Q 11-11-161JEE MainJEE Main 2018 (15 Apr, Shift 1)Easy
A Carnot's engine works like a refrigerator between $250\ \text{K}$ and $300\ \text{K}$. It receives $500\ \text{cal}$ heat from the reservoir at a lower temperature. The amount of work done in each cycle to operate the refrigerator is,
Answer: (B) $420\ \text{J}$
For a Carnot refrigerator,
$$\frac{Q_2}{W} = \frac{T_2}{T_1 - T_2} \quad\Rightarrow\quad W = Q_2\,\frac{T_1 - T_2}{T_2}$$
$$W = 500 \times \frac{300 - 250}{250} = 100\ \text{cal} = 100 \times 4.2 = 420\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics