In this problem, we have a real irreversible engine, not a reversible one. And I ask some questions about what is the entropy and work and other things related and how they relate to a reversible engine. So let's start with the first question that's asked for the change of entropy of the universe. Notice whatever is going on in here in this part of the system that's going through a cycle. And think of the carnot cycle and don't ask for cycle, reversible reversible does not matter. Got to ask for a psycho is zero. Entropy is a state function. All that matters is where you are. If you end up where you began, then we'll test this year. So all we gotta worry about is what's going on with these reservoirs and the heat transfers into and out of the reservoirs. So reservoirs meaning they don't change in temperature. So I think of these icy thermal transfers. So the first one I'm going to be taking Qh out Mattis qh or th plus now what QC is going in QC over to see. So that is answered apart a probably they want to know is discredited recorded zero. Now, just from the second law tells you that it's going to be greater than zero. But let's see if we can understand know more about that. What's happening is the following. Think about people walking around the room along the perimeter and you tell them you add heat by telling them to speed up. Now there's still along the perimeter. So it's very easy to tell him to slow down. They're still around the perimeter. You've got the same order or if you like, the same disordered state from one time to another, and you can easily get back. So the entropy changes zero between those states. Because if you like, the disorder is not changed now though, irreversible processes this QC. Like you see in this particular case it's got an added aspect to it. It's got a disorder piece to it. In addition, that's the guy who would tell people, you know you have the one part that says speed up. He's busy saying now move randomly into the room. You know, I've given you energy to move into the room. You tell tell them to slow down. You're not going to get back to that nice motion along the perimeter that that state of order. Or is that state of disorder? You've got a much more disordered state. So that's why entropy of the universe increases. You got this added aspect in this case for QC. Okay. Part C. They're looking for the relationship between the they are irreversible W. And the reversible. So doubt to you equals zero. We're just looking at this part now. Just a working aspect. You know the gas, whatever the system is here, it's interaction with the rest of the universe. Um The secret zero. It's a cycle right? As I talked about with Delta S. Just a minute ago. So this is Q. H minus QC. This is the first thought. So I get that. W You gotta qh ask you see this is true. Reversible. Irreversible. Does not matter now. I just mentioned that QC has got this added aspect. It's bigger. You know then if it was just reversible soak usin, it's greater than QC. Reversible. So if I were to replace this with the reversible form, this difference is going to be larger. Which means the work done reversible. E. Is going to be larger. So. And the second law tells us that also part D. S. Is there a way using what we have in front of us here? Delta. S. And T. Uh Two I said get the difference between these two works. The answer is yes. There is this quantity called the work unavailable which talks about the uh the degrading of energy inability. You know that that you lose the ability to do the same amount of work. And that by definition is the difference between what you could do reversible E. Minus what you are can do irreversibly. That's the meaning of it. So W. Mhm. A novel unavailable. W reversible Sw. That's the meaning. We lose something. If we did this process reversible E. We gain we have more work to do. Uh All right. Irreversible dan. We don't get as much. That's what we mean by the we still have the same energy involved in all this. The energy is not changed. But it can't do as much work. It's degraded. And this if you remember the formula is the lowest reservoir cases. TC. Time delta. S. Of the universe. That's that's that formula there kick. Uh For the problem we need these numbers th 852 Calvin T. c. Calvin. U. H. 12 85 jewels. Mm. Mhm. And W. To 64 jules. So we need those numbers. Now. The first thing we're gonna need is QC. You know, you can see from above. You can see from above. I can get QC from this you see is equal to uhh my S. W. And that's gonna be 102. Right? I mean 10 10 1021 jewels now for part A. They want delta. S. The universe. Nice. We have the formula -1285 jewels 852 Calvin Plus 10:21 chores Over 314 Calvin. This works out to be 1.74 choose per Calvin. That's him to be changed positive credit and zero as we expected. Or as we predicted from before. B. They want to know the work done reversible. Now let's go back. Here's daughter asked. Reversible. E. Q. H th. You see T. C. But now this is zero sarah. And so that gives me that this should be marked reversible. E. So that gives me accusing reversible. E. Is T. C. Q. H. Th. 314 Calvin 1285 jewels over 852 Calvin. And this comes out 3 474 jules. Now, I can use formula from before. Have you reversible? Uhh as qc reversible sequel to 811 jules. And lastly, lastly, they want to know the difference. Doesn't work out the T. V. The T. C. Delta S. So, and I'd see this. It's gonna be TC. You have to ask the universe. 314 Calvin 1.74 just for Calvin. And this works out to be 546 jules. I might say. There's a few did the 811-547. That's because we had around 1.74. And that is the whole problem.