteristic oscillator frequency

    This is the temperature at which translational mode kinetic energy will begin equipartitioning to and thus exciting rotational mode and vibrational mode quantum states.

    Given that in this context temperature is a measure of translational mode kinetic energy, and given that pressure and temperature are intimately connected…

    KE = (DOF / 2) k_B T
    T = (2 KE) / (DOF k_B)

    In statistical mechanics the following molecular equation is derived from first principles: P = n k_B T for a given volume.

    Therefore T = (P / (n k_B)) for a given volume.

    Where: k_B = Boltzmann Constant (1.380649e−23 J·K−1); T = absolute temperature (K); P = absolute pressure (Pa); n = number of particles

    If n = 1, then T = P / k_B in units of K / m³ for a given volume. We will cancel volume in a bit.

    We can relate velocity to kinetic energy via the equation:
    v = √(v_x² + v_y² + v_z²) = √((DOF k_B T) / m) = √(2 KE / m)
    As velocity increases, kinetic energy increases.

    Kinetic theory gives the static pressure P for an ideal gas as:
    P = ((1 / 3) (n / V)) m v² = (n k_B T) / V

    Combining the above with the ideal gas law gives:
    (1 / 3)(m v²) = k_B T

    See what I did there? I equated kinetic energy to pressure over that volume, thus canceling that volume, then solved for T.

    ∴ T = mv² / 3 k_B for 3 DOF

    ∴ T = 2 KE / k_B for 1 DOF

    ∴ T = 2 KE / DOF k_B

  40. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “Feynman says math is the language of physics, and if you are not careful with every step of the math, you are just saying words, which are not the language of nature.”

    “Klimate, I use U to mean total energy with units of J.”

    Total energy is not denoted by U. It is denoted by E_tot. Energy density is denoted by U or e, depending upon the field.

    You’re attempting to conflate kinetic energy density of an ideal gas (since it’s an ideal gas, only kinetic energy is taken into account, hence the enthalpy of an ideal gas is independent of its pressure or volume) with Stefan’s Law for radiation energy density. We’re dealing with radiative energy transfer in debunking CAGW (since CAGW proponents claim radiative energy transfer is the predominant energy transfer mechanism causing CAGW).

    The equation for the radiation energy density is Stefan’s Law and a is Stefan’s constant.
    e = aT^4 where a = 4σ/c, ∴ T = 4^√(e/4σ/c)

    Given that σ and c are constants, Stefan’s law states that temperature is a measure of energy density.

    In thermodynamics, Stefan’s formula (denoted by σ) states that the specific surface energy at a given interface is determined by the respective enthalpy difference. Enthalpy is the sum of the system’s internal energy and the product of its pressure and volume. H = U + pV, thus for a constant pressure and volume, temperature is a measure of energy density because U = T^4 4σ/c (the Stefan-Boltzmann relation between energy density and temperature).

    E_tot = E_tran + E_rot + E_vib + E_elec
    Sometimes E_grav is also used to account for adiabatic temperature change with altitude

    Temperature is not a measure of E_tot, because temperature only applies to E_tran. The energy stored in E_rot, E_vib, E_elec simply equipartitions into / out of E_tran (the basis for specific heat capacity) as energy is added/removed from the system, slowing the rate at which E_tran changes.

    Now calculate the specific enthalpy of your two boxes.

  41. LOL@Klimate Katastrophe Kooks says:

    When all else fails, do dimensional analysis.

    M = mass
    L = length
    T = time
    K = temperature
    I = electrical current
    N = number
    J = luminous intensity

    Temperature: [M0 L0 T0 K1 I0 N0 J0]

    Speed: [M0 L1 T-1 K0 I0 N0 J0]

    σ: [M1 L0 T-3 K-4 I0 N0 J0]

    σ/c: [M1 L0 T-3 K-4 I0 N0 J0] / [M0 L1 T-1 K0 I0 N0 J0] = [M1 L-1 T-2 K-4 I0 N0 J0]

    Radiation density constant (a): J m-3 K-4
    [M1 L2 T-2 K0 I0 N0 J0] [M0 L-3 T0 K0 I0 N0 J0] [M0 L0 T0 K-4 I0 N0 J0] = [M1 L-1 T-2 K-4 I0 N0 J0]

    Energy Density: [M1 L-1 T-2 K0 I0 N0 J0]

    e = aT^4 where a = 4σ/c, ∴ T = 4^√(e/4σ/c)
    [M1 L-1 T-2 K0 I0 N0 J0] = [M1 L-1 T-2 K-4 I0 N0 J0] [M0 L0 T0 K4 I0 N0 J0]

    Temperature is a measure of energy density.

  42. LOL@Klimate Katastrophe Kooks says:

    Check my math.

    Let’s take your example:
    …”take two boxes of monatomic ideal gases with T1=200, T2=100 . We know heat will flow from box 1 to box 2. Both have V=1. Box 1 has N1=100, box two has N2=1000. Then box 1 has energy density E1= 3/2 kB T N/V = 4.14e-19 J/m^3. Box 2 E2=2.07e-18 J/m^3″

    And change it a bit.

    Box 1 T = 100 K
    Box 1 N = 100
    Box 1 V = 1 m^3

    Box 2 T = 100 K
    Box 2 N = 1000
    Box 2 V = 1 m^3

    p = nRT / V

    R = 8.31446261815324 m^3 Pa K-1 mol-1 where:
    R = (F/A * V)/(NT) = FL/NT = Work / NT

    Box 1: (100 atoms / 6.02214076e23 atoms mol−1) * 8.31446261815324 m^3 Pa K-1 mol-1 * 100 K / 1 m^3 = 1.380649e-19 Pa

    Box 2: (1000 atoms / 6.02214076e23 atoms mol−1) * 8.31446261815324 m^3 Pa K-1 mol-1 * 100 K / 1 m^3 = 1.380649e-18 Pa

    It’s pretty easy to see that the energy you’re talking about is locked up in pressure in this case. Given that the two boxes are presumably coterminous but not able to exchange mass, nothing is going to flow… not mass, not energy.

    Now, going back to your example:
    …”take two boxes of monatomic ideal gases with T1=200, T2=100 . We know heat will flow from box 1 to box 2. Both have V=1. Box 1 has N1=100, box two has N2=1000. Then box 1 has energy density E1= 3/2 kB T N/V = 4.14e-19 J/m^3. Box 2 E2=2.07e-18 J/m^3″

    Box 1 T = 200 K
    Box 1 N = 100
    Box 1 V = 1 m^3

    Box 2 T = 100 K
    Box 2 N = 1000
    Box 2 V = 1 m^3

    Box 1: (100 atoms / 6.02214076e23 atoms mol−1) * 8.31446261815324 m^3 Pa K-1 mol-1 * 200 K / 1 m^3 = 2.761298e-19 Pa

    Box 2: (1000 atoms / 6.02214076e23 atoms mol−1) * 8.31446261815324 m^3 Pa K-1 mol-1 * 100 K / 1 m^3 = 1.380649e-18 Pa

    We can also use the Boltzmann constant, which helps to highlight that the energy is locked up mainly in pressure.

    k_B = 1.380649e−23 J K-1

    P = N k_B T / V

    = 100 atoms * 1.380649e−23 J K-1 * 200 K / 1 m^3 = 2.761298-19 Pa

    = 1000 atoms * 1.380649e−23 J K-1 * 100 K / 1 m^3 = 1.380649e-18 Pa

    Now, how much energy can be transferred from Box 1 to Box 2?

    = 1.380649e−23 J K-1 * 100 K = 1.380649e-21 J / 100 atoms = 1.380649e−23 J per atom

    How much energy is locked up in pressure vs. that which can be transferred from Box 1 to Box 2 (assuming, of course, that Box 2 remains at 100 K… you can do the math for yourself for Box 2 increasing in temperature as it receives energy from Box 1, but the percentage transferred below would be even lower)?

    1.380649e-21 J transferred

    100 atoms * 1.380649e−23 J K-1 * 100 K = 1.380649e-19 J

    1% of the energy can be transferred, 99% is locked up in pressure.

  43. LOL@Klimate Katastrophe Kooks says:

    Bah, gotta get more sleep. 50% of the energy is locked up in pressure, 50% transferred. Is that right? Somebody check my math.

    Off to bed, long day.

  44. LOL@Klimate Katastrophe Kooks says:

    A CO2 laser emits 10.6 μm, with a Wien Displacement Law peak equivalent temperature of 273.4 K. How does a CO2 laser cut or burn anything warmer than 273.4 K?

    E = n * h * f = P * t combined with a tight focal point, right? Power equals Work / Time, right?

    When all else fails, do dimensional analysis:

    M = mass
    L = length
    T = time
    K = temperature
    I = electrical current
    N = number
    J = luminous intensity

    Power: [M1 L2 T-3 K0 I0 N0 J0]
    Time: [M0 L0 T1 K0 I0 N0 J0]
    Area: [M0 L3 T0 K0 I0 N0 J0]

    Power * time (work) / area = [M1 L2 T-3 K0 I0 N0 J0] [M0 L0 T1 K0 I0 N0 J0] = [M1 L2 T-2 K0 I0 N0 J0]/[M0 L3 T0 K0 I0 N0 J0] = [M1 L-1 T-2 K0 I0 N0 J0]

    What are the dimensions of energy density?

    [M1 L-1 T-2 K0 I0 N0 J0], right?

    2LoT asserts that energy has quality as well as quantity… you’re looking only at quantity (ie: temperature) when claiming that energy flows higher temperature to lower temperature, but you’re including other forms of energy. Per your metric of including the internal energy locked up in pressure (which, being in separate boxes, isn’t allowed to flow), we should also be including the energy of the rest mass of the particles themselves (E^2 = m^2 c^4 + p^2 c^2). Of course, that energy is locked up in rest mass.

    Spontaneous processes occur in the direction of decreasing quality of energy (increasing entropy / increasing number of microstates).

    Higher energy density implies higher energy quality (lower entropy / lower number of microstates)… it wouldn’t spontaneously be in a higher energy density / lower entropy / lower number of microstates state if it were more highly entropied than the ambient (here I use the proper sign of entropy… lower entropy implies more order, higher entropy implies more disorder).

    dQ = T dS = du + P dV
    Where:
    dQ = energy input / loss
    T = temperature
    S = entropy per gram
    u = internal energy density per gram
    P = pressure
    V = 1/ρ (specific volume, the inverse of density)

    The equation above gives a balance between the energy input / loss, the change in gas energy, and the mechanical work. Entropy, energy density, and pressure are all functions of density and temperature: S(ρ, T) | u(ρ, T) | P(ρ, T) , but the energy input / loss Q is not a unique function of (ρ, T).

  45. LOL@Klimate Katastrophe Kooks says:

    “area” should be “volume” above.

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