A heat engine with an efficiency of 30% performs 2500 J of work.How much heat is discharged to the lower temperature reservoir?
A)5800 J
B)8300 J
C)750 J
D)1400 J
E)7100 J

Question

Quizplus · Accepted Answer

The efficiency ($e$) of a heat engine is given by the ratio of the work output ($W$) to the heat input ($Q_{in}$), i.e., $e = \frac{W}{Q_{in}}$. Given that the efficiency is 30% (or 0.3 when expressed as a decimal) and the work performed is 2500 J, we can find the heat input as $Q_{in} = \frac{W}{e} = \frac{2500}{0.3} = 8333.33$ J. The heat discharged ($Q_{out}$) to the lower temperature reservoir is the difference between the heat input and the work output, i.e., $Q_{out} = Q_{in} - W = 8333.33 - 2500 = 5833.33$ J. The closest value to this result is 5800 J.

A Heat Engine with an Efficiency of 30% Performs 2500