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Carnot cycle is a reversible cycle.

By definition a reversible cycle must have an infinitesimal temperature difference. That means ($T_{high}-T_{low}$) is very very small. Any finite temperature difference would increase the degree of irreversibility, hence reducing efficiency.

A Carnot Cycle has efficiency $n=1-\frac {T_{low}}{T_{high}}$). If ($T_{high}-T_{low}$) is very very small, as said above, then ($\frac {T_{low}}{T_{high}}$) must be close to 1. That would make efficiency $n$ close to zero. But Carnot efficiency is the maximum achievable efficiency for any heat engine. That requires the ($\frac {T_{low}}{T_{high}}$) to be as low as possible.

Can you see the same paradox? If I'm wrong, please show me why and how?

Many thanks for all the efficient brains out there.

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    $\begingroup$ "By definition reversible cycle must have infinitesimal temperature difference"...no, why do you think that? $\endgroup$ – ACuriousMind Nov 21 '15 at 22:04
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    $\begingroup$ No, you're mixing things up. In a reversible cycle, the temperatures of the system and the reservoir with which it's in contact are the same while they are exchanging energy via heat. That is what is being referred to when you say "infinitesimal temperature difference". Later in the cycle, the system is brought into contact with another reservoir at some other temperature, and the system must be at the same temperature as this new reservoir while they are exchanging energy via heat. $\endgroup$ – march Nov 21 '15 at 22:16

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