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Most of the physics textbooks I’ve read try to prove that the Carnot engine is the most efficient engine possible by showing that if you coupled a superefficient engine to a Carnot refrigerator, you’d get a perfect refrigerator, which violates the second law of thermodynamics.

An example from Physics for Scientists and Engineers: A Strategic Approach is shown below.

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Another example is this video by Khan Academy.

What I do not understand is what about this refrigerator identifies it as as Carnot refrigerator. Couldn’t you take any refrigerator, including one whose coefficient of performance is less than that of a Carnot refrigerator, couple it to a heat engine that is more efficient than the refrigerator driven in reverse (i.e. its engine “counterpart”), and get the same result that violates the second law of thermodynamics?

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No you cannot. It must be a Carnot refrigerator. This is because you want to prove that the super-efficient engine is impossible even if $<Q_H$ means infinitesimally smaller than $Q_H$ (and $<Q_C$ means infinitesimally smaller than $Q_C$).

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  • $\begingroup$ I understand that that is what you “want” to prove. I just don’t understand why this proof only holds for a Carnot refrigerator and not a refrigerator with a lower COP. What about its logic or its conditions restricts it to being valid only for Carnot refrigerator? $\endgroup$ – lightweaver Nov 25 '16 at 16:08

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