A Carnot engine operates with hot reservoir at 1000 K with efficiency 0.4. It extracts 150 J per cyc

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Question

A Carnot engine operates with hot reservoir at 1000 K with efficiency 0.4. It extracts 150 J per cycle from the hot reservoir. The work is fully used to run a heat pump with coefficient of performance (COP) 10. The hot reservoir of the heat pump is at 300 K. Which statements are correct? (ONE OR MORE)

Accepted Answer

Correct Answer: A, B, CStep-by-step Solution:Carnot Engine:Efficiency η = 1 − T_cold/T_hot = 0.4T_cold/T_hot = 0.6 → T_cold = 0.6 × 1000 = 600 K. ✓ Option B.Heat extracted Q_H = 150 J. Work done W = η × Q_H = 0.4 × 150 = 60 J. ✓ Option A.Heat rejected to cold reservoir: Q_C = Q_H − W = 150 − 60 = 90 J.Heat Pump:COP of heat pump = Q_hot_pump/W_pump = 10, where W_pump = W_engine = 60 J.Heat delivered to hot reservoir of pump: Q_hot_pump = COP × W = 10 × 60 = 600 J.Heat extracted from cold reservoir of pump: Q_cold_pump = Q_hot_pump − W = 600 − 60 = 540 J.Temperature of cold reservoir of heat pump:For Carnot heat pump: COP = T_hot/(T_hot − T_cold) = 10T_hot/(T_hot − T_cold) = 10 → T_hot − T_cold = T_hot/10 = 300/10 = 30 KT_cold = 300 − 30 = 270 K. ✓ Option C.Checking Option D: Heat supplied to hot reservoir of heat pump = Q_hot_pump = 600 J ≠ 540 J. ✗540 J is the heat extracted from the cold reservoir (Q_cold_pump), not heat supplied to the hot reservoir. Option D confuses these two quantities.Why Option D is incorrect:Q_hot (heat supplied to hot reservoir of pump) = 600 J, not 540 J. 540 J is Q_cold (heat absorbed from cold reservoir). These differ by the work input: Q_hot = Q_cold + W → 600 = 540 + 60 ✓. Option D mistakenly states 540 J for Q_hot.Key Concepts:Carnot engine efficiency: η = 1 − T_C/T_H = W/Q_H. The work output W = ηQ_H. Heat rejected Q_C = Q_H − W = (1−η)Q_H.Heat pump COP: COP_pump = Q_hot_delivered/W_input = T_hot/(T_hot − T_cold) (for Carnot heat pump). Compare with refrigerator COP = Q_cold_extracted/W = T_cold/(T_hot − T_cold).Energy conservation in heat pump: Q_hot = Q_cold + W. The work input goes into pumping heat from cold to hot reservoir, so the total heat delivered to the hot reservoir exceeds the work input.Relation between COP of heat pump and refrigerator: COP_pump = COP_refrigerator + 1. If COP_refrigerator = T_cold/(T_hot − T_cold) = 9 (since COP_pump = 10), then T_cold = 9(T_hot − T_cold) → 10T_cold = 9T_hot → T_cold = 0.9 × T_hot = 270 K for T_hot = 300 K.

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