Question

A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :

(1) 420 J (2) 2100 J (3) 2520 J (4) 772 J

Answer 1

Answer: 420 J

Explanation:

The efficiency of a Carnot's engine working as a refrigerator:

$\eta = \frac{T_c}{T_H-T_c}=\frac{Q_2}{W}$

$T_c=250K\\ T_H=300K\\ Q_2=500cal=2092 J$ (because 1 cal =4.184 J)

W is the amount of work done in each cycle.

$\frac{250K}{300K-250K}=\frac{2092 J}{W}$

$W=\frac{2092J}{250K}\times 50K=418.4J\approx 420 J$

Thus, the amount of work done in each cycle to operate the refrigerator is : option 1, 420 J