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I am 99% sure that I am wrong here, but here is my reasoning.

The efficiency of a Carnot engine can be written as:

$$ \eta_{carnot} = 1 - \frac{T_f}{T_i} $$

But since Effency's definition says $\eta = \frac{Q_{in} - Q_{out}}{Q_{in}}$, we can say:

$$ \eta_{carnot} = 1 - \frac{T_f}{T_i} = 1 - \frac{Q_f}{Q_i} \implies \frac{Q_i}{T_i} = \frac{Q_f}{T_i} $$

But since $\frac{Q}T$ is just entropy, that means the entropy for a carnot engine is constant after every cycle. How is that possible? Statistically shouldn't the entropy go up?


One place where I think I might have made a mistake is saying that $\frac{Q}T$ is entropy, when entropy is $\frac{\Delta Q}{T}$. Is that correct?

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  • $\begingroup$ For your last line, I would recommend you read the first few paragraphs of Entropy: Wikipedia and be very wary putting a Delta in front of Q, as you need to read the definition of Q carefully. The equation you have (Q/T) is not valid for free expansion of gases. $\endgroup$ – user184116 Feb 9 '18 at 19:18
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Remember that the Carnot cycle is for an ideal engine, one that operates as efficiently as possible given hot and cold bath temperatures $T_H$ and $T_C$. The only restriction that the Second Law of Thermodynamics gives is that entropy cannot decrease in an isolated system. It does not say that entropy must increase. Since the Carnot engine represents "the best you can do" in such a system, it makes sense that it would have the lowest possible allowed change in entropy (namely, zero). So the change in entropy of the reservoirs will be >0.

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If it’s a full cycle that returns the working fluid to its original state, it doesn’t matter how irreversible the process is. The change in entropy of the working fluid will be zero. Any entropy that is generated within the working fluid during the cycle will be transferred to the surroundings (the constant temperature reservoirs). So in an irreversible process like this, it is the entropy of the surroundings that will increase.

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