# Reversible vs Irreversible process

## A Hypothetical Situation:

Two thermally isolated identical systems have heat capacities which vary as:
$$Cv =βT^4$$($$β$$ is constant)

Initially one system is at $$300K$$ and the other at $$400K$$. The systems are then brought into thermal contact and the combined system is allowed to reach thermal equilibrium.

Note: I am skipping all the rigorous mathematical calculation because of typing issue.

Now, if I want to calculate (skipping Mathematical Calculation) the final Temperature of the combined system, I can do the calculation in two ways:

1. Through a reversible process.
2. Through an irreversible process.

As the whole system(both the systems) is thermally isolated no energy will be lost. But for the two different ways the final temperatures of the combined system will be different. I can understand it Mathematically but I am wondering what is actually happening physically!

For the irreversible case the change in entropy is greater than zero; So the heat exchange between the two systems will be same i.e.,
$$∫Cv dT1 = ∫Cv dT2$$(integration limit is according to the question and final temperature)

But, for the reversible case the change in entropy is equal to zero($$∆S1+∆S2=0$$). So, though the total energy is conserved, the heat exchange between the two systems is not same! I am wondering how is this possible?
I mean, if one system is releasing $$q$$ amount of heat and the other system is gaining the same amount of heat but the final temperature of the combined system is not in accordance with the consumed heat. At first, I thought that actually some heat is only used for gaining entropy but not for the temperature. But I don't know how it is possible to consume heat without rising temperature?(This is also not the phase transition stage so how is it possible?)

• Are you supposed to determine the change in entropy of the combined system, or are you just supposed to determine the heat exchanged and the final temperature? Jan 12, 2021 at 13:42
• Suppose you knew that initial and final temperatures of only one of the systems. Could you determine its change in entropy? If so, how? Jan 12, 2021 at 13:56
• I am trying to calculate the final temperature of the combined system after mixing! But as the mixing process can be done in either reversible or irreversible process that's why the final temperatures of the combined system will different for two different situations but I am wondering how is it possible? For your second question: as we know ds=dQ/T and dQ=Cv dT, here I know Cv so I can Integrate it! Jan 12, 2021 at 17:57
• The irreversible process determines the final temperature. The reversible path for each system separately determines the entropy change of each. Jan 12, 2021 at 18:19
• In other words to get the entropy change you need to first separate the two systems and then put each of them separately through a reversible path to get the entropy change of each one separately. notice I use the word separately three times toempasize it. Jan 12, 2021 at 18:23

## 1 Answer

The source of the confusion seems to be here:

But, for the reversible case the change in entropy is equal to zero($$∆S1+∆S2=0$$).

The above statement is responsible for the apparent inconsistency, but it's actually not true. Let me explain why. The explanation goes along the same lines as @ChetMiller’s in the comments.

In a reversible process the change in entropy is zero for the whole system undergoing the process. But what is this reversible process in this case? Importantly, the actual process of two systems at different temperatures exchanging heat is irreversible. But here's the key:

If all we care about is the initial and final state, then we can replace the actual process with any other fakey process we like as long as we don't alter the initial and final state.

So what fakey reversible process can we replace the actual process with? There are many options but there is one standard and simple reversible process that people use in situations like this. It entails connecting each of the two systems to a separate reservoir and slowly, reversibly, extracting or adding heat until the two systems reach the final state of having equal temperatures and the same total energy that they started with.

You might ask, how do we reversibly extract heat into a reservoir? Great question, because I lied a little bit, a single reservoir is not enough. For heat exchange to be reversible, the reservoir must be at (nearly) the same temperature as the system. So, as we extract heat from the system and its temperature gets lower and lower, we have to keep swapping out the reservoir for colder and colder ones, so that at each stage the system and the reservoir it's losing heat to are at the same temperature. But for simplicity I will continue the lie and use the word reservoir to denote this collection of reservoirs.

During the above mentioned reversible process which entropy change is zero? It's the entropy change of the first system plus its reservoir, and, separately, the entropy change of the second system plus its reservoir. It's not the entropy change of the first system plus the second system, those two systems do not participate in a reversible process together!

This shows that the entropy change of the first + second system does not have to be zero at all, which resolves the apparent inconsistency.

• there is no reservoir at all! Jan 21, 2021 at 8:23
• There is no reservoir in the actual process, which is irreversible. But what do you think the process is when you do the calculation through a reversible process? Jan 21, 2021 at 21:07
• I updated my answer with an explanation of where the reservoirs enter into the picture, I hope it makes things clearwe. Jan 22, 2021 at 23:23