Guys, let's take a look at this example problem. We have a heat engine that's shown as a cycle in a PV diagram. Remember heat engines are always going to be cycles. And what we want to do with part a is calculate the work that's done over a complete cycle in this process. Alright? So let's get started. With part a, we want to calculate w. Remember, w pops up in a bunch of different equations for heat engines, but probably the most fundamental one is going to be this Qh−Qc. The work that's done is the Qh, it's the heat that flows in minus the heat that comes out. That's just what comes from our energy flow diagrams. Right? So if you have a hot reservoir connected to an engine, connected to a cold reservoir here, this is your hot and cold. What happens is you have heat flowing in, that's Qh. You have heat flowing out, that's Qc, and then the work that's done is always going to be the difference between those two numbers. Right? So this work done is always QH−QC. Alright? So that's where that equation comes from, so let's see if we can try to use it. So what happens here is that we don't actually know from the energy flow diagram. We don't know actually what Qh and Qc are. All we have in this problem are just the Qs for some of the steps that are going on in this process here. So we're probably not going to use this equation over here. The other thing is that Qh, the heat transferred from the hot reservoir, is actually what we're going to calculate in part b. Right? So this is what we're going to calculate in part b. So we're going to have to use a different sort of method to calculate the work. And remember that for we actually do know how to do that from our discussion on cyclic processes. Remember that the work that's done over a cycle is always equal to the area that's enclosed within the loop. So the area enclosed gives you the work. So in other words, the work that's done is really just the area that is enclosed within this rectangle. And the good news is because it's a rectangle, there's a pretty easy way to calculate the area. So the work that's done here is really just going to be base times height. So in other words, this is the base over here and this is going to be the height, and I have all the numbers, you know, the pressures and volumes and things like that, so I actually can calculate this work this way. So this work that's done is going to be base times height. So in other words, we're going to use, the base, which is the difference between 0.45 and 0.60. So this base here equals 0.15. So this is going to be 0.15. And then this over here is going to be a height, which is equal to 300. So this h equals 300, and there's no need to convert or anything because all this stuff is in SI units. We have pascals and we'll have kilopascals or anything like that, so you're going to just multiply this by 300. If you work this out, what you're going to get is you're going to get 45 joules. So essentially, what happens is the area inside of this is going to be 45 joules. That's the work that's done. Alright? So that's the first part. Now let's move on to the second part here. In the second part, we want to calculate the total heat transfer from the hot reservoir. So in other words, what happens here is we've actually calculated in our energy flow diagram that this w here equals 45. So then if we have if we say, if this is 45 joules, then basically, what is Qh? That's what we're trying to find here. Alright? So can we use this equation again now that we actually have what QH is? Well, let's see. If we have w=Qh−Qc, then what happens is we can rearrange this equation here to solve for Qh. So what happens is I'm just going to move the Qc over to the other side. So in other words, the work that's done and I'm going to drop all the absolute values because they can get kind of annoying here. The work plus the heat that flows to the cold reservoir is equal to the heat that flows to the hot reservoir. Alright? So we have what w is. What is Qc? So in other words, what is this sort of heat? What is the total amount of heat that flows out of, this this sort of cycle here? And to do that, we're going to actually take a look at these two arrows over here. If you notice, what happens in this sort of cycle here is that there's 2 places where heat is flowing out of the, of the heat engine here. And that's indicated by these negative signs. This Q=−90 and Q=−25. What happens at a heat engine in general is that anytime you go up or_to...
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Heat Engines & PV Diagrams
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