Question 8

Analyze the Carnot engine first: Efficiency $\eta = \frac{W}{Q_H} \implies 0.4 = \frac{W}{150 \text{ J}} \implies W = 60 \text{ J}$. Thus, option (A) is correct. Temperature of cold reservoir $T_L$ is given by $\eta = 1 - \frac{T_L}{T_H} \implies 0.4 = 1 - \frac{T_L}{1000} \implies \frac{T_L}{1000} = 0.6 \implies T_L = 600 \text{ K}$. Thus, option (B) is correct.

Analyze the heat pump: The work input is $W = 60 \text{ J}$. Coefficient of Performance (COP) for a heat pump is $\text{COP} = \frac{Q_H'}{W} \implies 10 = \frac{Q_H'}{60} \implies Q_H' = 600 \text{ J}$. This is the heat supplied to the hot reservoir, so option (D) is incorrect. For a Carnot heat pump, $\text{COP} = \frac{T_H'}{T_H' - T_L'} \implies 10 = \frac{300}{300 - T_L'} \implies 300 - T_L' = \frac{300}{10} = 30 \implies T_L' = 270 \text{ K}$. Thus, option (C) is correct.