Question 8

MCQMEDIUM

The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of $1000 \text{ K}$ is $0.4$ . It extracts $150 \text{ J}$ of heat per cycle from the hot reservoir. The work extracted from this engine is being fully used to run a heat pump which has a coefficient of performance $10$ . The hot reservoir of the heat pump is at a temperature of $300 \text{ K}$ . Which of the following statements is/are correct:

(A)

Work extracted from the Carnot engine in one cycle is $60 \text{ J}$ .

(B)

Temperature of the cold reservoir of the Carnot engine is $600 \text{ K}$ .

(C)

Temperature of the cold reservoir of the heat pump is $270 \text{ K}$ .

(D)

Heat supplied to the hot reservoir of the heat pump in one cycle is $540 \text{ J}$ .

Detailed Solution

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.

Question 8

Detailed Solution

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