Carnot’s theorem tells us that "the most efficient heat engines operating between 2 reservoirs have the same efficiency as a Carnot engine".

This theorem assumes that the engines are operating between reservoirs; in essence, it assumes that the 2 temperatures stay constant. However, this is not the case when an engine is operating between materials with finite heat capacity, because the temperatures no longer stay constant when heat is transferred.

I've heard of explanation to treat the temperature as constant so that the materials behave as reservoirs instantaneously. However, if the heat transfer over one cycle is large enough then the temperature will not be constant, so I think that this approximation will not hold.

Why does Carnot’s theorem hold even when the temperatures are changing?

  • $\begingroup$ Talking about fixed temperature reservoirs is not important; you can rephrase Carnot's theorem as referring to entropy transfer at a given pair of maximum and minimum temperatures. The reservoirs come into play when you discuss how one could realize or approximate such transfer in practice. $\endgroup$ – hyportnex May 12 '18 at 16:23
  • $\begingroup$ @hyportnex How would you then prove that the instantaneous efficiency is given by 1-T1/T2, where T1 and T2 is the instantaneous temperature of the low and high temperature reservoir respectively? $\endgroup$ – user148792 May 12 '18 at 16:28
  • $\begingroup$ "instantaneity" does not play here, only the temperatures at which entropy is transferred in or out. If the temperature is time varying then Carnot's efficiency theorem refers to the maximum and the minimum temperatures. $\endgroup$ – hyportnex May 12 '18 at 16:38
  • $\begingroup$ @hyportnex Let's say there's 2 bodies with heat capacity C, one at temperature T1 and the other at a higher temperature T2. I would like to extract as much work as possible from this system. How would you prove that the work extracted by operating a Carnot engine between them will be the maximum possible work? $\endgroup$ – user148792 May 12 '18 at 16:41
  • $\begingroup$ that is a different question and I think @Chemomechanics answered that previously $\endgroup$ – hyportnex May 12 '18 at 16:53

In effect, @hyportnex statement that “ ‘instantaneity’ does not play here, only the temperatures at which entropy is transferred in or out” is correct, though technically entropy is not something that is transferred between a system and its surroundings. Only heat and work are transferred. Entropy is a state property of the system or surroundings like temperature, pressure, internal energy, enthalpy, etc.. Heat and work are not properties. They are the transfer energy. In any case let me attempt to answer in another way:

Carnot’s theorem is presented in various forms, but the bottom line is the most efficient heat engine possible is one that operates using the Carnot Cycle, which consists of two reversible isothermal (constant temperature) processes and two reversible adiabatic (constant entropy) processes. For any given temperatures TH (high temperature heat source) and TL (low temperature heat sink), in which the engine operates, the efficiency is given by:

η = 1 – TL/TH

If the heat sources are not thermal reservoirs, then for each subsequent cycle the temperature of the heat source will be lower and the temperature of the heat sink will be higher, making the ratio in the equation larger and the efficiency lower.

Bottom line, Carnot’s theorem does hold even as the temperatures are changing, but the value of the efficiency changes (in this case, decreases). Hope this helps.


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