Carnot engine problem?

Question

A carnot engine performs work at the rate of 440 kW while using 680 kcal of heat per second. If the temperature of the heat source is 570 C, at what temperature is the waste heat exhausted? Any help would be great, thanks!

Answer 1

As you probably know, for the Carnot engine, the ratio of both heats equals the ratio of both absolute temperatures:
Qc/Qh = Tc/Th (h stands for "hot source" from where the heat is transfered to the "cold sink"). The difference of the both heats is the mechanical work we get from the engine: W = Qh ─ Qc.
In your problem, Qh = 680 kcal = 680 x 4.184 kJ = 2845.12 kJ and W = Pt = 440 kW x 1 s = 440 kJ.
You're looking for the Tc, so you need Qc:
Qc = Qh ─ W = 2845.12 kJ ─ 440 kJ = 2405.12 kJ and
Tc = Th x Qc/Qh = 843 K x 2405.12 kJ / 2845.12 kJ = 713 K = 440 C

Answer 2

First thing, convert the input to the same units as the output.

1 kcal = 4184 J

Second, calculate the efficiency (output / input).

For a Carnot engine, the efficiency = [T(in) - T(out)]/T(in).  You can solve for T(out) once you have the other two numbers.