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?

His actual 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."

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.

Maybe it would help if we checked my calculations step by step. Start with conservation of energy just inside the chamber walls at equilibrium: power in = power out.

The plate is heated by constant electrical power flowing in. The 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?

Maybe it would help if we checked my calculations step by step. Start with conservation of energy just inside the chamber walls at equilibrium: power in = power out.

The plate is heated by constant electrical power flowing in. The 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?

Once again, if Dr. Latour understood the second law refers to net heat, he'd agree that adding a cold plate makes the heated plate lose heat slower. That's okay because net heat still flows from hot to cold, i.e. more heat moves from hot to cold than vice versa.

Again, he must have forgotten this nebulous correction which you still haven't linked. I linked to an archive of his blog post that I made yesterday, but here's another archive I just made showing that his blog post is still live today and still contains nonsense like this: "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."

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.

Apparently unlike you, sir, I have a basic understanding of math and physics. Please explain to us all where the Stefan-Boltzmann radiation law is in error. I am sure we would all love to know. [Jane Q. Public, 2012-11-20]

... just what part of the S-B law do you find controversial? [Jane Q. Public, 2014-07-29]

Again, the greenhouse effect is based on the Stefan-Boltzmann law. As I've explained: greenhouse gases re-emit some of [the upwelling long-wave IR], and it bounces around the troposphere until it gets to a height known as the "effective radiating level". Above this height (roughly 7km), there aren’t enough greenhouse gases to keep "most" of the IR from escaping to space altogether. This effective radiating level controls the outflow of heat from the Earth. Stefan-Boltzmann tells us that power radiated is proportional to temperature^4, and temperature decreases with height in the troposphere. Adding greenhouse gases raises the height of this effective radiating level, where it is cooler, which therefore decreases the outflow of heat from the Earth. This is the greenhouse effect, and it isn’t saturated because the effective radiating level can just keep getting higher (e.g. Venus).

Andrew Dessler also explains how the greenhouse effect depends on the Stefan-Boltzmann law. He even explains that an isothermal atmosphere wouldn't have a greenhouse effect: the Slayers' holy grail! Ironically, the greenhouse effect disappears if the upper troposphere isn't colder than the surface. The cold upper troposphere isn't a problem for the greenhouse effect. It's a fundamental requirement, along with the Stefan-Boltzmann law.

As a Hydra, I scoff at mortal concepts like flattery. I also seem to have grown a sixth head judging by Jane's claim that I quoted myself complimenting myself. If Jane's referring to these compliments then my sixth Hydra head also has a real name which is different than mine. We Hydras are powerful and tricksy, and certainly not paranoid delusions. Nope.

... You made the (quite incorrect) claim that Latour wasn't accounting for the fact that the subject at hand is net heat transfer. But that claim is simply incorrect. ... [Jane Q. Public]

Once again, if Dr. Latour understood the second law refers to net heat, he'd agree that adding a cold plate makes the heated plate lose heat slower. That's okay because net heat still flows from hot to cold, i.e. more heat moves from hot to cold than vice versa.

... You took a badly-worded sentence or two and jumped on them as though Latour made a mistake. But his only mistake was wording a couple of sentences badly. He does in fact NOT suggest that warmer objects absorb no radiation, and he has written as much many times. ... You have refuted NOTHING but a couple of unfortunately-worded sentences, which Latour himself publicly corrected shortly after that post appeared. ... [Jane Q. Public]

He must have forgotten this nebulous unlinked correction because his blog post is still live and still contains all these badly worded sentences:

"... the absorption rate of real bodies depends on whether the absorber T (radiating or not), is less than the intercepted radiation T, or not. If the receiver T > intercepted T, no absorption occurs; if the receiver T < intercepted T the absorption rate may be as great as proportional to (T intercepted – T absorber), depending on the amounts reflected, transmitted or scattered. What actually happens is the chiller radiates to the hot plate, but the plate cannot absorb any of it because it is too cold. The hot plate reflects, transmits or scatters colder radiation, just like my roof does for cold radio waves. ... Energy from colder cannot heat hotter further because the second law of thermodynamics says so, because nature says so; always and everywhere. ... Conclusion, the hot plate remains at 150. All physics I know supports it; no physics offered refutes it. Spencer mistakenly assumed the 150 plate absorbs incident 100 radiation ... The generalized claim that a cooler object placed near a warmer object cannot result in a rise in temperature of the warmer object stands. ..."

In fact, he did more than suggest that warmer objects absorb no radiation: "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."

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 infinity.

Explain to us what Venus vs. Mercury have to do with Pierre Latour's thermodynamic argument in regard to greenhouse warming? [Jane Q. Public]

Again, if Dr. Latour and the Slayers are right, why is Venus hotter than Mercury? Hint: the Slayers are wrong. Venus is hotter than Mercury because of the greenhouse effect.

... I have no desire (or any motivation, for that matter) to engage you in some ridiculous argument about whether Venus is proof of "greenhouse warming", as compared to Mercury or the Earth. There are many reasons why even if it were true, it is hardly relevant: Mercury has an extremely long day, almost no atmosphere, and a very eccentric orbit. Venus has a surface atmospheric pressure 92 times (give or take) Earth's, it's atmosphere is MOSTLY CO2 (around 96% or so), versus Earth's 0.04% or less, again give or take a bit. Not to mention the vast clouds of sulfuric acid. You seem to want to ignore all these other variables and argue about just CO2, when the degree to which CO2 in particular affects Venus' surface temperature is speculative, to say the least. ... [Jane Q. Public]

No, I didn't ignore those variables. In fact, I pointed out differences that should make Venus cooler than Mercury in the absence of Venus's greenhouse effect. For instance:

• I compared Mercury's daytime surface temperature to Venus's nighttime surface temperature because Venus's long night should be cooler than Mercury's long day.
• I mentioned Venus's high albedo which is due to its vast clouds of sulfuric acid, and mentioned that this should keep Venus cooler than dark Mercury by reflecting more sunlight.
• I mentioned that Venus is farther from the Sun than Mercury even when Mercury is at aphelion, which should make Venus cooler than Mercury.
• I pointed out that long-term equilibrium surface temperature is determined by conservation of energy, not the ideal gas law. So pressure only affects surface temperature by enhancing the greenhouse effect if and only if GHG's are present.

I've also explained that a planet with no atmosphere is a simple case where the effective radiating level is at the surface, so the equilibrium surface temperature can be determined using the planet’s albedo and distance from the Sun. The greenhouse effect modifies this simple case, which is why Venus is hotter than Mercury.

After I explained that Venus is hot because of its greenhouse effect, you replied by quoting a paper saying "Such an amount of CO2 causes greenhouse warming by 500 K there. On the other hand, the mere 0.006 bars of CO2 on Mars cause warming by 5.5K."

How do Slayers explain 500K of greenhouse warming on Venus, other than basketball player gloves and gray Oreos?

You also linked a crackpot website claiming that on Venus "the solar energy simply does not reach the surface."

I've explained that Venera 9 landed on the surface of Venus and found "surface light levels comparable to those at Earth mid-latitudes on a cloudy summer day." Check out the panorama.

Again, if the Slayers are right, why is Venus hotter than Mercury? Instead of regurgitating bad arguments you find in 30 seconds and which you don't even read carefully, please read carefully before regurgitating even more misinformation for me to debunk.

... Explain to us what Venus vs. Mercury have to do with Pierre Latour's thermodynamic argument in regard to greenhouse warming? ... [Jane Q. Public]

Again, if the Slayers are right, why is Venus hotter than Mercury? Instead of regurgitating bad arguments you find in 30 seconds and which you don't even read carefully, please read carefully before regurgitating even more misinformation for me to debunk.

... you have failed for 2 years to refute Latour. ... You know you can't refute Latour...

I refuted Dr. Latour's claim that mainstream physics predicts infinite warming, and explained how the greenhouse effect is based on the Stefan Boltzmann law and requires a cold upper troposphere. Again, a real skeptic would be checking my calculation that a completely enclosed heated plate would reach an equilibrium temperature of 235F (386K).