A Carnot engine with an efficiency of 70% is cooled by water at 10°C. What temperature must the hot reservoir be maintained at?

Hey, guys. So let's get some more practice here. So we're told that this is a Carnot engine, which means it operates with the maximum possible efficiency, and the efficiency here is 70%. We're told that this car no engine here is cooled by water at 10 degrees Celsius. So what does that mean? So I'm actually gonna draw out a quick little energy flow diagram. So remember, we have a hot reservoir that's connected to the engine. This is the Carnot engine, so I'm gonna write engine here. And this is connected to the cold reservoir. The cold reservoir here is basically where all the waste heat gets expelled. Now this engine this engine is gonna be cooled by water at 10 degrees Celsius. What that means here, when it's cooled by water, is that all the waste heat gets expelled into this water here. That's what cools down this engine and allows the thing to repeat itself. So what this cooled by water really means is that this is actually the temperature of the cold reservoir. This is the 10. So really, this is 10 degrees Celsius. And if you just add 273 to this, what you're gonna get is 283 Kelvin. Okay? So now what we have to do is figure out what the temperature of the hot reservoir is. Now remember, this is a heat engine and this is a Carnot engine, which means that the hot reservoir should be, hotter than the cold reservoir temperature. So already off the bat, you can kind of eliminate answer choices a and d. Those are colder temperatures than our cold reservoirs, so this can't be what's going on here. So it has to be either b or c. Now, how do we find this? We're basically just gonna use our E Carnot equation. So E Carnot is equal to 1 minus t c over t h. The variable that we're looking for is t h. So what's the efficiency of this engine? It's just 70%. But remember, we always have to plug in these numbers here as decimals. So this is gonna be 0.7. So we have 0.7 equals 1 minus Tc over t h. In fact, I'm actually just gonna replace this Tc because we already know what it is. It's the 283 Kelvin of our cold reservoir temperature. Alright. So how do I get this? Well, let's see. This t h is kind of at the bottom of this denominator here, so I'm gonna make this a little bit easier. I'm gonna move this to the left side, and I'm gonna move the 0.7 to the right side. What you end up with here is 283 divided by t h. This is gonna equal 1 minus 0.7. So we have 283 divided by t h equals 0.3. All you have to do now is you basically just have to trade places between these two variables. The t h goes up, the 0.3 comes down. And what you end up with here is 283 divided by 0 point 3 equals your temperature of your hot reservoir, and this is gonna be equal to 943.3 Kelvins. Alright? And just as expected, we got a number that was higher than our cold reservoir temperature. This is 943.3. It's always a good idea to sort of double check these numbers here. So what happens is if you plug these numbers into this equation, 283 and 943.3, you should get back to an efficiency of 0.7. Alright? So that's it for this one. Your answer choice is c. Let me know if you guys have any questions.

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A Carnot engine with an efficiency of 70% is cooled by water at 10°C. What temperature must the hot reservoir be maintained at?

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