JEE Challenger
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PhyMedAdv 2025 • Paper 2

Carnot Engine and Heat Pump Thermodynamic Analysis

The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K1000\text{ K} is 0.40.4. It extracts 150 J150\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 1010. The hot reservoir of the heat pump is at a temperature of 300 K300\text{ K}. Which of the following statements is/are correct:

Options

A

Work extracted from the Carnot engine in one cycle is 60 J60\text{ J}.

Correct
B

Temperature of the cold reservoir of the Carnot engine is 600 K600\text{ K}.

Correct
C

Temperature of the cold reservoir of the heat pump is 270 K270\text{ K}.

Correct
D

Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J540\text{ J}.

Topics & Concepts

ThermodynamicsCarnot EngineFirst Law of Thermodynamics

Step-by-Step Solution

To determine the correct options, we analyze the thermodynamics of the Carnot engine and the heat pump step-by-step.

1. Analysis of the Carnot Engine

We are given:

  • Temperature of the hot reservoir, T1=1000 KT_1 = 1000\text{ K}
  • Efficiency of the Carnot engine, η=0.4\eta = 0.4
  • Heat extracted per cycle from the hot reservoir, Q1=150 JQ_1 = 150\text{ J}

Work Extracted (WW):

The efficiency of an engine is defined as: η=WQ1\eta = \frac{W}{Q_1}

Substituting the given values: 0.4=W150 J  ⟹  W=0.4×150 J=60 J0.4 = \frac{W}{150\text{ J}} \implies W = 0.4 \times 150\text{ J} = 60\text{ J}

Thus, Option (A) is correct.

Cold Reservoir Temperature (T2T_2):

For a Carnot engine, efficiency is also given by: η=1−T2T1\eta = 1 - \frac{T_2}{T_1}

Substituting η=0.4\eta = 0.4 and T1=1000 KT_1 = 1000\text{ K}: 0.4=1−T21000 K0.4 = 1 - \frac{T_2}{1000\text{ K}} T21000 K=0.6  ⟹  T2=600 K\frac{T_2}{1000\text{ K}} = 0.6 \implies T_2 = 600\text{ K}

Thus, Option (B) is correct.

2. Analysis of the Heat Pump

The work input to the heat pump per cycle is Win=W=60 JW_{\text{in}} = W = 60\text{ J}.

We are given:

  • Coefficient of Performance of the heat pump, (COP)hp=10(\text{COP})_{\text{hp}} = 10
  • Temperature of the hot reservoir of the heat pump, TH=300 KT_H = 300\text{ K}

Heat Supplied to Hot Reservoir (QHQ_H):

By definition, the coefficient of performance for a heat pump is: (COP)hp=QHWin(\text{COP})_{\text{hp}} = \frac{Q_H}{W_{\text{in}}}

Substituting the values: 10=QH60 J  ⟹  QH=600 J10 = \frac{Q_H}{60\text{ J}} \implies Q_H = 600\text{ J}

Therefore, the heat supplied to the hot reservoir of the heat pump in one cycle is 600 J600\text{ J} (not 540 J540\text{ J}). Thus, Option (D) is incorrect.

Cold Reservoir Temperature of the Heat Pump (TCT_C):

For a Carnot heat pump operating between temperatures THT_H and TCT_C: (COP)hp=THTH−TC(\text{COP})_{\text{hp}} = \frac{T_H}{T_H - T_C}

Substituting (COP)hp=10(\text{COP})_{\text{hp}} = 10 and TH=300 KT_H = 300\text{ K}: 10=300 K300 K−TC10 = \frac{300\text{ K}}{300\text{ K} - T_C} 300 K−TC=30 K  ⟹  TC=270 K300\text{ K} - T_C = 30\text{ K} \implies T_C = 270\text{ K}

Thus, Option (C) is correct.

Conclusion

The correct statements are A, B, and C.

Carnot Engine and Heat Pump Thermodynamic Analysis | Physics PYQ Solution - JEE Challenger