# Which temperature does $T$ in Clausius inequality ($\oint \frac{\delta Q}T\le 0$) refer to?

I got a little confused about the temperature in Clausius inequality. As you can see in this answer of Luboš Motl, it seems that temperature is the temperature of the system.

But in some answers of Chester Miller e.g. this and this, he has apparently mentioned that,

$T$ is the temperature at the boundary interface (with the surroundings) at which heat transfer is occurring.

I think $T$ in $\oint \frac{\delta Q}T\le 0$ is the temperature of the system, because $\mathrm dS=\frac{\delta Q_{\textrm{rev}}}T$, and $S$ is a property of the system. So, it should be related to a property of whole of system not a property of system boundary.

On the other hand, Clausius inequality is valid for all cases either reversible or irreversible. And for some irreversible (e.g. non-quasi-static) processes, we cannot define a temperature for system.

May someone please explain clearly what is the temperature in Clausius inequality?

• – user279106
Commented Oct 27, 2021 at 1:54

The temperature appearing the the Clausius inequality is definitely the temperature of the "boundary interface (with the surroundings)", or simply the temperature of the sources. One of the best places I have seen this discussion is in Fermi's book, chapter 5, section 11. He is explicit about it. To see this you have to recapitulate the steps in obtaining the Clausius inequality. You start with a cycle and suppose each infinitesimal part of this cycle is exchanging heat $\Delta Q_i$ with a (external) source at a temperature $T_i$. Then you sum all contributions $\Delta Q_i/T_i$ and in the limiting case it gives the Clausius inequality.

This is in fact a point underestimated in many books. However it is crucial. For instance a way to find out whether a process is reversible or not is just to calculate the Clausius integral (using the temperatures of the source) and compare it with the entropy change (using the temperature of the system). Then one finds $$\Delta S=\int\frac{dQ}{T_{sys}}\geq\int\frac{dQ}{T_{sour}}.$$ The equal sign meaning it is reversible and the greater sign meaning it is irreversible.

It is worth to mention that the temperature in the expression for the entropy is then the temperature of the system. The process chosen to calculate $\Delta S$ is reversible which means the temperature of the system always equals the temperature of the source.

• Thanks! it makes sense. But, if it is the surrounding temperature, then it can always get out the integral. Because the surroundings temperature is constant. Commented Jul 22, 2016 at 16:04
• In the general case, the surrounding temperatures are not constant. Of course in the case they are constant they can be get out of the integral. Commented Jul 22, 2016 at 16:07
• Unfortunately Clausius himself was quite explicit that the temperature in the "Clausius inequality" was the temperature of the system. See The Mechanical Theory of Heat pp 141-142, esp. the description of T on page 141 and the extended footnote at the bottom. books.google.com/books?id=8LIEAAAAYAAJ&pg=PA141 Commented Feb 21, 2020 at 19:49
• Isn't it $\Delta S=\int\frac{dQ_{rev}}{T_{sys}}$? Commented Nov 19, 2023 at 11:18

This is how I think of it;

We are talking about a reversible process here, when the relation is an equailty: $$S_2-S_1=\int_1^2 \left( \frac{\delta Q}{T} \right)_\text{rev.}$$ it is valid only when the heat transfer (and the process itself, actually) is reversible. For this heat transfer to be reversible, it must be a quasi-equilibrium process. That means that if the reservoir temperature is $$T_L$$, the system temperature (or the component if it is a flow process) should be $$T_L \pm \delta T$$ at every instant, where $$\delta T$$ is an infinitesimal value. Hence, I consider the temperature of the system, surroundings or the boundary approximately the same; $$T_\text{sys.} \approx T_\text{boundary} \approx T_\text{surr.} \approx T$$

The whole confusion originates from picturing something like $$T_\text{sys.}=400K$$ and $$T_\text{surr.}=300K$$ (including me). However, that is not a reversible process by definition; there is a finite temperature difference. And indeed, that is when the formula turns into an inequality, because a net positive entropy generation occurs due to the irreversibility. Depending on the heat transfer direction and the temperature value, you get the following if you do the math: $$S_2-S_1>\int_1^2 \left( \frac{\delta Q}{T} \right)_\text{irrev.}$$ or $$S_2-S_1=\int_1^2 \left( \frac{\delta Q}{T} \right)_\text{irrev.}+\Delta S_\text{gen.}$$

where $$\Delta S_\text{gen.}$$ is strictly greater than zero. The formula does not explicitly tell you "just put that temperature into the integral as this function of $$\delta Q$$ and you will get the entropy difference" in case of an irreversible process. You can not integrate $$\delta Q=TdS$$ and find the heat transfer $$Q$$ when the process is irreversible, this is why reversible processes are drawn with a solid line and irreversible processes are drawn with dashed lines.

The formula is often given as $$S_2-S_1\ge\int_1^2 \frac{\delta Q}{T}$$ which I think is very vague without putting some thought into it.