How Much Work Can a Reversible Heat Engine Generate?

[Ayame17]

[/B]

Homework Statement

What is the maximum amount of mechanical work that can be got by a reversible engine working from a finite heat source of thermal capacity, C=1000J/K, if it is initially at 373K and the engine discharges to a bath of virtually infinite heat capacity at 273K?
[You may take C as a constant over the temperature range, 273K - 373K]

Homework Equations

Good question...!

The Attempt at a Solution

Well, I know that W=Q1-Q2, which is also =Q(T1-T2)/T1. I also know that W=(integral of)Pdv, and that Q=mc(deltaT). However, I can't see how to get the values for P or Q. To begin with, I'm not sure whether I should treat it as a Carnot engine, since most reversible ones are, but not all of them. And if so, should I treat the entropy (S) as 0, which might help? I worked out deltaS = Cln(T2/T1) = -312.1066, which I don't know if it is right and if it will help anyway!
Any help at all will be appreciated!

 

[Jhero]

I would like to clarify a few things before attempting to solve this problem. Firstly, I would like to confirm if the engine in question is indeed a Carnot engine, as this will affect the approach to finding the maximum work output. Additionally, I would like to confirm the units of the thermal capacity, C, as it is typically measured in J/K, not just J.

Assuming the engine is a Carnot engine and the units of C are J/K, we can proceed with the solution. The maximum work output of a Carnot engine is given by the Carnot efficiency, which is equal to (T1-T2)/T1, where T1 and T2 are the temperatures of the hot and cold reservoirs, respectively. In this case, T1=373K and T2=273K.

Using the formula for work, W=Q1-Q2, we can rewrite the Carnot efficiency as W/Q1. Substituting Q=mc(deltaT) and rearranging, we get W=C(T1-T2)/T1.

Plugging in the given values, we get W=C(373K-273K)/373K=1000J/K(100K)/373K=268.5J. Therefore, the maximum amount of work that can be obtained from this reversible engine is 268.5J.

I hope this helps in solving the problem. If there are any further clarifications needed, please let me know.