1
$\begingroup$

Introduction

In this question I want to know when calculating the entropy change where we take the temperature of the system or the reservoir's and if my thought process is sound.

Clausius Theorem

The state function of Entropy is defined by the Clausius's Theorem which in short is stated by the equation: $$ \int\limits_{\text{cycle}} \frac{\mathrm{d}Q}{T}\leq 0 \,.$$

And the equality stands only when the heat is transferred reversibly around the cycle.

Comments on the variable in the inequality

From the process of proving the theorem I understood that the temperature $T$ is the temperature of the reservoir between the Carnot Engine and the system that we study.

Entropy

Entropy is defined as $$ \mathrm{d}S= \frac{\mathrm{d}Q_{\text{rev}}}{T} \,,$$ so as to be an exact differential and therefore a state function.

Comments on Entropy

As I came to understand the only way to transfer energy through heat reversibly is for the temperatures of the two bodies in thermal contact to be the same. So the temperature $T$ in the above equation which is the temperature of the reservoir (or whatever) is to be the same as the system's one. Therefore the temperature $T$ in the equation is the temperature of the system. RIGHT?

Improve this question
$\endgroup$
6
  • $\begingroup$ In all cases, you are supposed to use the temperature at the interface between the system and it’s surrounings (I.e., the ideal reservoir temperature). For a reversible process, the system temperature is uniform, and equal to the reservoir temperature. $\endgroup$
    – Chet Miller
    Commented May 31, 2018 at 16:28
  • $\begingroup$ Hmm...since I dont seem to fully get it, in a case in which a system comes in thermal contact with a reservoir what is the entropy change of the "universe" and more importantly why? In my book , in the case of the reservoir the temperature that is inserted in the formula is the Temperature of the reservoir and as far as the system is concerned the temperature that we take is the temperature of the system. $\endgroup$
    – George Smyridis
    Commented May 31, 2018 at 20:11
  • $\begingroup$ Here is a link to an article I wrote on determining the change in entropy for systems and surroundings experiencing irreversible processes. It may answer many of your questions: physicsforums.com/insights/grandpa-chets-entropy-recipe If, after reading this, you have further issues, I would be pleased to discuss. $\endgroup$
    – Chet Miller
    Commented May 31, 2018 at 21:13
  • $\begingroup$ In the Clausius inequality the temperature you are supposed to use is the temperature at the interface between the system and the surroundings, where the heat transfer is occurring. In a case where an ideal constant temperature reservoir is being employed, this means that you use the reservoir temperature. $\endgroup$
    – Chet Miller
    Commented Jun 1, 2018 at 13:30
  • $\begingroup$ @ChetMiller But if the system and reservoir temperatures are equal then why heat transfer takes place? Thanks in advance. $\endgroup$
    – Antonios Sarikas
    Commented Feb 23, 2020 at 12:13

1 Answer 1

Reset to default
1
$\begingroup$

This answer is response to the question you asked in one of your comments. Suppose you have an ideal constant temperature infinite reservoir at temperature $T_R$ and a system (which does no work) of mass M, heat capacity C, and initial temperature $T_S$. You place the system in contact with the infinite reservoir and allow them to equilibrate spontaneously. The question is, what is the entropy change of the system, the reservoir, and the universe?

In the final thermodynamic equilibrium state, the temperature of the system will have changed from $T_S$ to $T_R$. The amount of heat transferred from the reservoir to the system will be $Q=MC(T_R-T_S)$. The entropy change of the reservoir will be $$\Delta S_R=-\frac{MC(T_R-T_S)}{T_R}$$The change in entropy of the system will be $$\Delta S_S=MC\ln{(T_R/T_S)}$$ The entropy change for the universe (system plus surroundings) will be: $$\Delta S_U=MC\left[\ln{(T_R/T_S)}-\frac{(T_R-T_S)}{T_R}\right]$$This is greater than zero for all unequal values of $T_R$ and $T_S$.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.