A non-Carnot heat engine operates between a hot reservoir at 610K and a cold reservoir at 320K. In a cycle, it takes in 6400 J of heat and does 2200 J of work. What is the total change in entropy of the universe over the cycle?

Hey, everybody. So in this practice problem, we have a heat engine that's operating between 2 temperatures, 610 and 320. And we ultimately want to calculate the total change in entropy of the universe over one cycle of this heat engine. So the first thing I want to do here is draw a diagram. Now remember the heat engine kind of looks like this. We've got the hot reservoir, the cold reservoir, and the engine itself. The hot reservoir, the temperature here is 610. And remember that a heat engine always takes in energy from the hot reservoir. In other words, the QH here, we're told, is 64 100 joules because that's what we're told the amount of energy that it takes in. And it does 22100 joules of work. So this engine extracts 220 joules of usable energy, and then the rest kinda gets gets discarded as waste heat out to the cold reservoir. Now this cold reservoir here is at a temperature of 320. We don't know what the waste heat is. So if you think about what's happening here, this is an entropy problem that we have multiple objects for. The objects that are losing heats and gaining heats are different temperatures, The hot reservoir and the cold reservoir. So in other words, there's not just one equation that we have to use for delta s. We're gonna have to do it for both of these objects. So in other words, the delta s total, my total change in entropy, is gonna be the change in entropy of the hot reservoir. Right? The hot reservoir is losing heat plus the change in entropy of the cold reservoir. The cold reservoir is gaining some heat. That's really what's going on here. So in order to calculate delta s total, we just have to figure out, well, what are the changes in entropies for the 2 hot and cold reservoirs? That's basically what we have to do. So now that we've written our second, done the second step, we've written our delta s total equation, All we have to do now is just solve for the target variable. So what's going on here with the delta s? Well, if you think about what's happening here is we have this temperature that's 610, and usually what happens in a heat engine is that these reservoirs here are constant temperatures. So in other words, we can use our q over t equation. So if I just replace my delta s, it just becomes qh over t. However, what's happening here is that if you remember in a heat engine, we always use these numbers as positive, just to not get confused with the signs and things like that. But what's happening with this cold or this hot reservoir is that it's actually losing heat to the engine. So what hap what we have to do here is we actually have to insert a negative sign. The hot reservoir loses 64100 joules. And so in other words, the heat the heat sort that gets gained is actually negative. Right? It's negative 64100 joules. Alright. And then what we have to do is we have to add the heat that gets absorbed into the cold reservoir. Right? This QC is positive here and this is divided by TC. Alright. So we have with QH, TH, and TCR. All we have to do is actually just go ahead and figure out what is the QC. Now remember if you remember, this is actually pretty straightforward to do because we have both of the values of qh and w. So we can just use an old equation here. That's this absolute value of w is qh minus qc. Alright? And so what we can do here is just start plugging in some numbers. Basically, when you rearrange this equation, you move this to the other side, you move the w to the right. What you're gonna get here is that qh is equal to qh sorry. Sorry. Qc is equal to qh, which is 64100 minus the w, which is 22100. Notice how I used everything's absolute values so all these numbers here become positive. I don't have to worry about negative 64100 or anything like that. Alright. So then this QC just becomes 42100, and that's what I'm gonna plug into this equation. So now we just go ahead and solve. This delta s total is gonna be negative, this is gonna be 6,400 divided by 610 and then plus the 42100 divided by 310. Or sorry. This is 320. Alright? So if you work this out, what you're gonna get here is a delta s total of 2.6 joules per Kelvin, and that is the final answer. Alright? So you look through your answer choices and it's answer choice a. Let me know if you guys have any questions with this.

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A non-Carnot heat engine operates between a hot reservoir at 610K and a cold reservoir at 320K. In a cycle, it takes in 6400 J of heat and does 2200 J of work. What is the total change in entropy of the universe over the cycle?

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