Hey everybody. So let's check out our example problem. We have a Carnot heat engine that's taking in some heat from the hot reservoir, expelling it to a cold. And we want to calculate the total change in entropy of the universe. Now we already know this is gonna be a entropy problem that involves multiple objects because the hot and the cold reservoirs are both exchanging heats at different temperatures. So let's go ahead and draw our quick little diagram here. That's my hot, that's my engine, that's my cold. We've got the temperature of the hot is 500 the temperature of the cold is 3 50. All this stuff is in kelvin and so we have the heat that's taken in from the hot reservoir is 2000 jewels. So that's my cue. H. right? That's Q. H. equals 2000 jewels. So then what happens is expel some heat to the cold reservoir but we actually don't know what that Q. C. Is. So the next thing I wanna do is just set up our delta S total equation. The total change in entropy for the universe is going to be adding up the two n. Trapeze entropy changes for the hot and cold reservoirs. In other words, DELTA S. H. Plus delta S. C. Alright, so what happens here is that this delta S. H. Because we can assume that this happens at constant temperature is going to be qh over th but remember we have to add a negative sign because the hot reservoir loses some heat to the engine. Whereas this delta S. C. Here is gonna be plus que si over T. C. And that's because this cold reservoir is absorbing heat from the engine. Right? So one's positive, one's negative. Okay, So basically this question figure this out, right? So, we've got delta or so you have negative QH which is going to be negative Divided by the 500. Now, we have the T. C. Here, we have this 3 50 but we don't actually have what the Q. C. Is. So, we're gonna have to go figure that out now. The way that we did this in another problem was that we actually had the work done, and so if we used the work equation, right? This w equation, we calculate Q. C. But we actually don't have that here. So, in order to figure out what this Q. C. Here is, I'm gonna need another equation. So, let's go ahead and do that. So, which equation can we use for the Carnot cycle? We actually have a couple of them. If we need the efficiency we would use this equation, but we don't have the efficiency. We can't use this equation here. So, the only one that that I can use that sort of special to Carnot engine is going to be this one over here. This is the ratio of the heats. Does he go to the ratio of the temperatures? Remember, we can only use this for a carnot cycle. So, let's set this up here. So, we've got Q. C over Q. H. Is equal to T. C over th All right. So, if you look at this, I have three out of four variables. Right? I've got all the heats and the temperatures. So I can calculate this real quick here. This Q. C. Is really just gonna be well, T. C over th is 3 50 over 500. So, we're going to multiply that times the qh which was 2000. If you go ahead and work this out, what you're gonna get here is you're gonna get um Let's see, I get 1400. So, this is 1400 jewels. That's what we plug into this equation now. Alright, so, basically this just becomes negative. 2000 over 500 plus 1400 over 3 50. Now, when you work this out, what you're gonna get is that this delta s total here is actually equal to zero. So, essentially happens is these two things will cancel each other out. So, what's going on here is that because carnot heat engines are ideal and they are perfectly reversible, then, essentially what that means is that the change in entropy for a card lifecycle is always equal to zero. So, in other words, carnot engines are perfect and perfectly reversible, then, basically what happens is that the best outcome happens, you have no total change in entropy of the universe. All right. So, that's sort of like the best case scenario that you could hope for. All right, guys. So that's it for this one.