First slide

Heat engine

Question

A Camot engine having an efficiency of $\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is l0 J, the amount of energy absorbed from the reservoir at lower temperature is:

Moderate

A

90 J
B

99 J
C

100 J
D

1 J

Solution

$Efficiency = \frac{W}{q_{1}} = \frac{10}{q_{1}} = \frac{1}{10} \Rightarrow q_{1} = 100 J$
$\Rightarrow q_{2} = q_{1} - W = 100 - 10 = 90 J$

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