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I read the discussion of the same problem in another problem posted here Carnot Total Reversibility, but still I can't get the reason why Carnot cycle is stated as "Totally Reversible", it consists of two totally reversible processes namely; the adiabatic compression and expansion, but there are also two isothermal heat transfer processes in which the corresponding change in entropy is calculated and not equal to zero:

$$\Delta S = Q/T$$

so why is it proposed to be totally reversible?

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  • $\begingroup$ I think you are only a little confused about the meaning of "reversible". Reversible means that $\Delta S = Q/T$, irreversible means that $\Delta S > Q/T$ (strict inequality). So reversible does not mean that the entropy doesn't change $\endgroup$
    – pglpm
    Commented Dec 11, 2020 at 13:12
  • $\begingroup$ I understand your confusion, because in some texts the second law and reversibility are summarized in imprecise and misleading ways. $\endgroup$
    – pglpm
    Commented Dec 11, 2020 at 13:13
  • $\begingroup$ For precise definitions and great explanations about such matters I always recommend Astarita's book Thermodynamics: An Advanced Textbook for Chemical Engineers. It's a little gem. $\endgroup$
    – pglpm
    Commented Dec 11, 2020 at 13:17
  • $\begingroup$ But is not total reversibility has the conditions of both internal and external reversibility , then entropy generated(Sg) and heat tranfer both must be zero ? Plus based on your comment can I say that a process can be reversed if Delta entropy is generated by heat transfer only ? $\endgroup$
    – noor 1
    Commented Dec 11, 2020 at 13:19
  • $\begingroup$ I don't know how your text defines "total reversibility". You can have two systems $1$ and $2$, one of which gives heat to the other, in such a way that $\Delta S_1 = Q/T$ and $\Delta S_2 = -Q/T$. This means that both are undergoing a reversible process and that $\Delta S_1 + \Delta S_2 = 0$. "A process can be reversed if $\Delta S$ is generated by heat transfer only" – if you mean $\Delta S = Q/T$, then yes when this happens the process (for the body of which $S$ is the entropy) is reversible. $\endgroup$
    – pglpm
    Commented Dec 11, 2020 at 13:24

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You need to understand the difference between entropy transfer and entropy generation or production.

In the two reversible isothermal processes there is a transfer of entropy to the system during the isothermal expansion and a transfer of entropy out of the system of an equal amount during the isothermal compression. Consequently the change in entropy of both the system and surroundings is zero. The processes are reversible because they are carried out very slowly (quasi-statically) transferring heat over an infinitesimal difference in temperature between the system and surroundings.

An example of an irreversible isothermal process is one where the heat transfer occurs over a finite temperature difference producing a temperature gradient at the boundary between the system and surroundings. Entropy is generated at the boundary and passed to the system. So the change in entropy of the system involves two components, the entropy transferred plus the entropy generated due to the irreversible process. In order to return the system to its original state over the cycle (original entropy), the additional entropy generated has to be transferred to the surroundings in the form of additional rejected heat, resulting in less net heat available to perform work.

Hope this helps.

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