# Thermal efficiency

#### fish

A heat engine has an efficiency of 25.0% and its heat input is 600J per cycle from the high-temperature reservoir. What is the rate of heat output to the low-temperature reservoir per cycle?

E=(Qh-Qc)/Qh

.25=(600J-Qc)/600J
Qc=450J

book has the answer as 800J.
How are they getting 800J and not 450J?

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#### Doc Al

Mentor
Originally posted by fish
book has the answer as 800J.
How are they getting 800J and not 450J?
The book's answer makes no sense. Your reasoning is correct. Are you sure you're not mixing up input and output heats? That would explain the book's answer.

#### fish

yes, your're right. It looks like I mixed up input and output heats.
Qc=600J (heat flow into cold res.)

find heat output from high-temp res. which would be
Qh, so solving for Qh you get 800J

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