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For change in entropy dS = dqrev/T, is T the temperature of system or surrounding or both?

I am confused about Thot, Tcold, Tsys and Tsurr. If qrev, are we talking about the reversible cycle such as the carnot engine?

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  • $\begingroup$ Of the particular system of which you are trying to ascertain the entropy, of course. $\endgroup$
    – Danu
    Sep 27 '14 at 18:11
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If the heat is transferred reversibly, the temperatures of the two bodies have to be the same. Transfer of heat from hotter to colder body is irreversible.

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  • $\begingroup$ Thank you for your answer. But in a Carnot engine, it involves transfer of heat from a hot reservoir to a cold reservoir, and the process is reversible. Will you please explain a bit more about that? Thanks! $\endgroup$
    – PhysC
    Sep 27 '14 at 18:06
  • $\begingroup$ The heat goes from the hot reservoir to the working medium of the engine at the same temperature. The working medium is then let to work and its temperature is let to decrease to that of the cold reservoir. Only then is the waste heat given to the cold reservoir. So all heat transfers in the Carnot engine occur over zero temperature difference. $\endgroup$ Sep 27 '14 at 18:26
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I can't comment, so to reply to your question:

But in a Carnot engine, it involves transfer of heat from a hot reservoir to a cold >reservoir, and the process is reversible. Will you please explain a bit more about that? >Thanks!

Carnot engine involves isothermal and isentropic processes. For the isentropic process, $dS = 0$. For the isothermal process, because $T_{hot} = T_{cold}$, so $$dS_{tot} = dS_{sys} + dS_{surr}$$ $$dS_{tot} = \frac{Q_{rev}}{T} - \frac{Q_{rev}}{T} = 0$$ Since the same amount of heat gained by system is obtained from surroundings.

For a reversible process, $dS_{tot} = 0$.But it does not mean that no heat transfer has occured.

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