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In the definition of entropy

$$\mathrm d S=\left(\frac{ đQ}{T}\right)_\textrm{rev}$$

is $T$ the temperature of the system or of the environment (reservoirs)?

In Clausius' Theorem,

$$\oint \frac{đQ}{T}\leq 0$$

$T$ is the temperature of the reservoirs

But since to calculate $\mathrm dS$ we consider a reversible transformation, the temperature of the system and of the reservoirs should be always the same.

So can I say that $T$ is the temperature of the system in the definition of entropy?

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As you said, in any reversible transformation the system and the reservoir have the same temperature. So, since the definition of entropy needs that you take the system through a reversible path, you can use the system's temperature or the reservoir's temperature alike.

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  • $\begingroup$ Thanks for the answer! If I may ask, I read that from the second law we can write for the heat exchanged $$\delta Q \leq T \mathrm{d} S$$ So is $T$ the temperature of the system also in this case? I'm not sure about this because we are considering a generic process (so also irreversible) here, hence it is not necessarily true that the temperatures of system and environment are the same. But then why is it possible to say so? $\endgroup$
    – Sørën
    Sep 18, 2016 at 21:49
  • $\begingroup$ That inequality is the differential form of the Clausius inequality and in that case T is the reservoir's temperature, which can be, in general, different from the system's. The "=" holds for reversible processes and in that case, again, T is the same for the system and the reservoir. $\endgroup$ Sep 18, 2016 at 22:12
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    $\begingroup$ For both reversible processes and irreversible processes, the system temperature matches the reservoir temperature at the boundary with the reservoir (where the heat transfer is occurring). But, for an irreversible process, the temperature within the system is typically not spatially uniform, so what temperature do you use? For the Clausius inequality, you use the temperature at the system boundary with the reservoir where the heat transfer is occurring. For a reversible process, it doesn't matter because the system temperature is uniform. $\endgroup$ Sep 18, 2016 at 22:42

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