.. physicists are now saying "Climate scientists should start listening to physicists about physics." [Jane Q. Public, 2012-04-14]

Does Jane listen to physicists about physics?

.. A cooler object cannot increase the temperature of a warmer object via thermal radiation. It just doesn't happen. Ask any physicist. .. [Jane Q. Public, 2012-04-17]

.. An article by Spencer linked to elsewhere in this discussion (look for "Yes, Virginia") describes this concept of back-radiation, which is central to many of the AGW models. The article that I linked to above is by a Ph.D. physicist, refuting the first article. .. [Jane Q. Public, 2012-04-18]

Does Jane think PhD physicists are credible regarding physics?

And yet the "climate scientists" themselves have not been asking the statisticians about the math or physicists about the physics. [Lonny Eachus, 2012-04-20]

Does Lonny Eachus ask physicists about the physics?

.. climate scientists themselves have not been consulting .. physicists about the physics! [Jane Q. Public, 2012-05-02]

Does Jane consult physicists about the physics?

.. How many of the CO2 models rely on the concept of "back radiation" to explain the radiative forcings? There's a bit of a problem with that: "back radiation" is physically impossible. Again see that link to the article by Latour (a physicist) who shows very clearly exactly why that is so. .. [Jane Q. Public, 2012-05-10]

So anyway, here is physicist Pierre Latour, refuting Spencer's explanation: bit.ly/JV9XmI [Lonny Eachus, 2012-05-21]

.. the CO2-warming model rely on the concept of "back radiation", which physicists (not climate scientists) have proved to be impossible. I'm happy to leave actual climate science to climate scientists. But when THEIR models rely on a fundamental misunderstanding of physics, I'll take the physicists' word for it, thank you very much. .. [Jane Q. Public, 2012-07-05]

Does Jane actually take the physicists' word for it?

.. now it's physicists saying that they've got the physics wrong. .. [Jane Q. Public, 2012-07-05]

.. They have been accused of getting the physics of their models wrong by professional, well-respected physicists. [Jane Q. Public, 2012-07-05]

Actually, the rules aren't even well-known. The majority of CO2 warming models rely on a concept of "back radiation" that (according to physicists) does not even exist.. [Jane Q. Public, 2012-07-15]

After Namarrgon notes that Dr. Latour is actually a chemical process engineer, Jane admits his mistake:

By the way: Latour is a process engineer with particular expertise in thermodynamic control systems. If I were in a room in which you challenged him over thermodynamics, I'd probably want to go outside to avoid the bloodbath. Good luck with that whole argument. To say it's weak is just.. well.. weak. [Jane Q. Public, 2012-11-20]

A large body of scientists who are PHYSICISTS agree with me. A large body of scientists who are CLIMATE RESEARCHERS disagree. .. which group should I listen to? The ones whose SPECIALTY it is, or the tyros? Go learn a little humility yourself. Like for example learning to admit when you're wrong. [Jane Q. Public, 2013-05-30]

I showed Jane statements from the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society. Spoiler alert: mainstream physicists don't agree with the Slayers.

Maybe Jane doesn't actually take the physicists' word for it?

.. To the best of my knowledge -- and I have been following the issue -- not one physicist has even attempted to refute LaTour's analysis, while a number of physicists have backed him up. .. [Jane Q. Public, 2013-05-30]

rgbatduke is Prof. Brown, a physicist who'd refuted Dr. Latour's analysis directly to Jane, but as usual Jane just doubled down. On a Slayer blog post about Prof. Brown, Lonny Eachus even repeated Jane's arguments to physicist Joel Shore, who refuted Lonny.

Maybe Jane/Lonny Eachus doesn't actually take the physicists' word for it?

.. I consult "the experts". When it's a question of physics, for example, I look to references from physicists, not climatologists. After all, physicists are "the experts" when it comes to physics. [Jane Q. Public, 2013-11-15]

Does Jane really think physicists are "the experts" when it comes to physics?

.. First, they mention the theory of AGW "radiative" forcing, which as I stated earlier is probably myth, according to physicists and experts in radiative heat transfer. .. [Jane Q. Public, 2013-12-23]

Funny.. it's physicists and recognized experts in radiative heat transfer who are disagreeing with the concept. Since the concept involves physics and radiative heat transfer, I am rather inclined to believe them over "climate scientists". [Jane Q. Public, 2013-12-24]

.. Jesus, man. This guy designed heat transfer control systems for NASA. Do you really think he's going to make that kind of mistake? [Jane Q. Public, 2014-02-11]

Hopefully he just made elementary mistakes, rather than deliberately spreading civilization-paralyzing misinformation. Sadly, the result isn't too different either way.

.. Further, why do you imply that climate scientists are experts on thermodynamics? That's an area of physics, not climatology, and I know some physicists who very much disagree with today's mainstream "climate science". [Jane Q. Public, 2014-03-07]

.. I repeatedly linked you and others in the past to PHYSICISTS who say otherwise. .. [Jane Q. Public, 2014-03-21]

.. Latour is a control engineer for chemical processes and he has designed heat-transfer systems for NASA. [Jane Q. Public, 2014-03-22]

Latour designs heat-transfer control systems for a living. He did it for NASA, among other notables. [Jane Q. Public, 2014-03-24]

.. There are also physicists who worked for NASA, and other science professionals, currently challenging the very foundations of AGW theory. .. [Jane Q. Public, 2014-03-31]

Does Jane think physicists who work for NASA are credible regarding physics?

.. Even if you did not take his word for it, his career building control systems precisely for the purpose of managing heat transfer would strongly suggest that this is hardly something he is likely to neglect. .. [Jane Q. Public, 2014-08-01]

.. Dr. Latour did heat-transfer work for NASA, and has made a career of building control systems for chemical processes involving heat. I daresay he is more of an expert on the subject than "Khayman80". .. [Jane Q. Public, 2014-07-25]

Did heat-transfer work for NASA, or managed NASA's Apollo Docking Simulator development? Doesn't seem to matter, as long as he did it for NASA. If having worked for NASA gives Dr. Latour credibility, shouldn't Jane find climate.nasa.gov credible?

Jane's repeatedly implied that working for NASA gives one credibility, that physicists are "the experts" when it comes to physics, and that Jane "takes the physicists' word for it." I'm skeptical.

.. why can a layman so easily poke holes in your "physics" arguments? I'm not a physicist, and haven't claimed to be one. [Jane Q. Public, 2014-08-01]

If Jane thinks he's poking holes, maybe he's the Black Knight. Jane has claimed to "take the physicists' word for it," but I'm skeptical.

".. non-person.. disingenuous and intended to mislead .. he is either lying .. dishonest .. intellectually dishonest .. intellectually dishonest .. Khayman80's intellectual dishonesty .. Pathetic. .. you've come out the loser in every case.. you can't win a fucking argument. You don't know how. You don't understand logic. You've proved this many times. Get stuffed, and go away. The ONLY thing you are to me is an annoyance. I have NO respect for you either as a scientist or a person. .. cowardice .. odious person .. you look like a fool .. utterly and disgustingly transparent .. Now get lost. Your totally unjustified arrogance is irritating as hell. .. You are simply proving you don't know what you're talking about. .. Jesus, get a clue. This is just more bullshit. .. spewing bullshit .. You're making yourself look like a fool. .. Hahahahahaha!!! Jesus, you're a fool. .. a free lesson in humility.. you either misunderstand, or you're lying. After 2 years of this shit, I strongly suspect it is the latter. .. Now I KNOW you're just spouting bullshit. .. if we assume you're being honest (which I do not in fact assume) .. I wouldn't mind a bit if the whole world saw your foolishness as clearly as I do. .. stream of BS.. idiot .. Your assumptions are pure shit. .. I'm done babysitting you.." [Jane Q. Public]

Clicking "Try Sage Online" doesn't require downloading or installing Sage on your computer. But more importantly:

More importantly, can we agree that in equilibrium, power in = power out?

No. I am not aware of any "conservation of power" law. [Jane Q. Public]

Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out.

That's the basis of all these calculations, which is why I've repeatedly asked if we could agree on it.

Once again, can we agree that in equilibrium, power in = power out?

For the moment, I'll assume we can. If not, please explain why you don't agree that in equilibrium, power in = power out.

I'm sorry that I didn't realize earlier that we have such a fundamental disagreement. I should've been building a common understanding of equilibrium and conservation of energy rather than solving increasingly complicated thought experiments. So let's take this step by step and see if we can agree on anything.

Let's start with conservation of energy just inside the chamber walls at equilibrium: power in = power out.

A blackbody plate is heated by constant electrical power flowing in. Blackbody cold walls at 0F (T_c = 255K) also radiate power in. The heated plate at 150F (T_h = 339K) radiates power out. Using irradiance (power/m^2) simplifies the equation:

electricity + sigma*T_c^4 = sigma*T_h^4 (Eq. 1)

(Eq. 1 looks better in LaTeX, but hopefully this version is legible.)

Yes/No: can we agree that Eq. 1 is based on the Stefan-Boltzmann law and correctly describes conservation of energy just inside the chamber walls at equilibrium?

If yes, the next step is to solve Eq. 1 for the constant electrical input using a calculator or the Sage worksheet I provided.

If no, could you please write down the equation you think correctly describes conservation of energy just inside the chamber walls at equilibrium?

Earlier I made an offhand remark that enclosing the heated plate is like suddenly warming the chamber walls. This simpler scenario might be more helpful. Suppose the chamber walls are suddenly warmed from 0F to 149F. What will happen to the heated plate if the electrical power heating the plate remains constant? If you claim it would remain at 150F, think carefully about energy conservation at equilibrium. When the walls were at 0F, the plate was in equilibrium because power in = power out. But now the net power radiating out is much smaller, which means power in > power out. So what happens to the heated plate?

Instead of saying "an aluminum plate warms the inner plate" perhaps I should've said "an aluminum plate warms the enclosed heated plate." Maybe this will help distinguish between the inner surface of the enclosing plate and the enclosed heated plate. I'm sorry for any confusion this caused, and corrected it at Dumb Scientist.

Nice link. Do you really expect me to read that .sws file? How about something human-readable? [Jane Q. Public]

These open source Sage worksheets show my work for these thought experiments. Clicking "Try Sage Online" would let you upload my third worksheet, and hitting shift-enter a few times would recalculate all its answers. But in case you don't want to do that, here's a formatted copy of that worksheet and its answers:

#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()

ANSWER = 29.3986743761843

6379^2/6371^2.n()

ANSWER = 1.00251295644620

338.706*1.00251295644620^(-.25).n()

ANSWER = 338.493545219805

#Completely surrounded by 2nd plate
soln2 = solve(eq1.subs(T_c=338.493545219805,electricity=29.3986743761843,sigma=5.670373e-8,epsilon_h=0.11,epsilon_c=0.11),T_h)
soln2[0].rhs().n()

ANSWER = 385.286813818721*I

This could also be done on a calculator, which is why I explained how to derive the equations using the principle that at equilibium, power in = power out.

... its outer temperature is 149.6F ... pretend the enclosing shell is a thermal superconductor, so its inner temperature is also 149.6F ... [Dumb Scientist]

So, first you postulate a thermal superconductor, and then assert that it has a far higher temperature on one side than on the other? What a magical world you must live in. [Jane Q. Public]

No, I said both sides of a thermal superconductor enclosing shell are at 149.6F. Accounting for aluminum's finite conductivity would mean its inner temperature would be higher than its outer temperature. If you'd like, we could see how an aluminum plate warms the inner plate higher than the 233.8F it would be at with a superconducting plate. Just let me know, and I'll do the calculations.

But I don't think that would be helpful yet, because I didn't realize we have a fundamental disagreement:

More importantly, can we agree that in equilibrium, power in = power out?

No. I am not aware of any "conservation of power" law. [Jane Q. Public]

Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out.

That's the basis of all these calculations, which is why I've repeatedly asked if we could agree on it.

Once again, can we agree that in equilibrium, power in = power out?

For the moment, I'll assume we can. If not, please explain why you don't agree that in equilibrium, power in = power out.

I'm sorry that I didn't realize earlier that we have such a fundamental disagreement. I should've been building a common understanding of equilibrium and conservation of energy rather than solving increasingly complicated thought experiments. So let's take this step by step and see if we can agree on anything.

Let's start with conservation of energy just inside the chamber walls at equilibrium: power in = power out.

A blackbody plate is heated by constant electrical power flowing in. Blackbody cold walls at 0F (T_c = 255K) also radiate power in. The heated plate at 150F (T_h = 339K) radiates power out. Using irradiance (power/m^2) simplifies the equation:

electricity + sigma*T_c^4 = sigma*T_h^4 (Eq. 1)

(Eq. 1 looks better in LaTeX, but hopefully this version is legible.)

Yes/No: can we agree that Eq. 1 is based on the Stefan-Boltzmann law and correctly describes conservation of energy just inside the chamber walls at equilibrium?

If yes, the next step is to solve Eq. 1 for the constant electrical input using a calculator or the Sage worksheet I provided.

If no, could you please write down the equation you think correctly describes conservation of energy just inside the chamber walls at equilibrium?

Now I KNOW you're just spouting bullshit. Because you well know that even "arbitrarily small" is not zero. And any deviation from zero is enough to make the difference between T and T0 (or however you want to designate them) non-zero. A non-zero difference is all we need, no matter how "arbitrarily small" you try to make it. ... he USES "mainstream" physics to show that in reality it does not warm infinitely, and therefore Spencer's argument was wrong. I mean you're just absolutely trying to reverse the real argument here. But I really don't expect you to see that, because if we assume you're being honest (which I do not in fact assume), you don't even see the enormous gaping holes in your own argument which you made above. IN REALITY, the enclosing plate will be somewhat cooler. That's not even advanced physics, it's simple math. Repeat: why can a layman so easily poke holes in your "physics" arguments? I'm not a physicist, and haven't claimed to be one. [Jane Q. Public, 2014-08-01]

Jane's insistence that "a non-zero difference is all we need" between the heated plate's initial temperature of 150F and the enclosing plate's final temperature of ~150F was interesting. In this thought experiment, the enclosing plate was initially cooler than 100F.

... your analysis above actually verifies what I stated earlier. I've been wasting my time with (my opinion) an idiot. 0.2% is not zero. Therefore T0 (if that is the outward extent of earth system) has a surface area of T * 1.002, and its temperature will be measurably lower than that of heat source T. Therefore we have a net heat transfer proportional to T - T0, which is a non-zero quantity. You've proved nothing here except to verify my point. ... [Jane Q. Public, 2014-08-01]

The outward surface S0 (if that is the outward extent of earth system) has a surface area of 1.002 * Earth's non-atmosphere surface), and therefore its temperature will be measurably lower than that of heat source T. And therefore we have a net heat transfer proportional to T - T0, which is a non-zero quantity. [Jane Q. Public, 2014-08-02]

Jane continues to focus on the difference between the heated plate's initial temperature of 150F and the enclosing plate's final temperature of ~150F, while the enclosing plate's initial temperature was below 100F. For all the thought experiments we've discussed, the heated plate at time "t" has always been warmer than the enclosing plate at the same time.

As long as it's warmer than the chamber walls, the exact final equilibrium temperature of the enclosing plate is completely irrelevant to the fact that enclosing the heated plate warms it.

The experiment requires the passive plate to be some unspecified distance from the heat source. (This is a condition of the experiment; no contact is allowed since this is only about radiative transfer.) ... We also know that the passive plate P will always be at a temperature less than that of the source... T(p) will be lower than T(s). The amount is of no consequence, as long as it is non-zero (and it is). ... An elementary, obvious, and perfectly sound conclusion from this is that the source is not made hotter under any of these conditions. Even if plate P completely encloses source S, we know for at least two different reasons (greater radiative area, and the simple fact that it does radiate outward to the wall, which cools it) that T(p) will always be less than T(s), even at equilibrium. Since the temperature of every other object is less than that of the heat source, there is no net heat flow TO the heat source, therefore the heat source does not become hotter. This is, and has been, the whole of Latour's argument, and it is valid. It is not crazy speculation by some nitwit, it is straightforward application of Stefan-Boltzmann law. Q.E.D., indeed. If the above inequalities hold (and they do), Latour's conclusion is the only one that is mathematically valid. [Jane Q. Public, 2014-08-02]

Nonsense. T(p) (which I call T_c) at time "t" has always been less than T(s) (which I call T_h) at the same time in all the thought experiments we've discussed. That's why I called them T_c and T_h! Constant electrical power flows in, and at equilibrium equal power radiates out. Enclosing the heated plate with a plate that's warmer than the chamber walls initially decreases that power out. Just as if the chamber walls were suddenly warmed. So the plates are no longer in equilibrium and build up heat until they radiate enough power via Stefan-Boltzmann to reach a warmer equilibrium.

That's why I solved for equilibrium temperatures using the principle that power in = power out. Once again, can we agree that in equilibrium, power in = power out?

Jane's still talking about plate areas, but he's definitely not concerned.

Let's see how a 0.2% larger enclosing plate affects equilibrium temperatures. The heated plate is a sphere with radius 6371 mm and surface area A_h. The enclosing plate is a 1 mm thick concentric shell with an inner radius of 6378 mm, surface area A_c1 on the inside, and A_c2 on the outside. The chamber is also a concentric sphere with inner radius 6386 mm, so there's a 7 mm gap on both sides of the enclosing shell. Again, the plates and walls are oxidized aluminum.

At equilibrium, the enclosing shell radiates the same power out as the heated plate did before it was enclosed. But its area is 1.0025 times larger, so its outer temperature is 149.6F (338.5K) instead of 150.0F (338.7K):

A_h*T_h^4 = A_c2*T_c2^4 (Eq. 3)

For the moment, let's pretend the enclosing shell is a thermal superconductor, so its inner temperature is also 149.6F (338.5K). Energy conservation at equilibrium just inside the enclosing shell shows that the heated sphere will warm to an equilibrium temperature of 233.8F (385.3K)

Note that 233.8F is warmer than the heated sphere's original 150.0F equilibrium temperature.

We could keep making this thought experiment more realistic, but that wouldn't change the fact that enclosing the heated plate makes it warmer. For instance, instead of correcting the temperature manually as I did in Eq. 3, we could use Wikipedia's equation which includes areas. Or we could account for the enclosing shell's finite conductivity, but that would just make the heated plate even hotter.

Again, Dr. Latour and the Sky Dragon Slayers are wrong.

Again, can we agree that in equilibrium, power in = power out?

More importantly, can we agree that in equilibrium, power in = power out?

No. I am not aware of any "conservation of power" law.

Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), the power going in minus the power going out must equal the rate at which energy inside that boundary changes. At equilibrium, the system isn't changing so its energy is constant. Therefore, at equilibrium power in = power out.

That's the basis of all these calculations, which is why I repeatedly asked if we could agree on it.

... therefore we have a net heat transfer proportional to T - T0

Don't you mean proportional to T^4-T0^4?

More importantly, can we agree that in equilibrium, power in = power out?

If so, the only way the heated plate would stay at 150F after it's enclosed is if the enclosing plate is at the same temperature as the chamber walls (0F).

Again. Yes/No: do you claim the heated plate will remain at 150F after the second plate is added?

Again, the only way the heated plate would stay at 150F after it's enclosed is if the enclosing plate is at the same temperature as the chamber walls (0F).

Again. Yes/No: do you claim the heated plate will remain at 150F after the second plate is added?

We've determined equilibrium temperatures in a simple example, so let's solve a more general example.

Jane's concerned that the enclosing plate is bigger than the heated plate. But Earth's mean radius is 6371 km, and the effective radiating level is ~7 km higher, so these surface areas are only ~0.2% different. Of course, in a thought experiment this difference can be made arbitrarily smaller. Despite Jane's protests, this doesn't change the fact that enclosing the heated plate makes it warmer.

More importantly, I treated the plates as blackbodies where absorptivity alpha = 1 and emissivity epsilon = 1. This is a reasonable approximation for plates made of carbon nanotube arrays (PDF) which have alpha = ~0.99955. But more conventional plates have alpha and epsilon considerably less than 1.

The next step is to treat the plates as graybodies where absorptivity and emissivity are independent of wavelength, so they appear gray. Kirchoff's Law states that absorptivity = emissivity for graybodies.

MIT calculates heat transfer between graybody plates using an infinite sum of emission, reflection and absorption. Using my variable names, their final expression is:

net heat flow = sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1) (Eq. 2)

(Again, Eq. 2 looks better in LaTeX, but hopefully this version is legible.)

At equilibrium, net heat flow equals the electrical input. Note that MIT's Eq. 2 reduces to my Eq. 1 for blackbodies where epsilon_h = epsilon_c = 1.

Suppose the plates and chamber walls are made of oxidized aluminum with emissivity = 0.11. In this case, Sage solves Eq. 2 for a constant electric input of 29.6 W/m^2, which is lower than before because aluminum doesn't radiate as well as a blackbody.

Using Eq. 2 and the same reasoning as before, fully enclosing the heated plate warms it to the same equilibrium temperature of 235F (386K). Fully exposing the plate to the cosmic microwave background radiation cools it to 13F (263K), which is lower than before because the CMBR is a blackbody and aluminum chamber walls aren't.

So even for graybody plates, MIT's mainstream physics refutes Dr. Latour's nonsensical claim that the enclosed heated plate remains at 150F. They also use this equation to explain how thermos bottles insulate drinks, and describe the same radiation shields used since at least Daniels 1968 (PDF). Again, these shields work like Dr. Spencer's insulating plate.

If you won't listen to MIT physicists, note that their final expression is consistent with these equations and Eq. 1 in Goodman 1957.

A non-zero difference is all we need, no matter how "arbitrarily small" you try to make it. ... IN REALITY, the enclosing plate will be somewhat cooler.

All we need to show that the heated plate stays at 150F after it's enclosed? No. The only way the heated plate would stay at 150F after it's enclosed is if the enclosing plate is at the same temperature as the chamber walls (0F).

Again. Yes/No: do you claim the heated plate will remain at 150F after the second plate is added?

Dr. Latour's argument is that "k is the fraction of re-radiation from the second bar absorbed by the first hotter bar... k must be identically zero, so no cold back-radiation is absorbed and T remains 150. Quod Erat Demonstrandum, QED."

And within the last year you've claimed that:

"... in Spencer's thought experiment, the passive body that is inserted into the system cannot make the source warmer than it already is. That is Latour's whole point. ..." [Jane Q. Public, 2014-02-13]

Yes/No: do you claim the heated plate will remain at 150F after the second plate is added?

Again, he's completely wrong. The hotter bar absorbs cold back-radiation, and T does not remain 150F. That's why I refuted Dr. Latour by showing that a completely enclosed heated plate reaches an equilibrium temperature of 235F (386K), which is less than the infinite temperature he claimed.

Hahahahahaha!!! Jesus, you're a fool. THAT ISN'T WHAT HE CLAIMED. Quite the contrary. He claimed that a completely enclosed plate DOES NOT reach infinite temperature, which of course agrees with observations. [Jane Q. Public, 2014-08-01]

Again, Dr. Latour claimed that mainstream physics, which includes absorption of cold back-radiation, "would constitute creation of energy, a violation of the first law of thermodynamics." You've even repeated his claim:

... the temperature would go up until it outshone the rest of the universe, or it would cool down to zero. ... [Jane Q. Public, 2013-12-24]

"I told you what I think happens to that 10 Joules. By the first law of thermodynamics it doesn't just disappear so what happens to it?"

Yes, I know you told me but that doesn't happen. It would be a violation of the First Law of Thermodynamics. ... The system would never reach equilibrium, but would continue warming to infinity (if such a thing as infinite temperature existed). It would soon destroy itself from all this extra energy that is coming from nowhere. ... [Jane Q. Public, 2014-02-17]

That's why I've repeatedly told you that:

"Dr. Latour was wrong to claim that mainstream physics predicts the heated plate warms infinitely."

"I refuted Dr. Latour's claim that mainstream physics predicts infinite warming..."

If you're retracting your claim that absorbing cold back-radiation (i.e. mainstream physics) would violate the first law and "continue warming to infinity" then that's great news!

Here's one way you are wrong. In any realistic system, the enclosing plate would be of larger dimensions than the internal source, however slightly. So while the total re-radiated energy might be the same, it is spread over a larger area, so the energy density (and therefore temperature) would be lower. How did you allow a layman to catch you in such an elementary error? Not that I had any obligation to do so. Your argument is with him, not me. Just consider it a free lesson in humility. [Jane Q. Public, 2014-08-01]

The key phrase is "however slightly" because that difference can be made arbitrarily small. Since the only objection you've raised is arbitrarily small, does that mean you now see that Dr. Latour is wrong to claim that the heated plate will stay at 150F after the second plate is added, because he wrongly claims that absorbing cold back-radiation would violate the first law?

If not, maybe it would help if we kept checking my calculations step by step. For the simplest case of blackbody plates with arbitrarily similar areas, this equation represents conservation of energy at equilibrium:

electricity + sigma*T_c^4 = sigma*T_h^4 (Eq. 1)

Sage solves Eq. 1 for a constant electric input of 509 W/m^2.

As Dr. Spencer said, now imagine that the second plate completely surrounds the heated plate. This simpler problem is a closer analogy for the greenhouse effect which completely surrounds Earth.

Electric input of 509 W/m^2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).

But the second plate also radiates the same power in, toward the enclosed heated plate. Just like the cold chamber walls do. Now consider conservation of energy just inside the second plate (but outside the first) at equilibrium. We can solve for the insulated heated plate's temperature using Eq. 1 by setting T_c = 150F (339K). That yields an insulated heated plate temperature of 235F (386K).

So Dr. Latour was wrong to claim that mainstream physics predicts the heated plate warms infinitely. In reality, insulating the heated plate only warms it by a finite amount. Energy is conserved, and the second law is satisfied because net heat flows from hot to cold.

Why do Slayers think this is a problem? If we kept the same electric input but took those walls (and everything else) away to reveal the cosmic microwave background radiation at -454.8F (2.7K), the heated plate would cool to 95F (308K). Even before the heated plate was surrounded by a second plate, it was still (finitely!) heated by radiation from the cold walls. Again, this is okay because net heat flows from hot to cold.

This is an important point. Greenhouse gases can insulate Earth's surface because they're warmer than the cosmic microwave background radiation.

When we can agree on this simplest example, we can move on to a more general example.

I'm refuting his whole point:

"... in Spencer's thought experiment, the passive body that is inserted into the system cannot make the source warmer than it already is. That is Latour's whole point. ..." [Jane Q. Public, 2014-02-13]

The first step to understanding this thought experiment is determining the constant electrical power needed to keep the heated plate at 150F before the cool plate is added. Since you've done your due diligence, what electrical power did your research reveal?