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Perhaps counter-intuitively, radiative heat transfer equations show that if you have an internally-heated plate with constant energy input (eg electric resistance heating with constant power, or a constant light source) in a vacuum, it will reach a certain equilibrium temperature $T$, and if you then bring a cold plate (i.e. one with no heat source) near it but not touching it, the equilibrium temperature of the hot plate will be higher than $T$!

Indeed the closer the plates are together, the hotter the hot plate gets, to a point ($F$ is the view factor). Finally, when they touch, if conductivity is good they will both equalize to close to the initial equilibrium temperature.

This seems counter-intuitive and that it would violate the second law of thermodynamics. Calculations and models are all well and good, but my question is, have any experiments been done that actually demonstrate this in action?

This paper reproduces the set-up, but they don't show the result of running just the hot plate alone to compare. Plus it isn't in a vacuum so it couldn't fully discount effects of prevention of loss due to convection.

It seems easy enough to try at home in a vacuum chamber but I don't have experience rigging things like this up so it'd be a bit of a learning curve.

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    $\begingroup$ What are the distances involved here? Micrometers? $\endgroup$
    – Mauricio
    Commented Apr 7, 2023 at 9:43
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    $\begingroup$ @Mauricio: The example at the link is with 1 meter x 1 meter plates with thickness 1 cm. If they are spaced 1 cm apart the view factor is ~0.98 $\endgroup$
    – Cloudyman
    Commented Apr 7, 2023 at 9:47
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    $\begingroup$ Reminder from last time last time we discussed this (link) that if you account for the 300K blackbody walls of your experiment, this is not so counterintuitive. Your "cold body" will need to be hotter than 300K, then it will play the role of insulation. Every experiment involving warm (or cold) samples uses insulation, and many of them model their heat. None of them have found that insulation fundamentally doesn't work. physics.stackexchange.com/questions/756492/… $\endgroup$
    – AXensen
    Commented Apr 7, 2023 at 10:08
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    $\begingroup$ Yes the (originally) 300K "cold plate" will be warmed above 300K by the warm body, then it will act as insulation, reducing how much solid angle from the internally heated plate is exposed to 300K radiation. In fact this is always how insulation works - the insulation isn't actively heated, it is first heated by the body it's trying to insulate, then the thing its trying to insulate is seeing higher temperature radiation as it's now looking at something hotter. Indeed, I'm not presently aware of a simple experiment that shows just this effect, which is why I haven't submitted an answer. $\endgroup$
    – AXensen
    Commented Apr 7, 2023 at 10:27
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    $\begingroup$ If the hot plate is internally heated, wouldn't it start to get colder as soon as its temperature rises above the temperature of its reservoir? $\endgroup$
    – Mauricio
    Commented Apr 7, 2023 at 11:43

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This seems counter-intuitive and that it would violate the second law of thermodynamics. Calculations and models are all well and good, but my question is, have any experiments been done that actually demonstrate this in action?

What might make it "counterintuitive" and seemingly contradictory is neglecting the radiation. Once we consider radiation as a matter having energy of its own, we have essentially a heat transfer via a third constituent of the system: the hot plate is in contact with much colder photon gas in vacuum, which is heated and gives some of its energy to the cold plate. The photon gas is however never reaching equilibrium (unless we surround the whole system by reflecting borders), so that the energy is eventually lost and both plates cool down.

The phenomenon of radiative transfer is not uncommon - e.g., the Earth is heated by the Sun, and as it loses its energy (also via radiation) slower than it is being heated, its average temperature rises (aka global warming.)

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    $\begingroup$ I understand the 1st plate would heat the 2nd due to radiation, and that as the 2nd emits radiation it also causes 1st to be warmer (really the 1st is losing less heat). I'm not the only one that finds it counterintuitive, one commenter and the answerer on the linked question both initially thought it wouldn't work out to what the equations pointed to. Note the 1st plate also has an internal heat source so it won't be cooling down. Basically I'm looking for a simple experiment I can just point people to and say "Look it really does work that way, here is incontrovertible proof." $\endgroup$
    – Cloudyman
    Commented Apr 7, 2023 at 10:16
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    $\begingroup$ Also see 20 pages of comments here - skepticalscience.com/… . I think a simple experiment would help the debate! $\endgroup$
    – Cloudyman
    Commented Apr 7, 2023 at 10:18
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    $\begingroup$ Perhaps you need to formulate a more specific question - I do not see how this is different from getting warm, while sitting near a fireplace. I admit that I didn't follow the links in your question... but if this is necessary to understand it, it confirms that the question must be improved (or asked again in a more specific form.) $\endgroup$
    – Roger V.
    Commented Apr 7, 2023 at 11:10
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    $\begingroup$ Ah it’s equivalent to saying that if you sit near a fireplace, the fireplace gets warmer. Apparently many people have trouble with this! $\endgroup$
    – Cloudyman
    Commented Apr 7, 2023 at 11:12
  • $\begingroup$ @Cloudyman if you keep throwing wood into the fireplace it might get warmer. $\endgroup$
    – Roger V.
    Commented Apr 7, 2023 at 11:35

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