First slide

Heat Engine; Carnot Engine and refrigerator

Question

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

Moderate

A

99 J
B

100 J
C

90 J
D

1 J

Solution

In heat engine,
$η = \frac{w o r k d o n e}{Q_{a b s o r b e d}} \Rightarrow \frac{1}{10} = \frac{10}{Q_{a b s}}$
$\begin{array}{l} \Rightarrow Q_{a b s} = 100 J \\ N o w, Q_{r e j} = Q_{a b s} - W = 100 = 10 = 90 J \end{array}$
In refrigerator,
$\begin{array}{l} Q_{a b s} = Q_{r e j} i n h e a t e n g i n e \\ ∴ Q_{a b s} = 90 J \end{array}$

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