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JEE Advanced 2025 · Thermodynamics

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:

Answers: A. Work extracted from the Carnot engine in one cycle is 60 J60\,\text{J}. · B. Temperature of the cold reservoir of the Carnot engine is 600 K600\,\text{K}. · C. Temperature of the cold reservoir of the heat pump is 270 K270\,\text{K}.

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Written solution

Step 1: Analyze the Carnot engine.

The efficiency η\eta of a Carnot engine is related to the temperatures of its hot reservoir THT_H and cold reservoir TCT_C.

η=1−TCTH\eta = 1 - \frac{T_C}{T_H}

Substitute the given values TH=1000 KT_H = 1000\text{ K} and η=0.4\eta = 0.4 to find TCT_C.

0.4=1−TC10000.4 = 1 - \frac{T_C}{1000} TC=600 KT_C = 600\text{ K}

The work output WW per cycle is calculated from the heat extracted from the hot reservoir QH=150 JQ_H = 150\text{ J}.

W=ηQHW = \eta Q_H W=0.4×150=60 JW = 0.4 \times 150 = 60\text{ J}

Step 2: Analyze the heat pump's energy transfer.

The work output W=60 JW = 60\text{ J} runs the heat pump. Use the coefficient of performance (COP=10\text{COP} = 10) to find the heat delivered to the hot reservoir QH′Q_H'.

COP=QH′W\text{COP} = \frac{Q_H'}{W} 10=QH′6010 = \frac{Q_H'}{60} QH′=600 JQ_H' = 600\text{ J}

The heat extracted from the cold reservoir of the heat pump QC′Q_C' is given by conservation of energy.

QC′=QH′−WQ_C' = Q_H' - W QC′=600−60=540 JQ_C' = 600 - 60 = 540\text{ J}

Thus, 540 J540\text{ J} of heat is taken from the cold reservoir, while 600 J600\text{ J} is delivered to the hot reservoir.

Step 3: Find the cold reservoir temperature of the heat pump.

For a Carnot heat pump, the COP is related to the reservoir temperatures TH′T_H' and TC′T_C'.

COP=TH′TH′−TC′\text{COP} = \frac{T_H'}{T_H' - T_C'}

Substitute TH′=300 KT_H' = 300\text{ K} and COP=10\text{COP} = 10.

10=300300−TC′10 = \frac{300}{300 - T_C'} 300−TC′=30300 - T_C' = 30 TC′=270 KT_C' = 270\text{ K}

Step 4: Evaluate the statements.

Comparing the calculated values with the statements:

  • The work extracted from the engine is 60 J60\text{ J} (Correct)
  • The temperature of the cold reservoir of the engine is 600 K600\text{ K} (Correct)
  • The temperature of the cold reservoir of the heat pump is 270 K270\text{ K} (Correct)
  • The heat supplied to the hot reservoir of the heat pump is 540 J540\text{ J} (Incorrect, it is 600 J600\text{ J})

Answer: The correct statements are:

  1. The work extracted from the engine is 60 J60\text{ J}.
  2. The temperature of the cold reservoir of the engine is 600 K600\text{ K}.
  3. The temperature of the cold reservoir of the heat pump is 270 K270\text{ K}.

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