Rate of Heat Output to Low-Temp Res per Cycle: 450J

AI Thread Summary
The heat engine operates with an efficiency of 25% and a heat input of 600J per cycle. Using the formula E=(Qh-Qc)/Qh, the correct calculation shows that the heat output to the low-temperature reservoir (Qc) is 450J. However, confusion arose regarding the book's answer of 800J, which resulted from mixing up input and output heats. Upon clarification, it was determined that Qc is indeed 600J, leading to a heat output from the high-temperature reservoir (Qh) of 800J. This highlights the importance of accurately distinguishing between heat input and output in thermodynamic calculations.
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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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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.
 
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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