It seems to me that you could use thermal radiation to drive an engine that does not require a cold reservoir. Carnot's theorem states that the only way to convert heat into work is with a temperature difference, but something as simple as glowing hot steel next to a solar panel seems to violate this. What am I missing?
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$\begingroup$ The efficiency of the Carnot engine is a function of the temperature gradient across the engine. By definition it requires reservoirs at different temperatures. Can you ask a specific question/concept that you are finding difficult? $\endgroup$– boyfarrellCommented Dec 8, 2015 at 22:30
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$\begingroup$ @boyfarrell I think it's possible to generate electricity in an environment that has an equally distributed temperature above 0K, by converting heat directly into electricity, but all examples of thermoelectric generators I can find rely on temperature gradients. The more exotic examples I can find like nanoscience.gatech.edu/paper/2012/12_NL_05.pdf, and ece.tamu.edu/~noise/research_files/Therm_noise_eng_ICNF_.pdf are either not cyclic, or rely on a "hidden" temperature difference. Is the conversion of heat to electricity (or work) strictly limited to temperature differences? $\endgroup$– user1982116Commented Dec 8, 2015 at 22:52
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$\begingroup$ Yes, photovoltaics convert 'heat' in the form of optical radiation directly to electricity and because the solar cell has a constant temperature I can see were confusion might arise. Endoreversible thermodynamic provides a nice way of viewing this problem: the solar cell is connected to two temperature reservoirs via resistors that introduce irreversibilities (look up the work of DeVos). In photovoltaics the gradient across the Carnot engine is a chemical potential and not thermal in nature. But this gradient is ultimately driven by the temperature difference between the sun and the Earth. $\endgroup$– boyfarrellCommented Dec 8, 2015 at 23:53
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$\begingroup$ Let me simplify: A negatively charged mirror next to a gloving hot steel is blown away by radiation pressure. It's an electricity generator. Wheres' the heat sink? $\endgroup$– stuffuCommented Dec 9, 2015 at 6:34
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3 Answers
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How do you want to convert thermal radiation to heat? I imagine this is what you have in mind: even when there is no temperature gradient, every object with a nonzero temperature radiates. This is electromagnetic radiation, so if I catch it, I can drive a current and thereby produce energy.
Here is why this cannot work: We assume here that there is no temperature gradient, which means that your engine has the same temperature as the object. This means that it, too, radiates quite a bit. In fact, it will radiate at the exact same amount that it will absorb radiation from the object - otherwise the temperature would either increase or decrease which it can't as the whole system has the same temperature. This is the definition of thermodynamic equilibrium in this case: Since the only heat transfer is by radiation, absorption and emission rate must be the same. But then, once you would switch the device on and take some of the radiation and convert it into an electric current, the temperature in your engine must decrease - and suddenly you have a gradient where before, you did not. This is prohibited by the 2nd law of thermodynamics.
To take your example: Your glowing hot steel next to a solar panel works perfectly fine. And it doesn't violate anything, because you actually have a temperature gradient: You have the hot steel and the cold solar panel. Converting the heat of the steel into electric energy, the solar panel heats up. Once it reaches the temperature of the steel (well, probably before that because it melts) it will stop producing energy.
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$\begingroup$ And by this you mean that there is likely not some theoretical solar panel that can convert thermal radiation into work independent of the temperature of the panel and the temperature of the object emitting thermal radiation? $\endgroup$ Commented Dec 8, 2015 at 23:45
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$\begingroup$ Yes - at least if the first and second law of thermodynamics hold for which there is strong evidence. $\endgroup$– MartinCommented Dec 9, 2015 at 8:30
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Yes, usern (where n is a large number), that's what Martin means. When he writes "absorption and emission rate must be the same", you can picture it like this: For each absorption of a photon, an electron in the solar panel has to move to a higher energy level. Now, if the solar panel is already at the same energy level as your heat source, most electrons will already be at that level. So before they can absorb a photon, they have to move down. Since per your condition, both have the same temperature, there are equal amounts of electrons moving up and down - that's the "thermodynamic equilibrium" Martin talks about.
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Let's say we have an air tank filled with cool air at atmospheric pressure. What happens when we heat the air in the tank a little bit? The air gains heat energy, that's the only thing that happens, although some people might say something about pressure energy, but "pressure energy" is just heat energy, nothing else.
Now with that heat energy in the tank we can drive pneumatic tools, which do not have to get hot in the process.
Can we find a heat sink in this scenario? At the end the air is a little bit warmer than originally, the air is some kind of heat sink.
Why did I write the above? Well I planned to show that you can't find any heat sink in the above scenario, but I failed.
Now let's say we have some radiation in a tank, and that tank is inside a big tank where radiation pressure is lower than in the small tank. Now we can use the radiation in the small tank to power some tools that work by radiation pressure. Those tools don't have to get hot in the process - the tool may be a turbine, in which case the big tank is the heat sink.
