If a hot and a cold container are placed thermo sealed system, and a thermoelectric device is used between them and electricity is withdrawn that way, will the resulting entropic temperature in both containers be lower than the average of hot and cold before the withdrawal?
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$\begingroup$ Thanks so far. Indeed, the thermodynamic temperature. The underlying question is if heat is actually transformed into electricity here or just temperature difference. $\endgroup$– Freud ChickenCommented Aug 29, 2021 at 13:47
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$\begingroup$ I’m glad this is on track with what you’re asking. I’ve transferred my comment to an answer. $\endgroup$– ChemomechanicsCommented Aug 29, 2021 at 16:30
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1 Answer
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I’m not sure what you mean by “entropic temperature” If you mean the thermodynamic temperature (i.e., the container absolute temperatures $T_1$ and $T_2$ at equilibrium), then yes; the final temperature can be as low as the geometric mean $\sqrt{T_1T_2}$ of the two absolute temperatures, which is lower than the arithmetic mean $\frac{T_1+T_2}{2}$.
See here for the math; briefly, instead of the two temperatures combining irreversibly under conservation of energy to give a final temperature of $\frac{T_1+T_2}{2}$, we maintain constant entropy with a perfect-efficiency engine, transferring entropy $dS=\frac{C\,dT}{T}$ from the hotter object and to the colder object while extracting all available energy. ($C$ is the heat capacity, assumed to be temperature independent.)
Integration from the initial to the final temperature yields $\frac{T_\mathrm{final}}{T_1}=\frac{T_\mathrm{final}}{T_2}$, or $T_\mathrm{final}=\sqrt{T_1T_2}$.
Some of the heat is transformed to current flow (the rest goes to heat the cooler object), and the temperature difference in conjunction with the heat capacity is converted to electrical energy $C\left(\frac{T_1+T_2}{2}-\sqrt{T_1T_2}\right)$.
answered Aug 29, 2021 at 16:29