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Entropy change for a system is defined classically as:

$$dS=\frac{\delta Q_{rev}}{T}$$

where $\delta Q_{rev}$ is infinitesimal reversible heat that flows in a system.

I don't understand whether $T$ refers to the temperature of a system or its surroundings. Many sources said $T$ is the temperature of a system. However since this entropy definition applies to reversible processes and most reversible processes have a system in thermal equilibrium with its surroundings, so $T$ refers to either system and surroundings because temperatures in equilibrium are equal.

To calculate the entropy change when a system changes its temperature from $T_i$ to $T_f$:

$$\Delta S=\int_{T_i}^{T_f} \frac{\delta Q_{rev}}{T}$$

This integral assumes that the path we are integrating is quasistatic and reversible. If $T$ refers to either system and surroundings, then that means in the selected quasistatic and reversible process, the temperature of both system and surroundings change from $T_i$ to $T_f$?

Also, when the definitions of Helmholtz and Gibbs free energy are discussed

$$F=U-TS$$

$$G=H-TS$$

Schroeder wrote that the term $TS$ refers to heat that flows in a system where $S$ is a system's final entropy and $T$ is the temperature of surroundings. Can $T$ refers to the temperature of a system?

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    $\begingroup$ The temperature in the denominator is the temperature of the thermal reservoir that supplies the entropy. In case the entropy transport is reversible the temperature of the reservoir is the same or infinitesimally close to the temperature for the system. $\endgroup$
    – hyportnex
    Commented Mar 4, 2023 at 22:59
  • $\begingroup$ > "Schroeder wrote ..." Reference? $\endgroup$ Commented Mar 5, 2023 at 0:37
  • $\begingroup$ @JánLalinský chapter 5 in An Introduction to Thermal Physics $\endgroup$
    – Jimmy Yang
    Commented Mar 5, 2023 at 1:25
  • $\begingroup$ Sometimes when discussing various free energies like $F,G$, temperature $T_0$ instead of $T$ is used, where $T_0$ refers to temperature of the environment, or reservoir. This is because a process may be discussed where $T$ of the system changes uncontrollably or isn't defined during the process (e.g. a violent chemical reaction), but at both endpoints of the process, the system has temperature of the environment $T_0$. $\endgroup$ Commented Mar 5, 2023 at 1:28

2 Answers 2

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I don't understand whether 𝑇 refers to the temperature of a system or its surroundings.

$T$ refers to the temperature at the boundary between the system and the surroundings. If the surroundings is a thermal reservoir, then $T$ is the temperature of the surroundings. If the system is in thermal equilibrium with the surroundings, then $T$ is the temperature of both the system and the surroundings, not just at the boundary.

If $T$ refers to either system and surroundings, then that means in the selected quasistatic and reversible, the temperature of both system and surroundings change from $T_i$ to $T_f$?

Yes they do, but for the process to be reversible the system has to be in contact with an infinite series of thermal reservoirs ranging from $T_i$ to $T_f$ with each subsequent reservoir having a temperature infinitesimally greater than (or less than) the previous reservoir, so that the system is constantly in thermal equilibrium with its surroundings.

Hope this helps.

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For a reversible process, the temperature of the system is spatially uniform, as is the temperature of the surroundings. So the temperature T refers both to the system temperature and the surroundings temperature. If the system temperature is changing between the beginning and end of a reversible process, so also does the surroundings temperature change. In the case of the surroundings, we typically envision what is happening is the system being placed in contact with a continuous sequence of constant temperature reservoirs with temperatures running from $T_i$ to $T_f$.

For an irreversible process, the temperature of the system may not be spatially uniform, and the temperature of the surroundings may not be spatially uniform either. But, at their interface through which the heat flows, their temperatures will match. In the Clausius inequality, we use this temperature at the interface $T_I$ to express the condition that $$\Delta S>\int{\frac{dQ}{T_I}}$$

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  • $\begingroup$ Thank you. I don't understand what exactly "interface" refers to. If two objects are in thermal contact with each other, where is the interface? On the surface of contact? $\endgroup$
    – Jimmy Yang
    Commented Mar 4, 2023 at 23:34
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    $\begingroup$ Precisely. So the temperature of A near the contact surface will be a little closer to the bulk temperature of B, and the temperature of B near the contact surface will be a little closer to the bulk temperature of A, and presumably this creates a continuous temperature across the interface itself so that you could speak of its temperature unambiguously. $\endgroup$
    – CR Drost
    Commented Mar 4, 2023 at 23:40
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    $\begingroup$ The interface is the surface between the system and surroundings. $\endgroup$ Commented Mar 5, 2023 at 0:10

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