The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is 0.4 . It extracts 150 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 K . Which of the following statements is/are correct :

Work extracted from the Carnot engine in one cycle is 60 J., Temperature of the cold reservoir of the Carnot engine is 600 K., Temperature of the cold reservoir of the heat pump is 270 K.

Work extracted from the Carnot engine in one cycle is 60 J.

Temperature of the cold reservoir of the Carnot engine is 600 K.

Temperature of the cold reservoir of the heat pump is 270 K.

Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J.

Heat and Thermodynamics PYQ — JEE Advanced 2025 · Shift 2 · Q40 with Solution

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Correct answer: A, B, C

Carnot engine data

Given:

Hot reservoir temperature of engine: $T_H=1000\,\text{K}$
Efficiency: $\eta=0.4$
Heat absorbed per cycle: $Q_H=150\,\text{J}$

For a heat engine,

\eta=\frac{W}{Q_H}

So,

W=\eta Q_H=0.4\times 150=60\,\text{J}

Hence Option A is correct.

Cold reservoir temperature of the Carnot engine

For a Carnot engine,

\eta=1-\frac{T_C}{T_H}

Given $\eta=0.4$ and $T_H=1000\,\text{K}$ :

0.4=1-\frac{T_C}{1000}

\frac{T_C}{1000}=0.6

T_C=600\,\text{K}

Hence Option B is correct.

Heat pump data

The work produced by the engine is fully used to run the heat pump, so

W_{\text{pump}}=60\,\text{J}

Coefficient of performance (COP) of heat pump is 10. For a heat pump,

\text{COP}=\frac{Q_H}{W}

where $Q_H$ is the heat delivered to the hot reservoir.

Thus,

10=\frac{Q_H}{60}

Q_H=600\,\text{J}

So the heat supplied to the hot reservoir of the heat pump per cycle is $600\,\text{J}$ , not $540\,\text{J}$ . Hence Option D is incorrect.

Cold reservoir temperature of the heat pump

For an ideal (Carnot) heat pump,

\text{COP}=\frac{T_H}{T_H-T_C}

Given:

$\text{COP}=10$
Hot reservoir temperature of pump: $T_H=300\,\text{K}$

So,

10=\frac{300}{300-T_C}

300-T_C=30

T_C=270\,\text{K}

Hence Option C is correct.

Final evaluation of options

A: Correct
B: Correct
C: Correct
D: Incorrect

Therefore, the correct options are:

\boxed{A,\ B,\ C}

Comparison with stored correct answer

Stored correct answer: $A, B, C$

Our derived answer matches the stored answer exactly.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: A, B, C

Carnot engine data

Given:

Hot reservoir temperature of engine: $T_H=1000\,\text{K}$
Efficiency: $\eta=0.4$
Heat absorbed per cycle: $Q_H=150\,\text{J}$

For a heat engine,

\eta=\frac{W}{Q_H}

So,

W=\eta Q_H=0.4\times 150=60\,\text{J}

Hence Option A is correct.

Cold reservoir temperature of the Carnot engine

For a Carnot engine,

\eta=1-\frac{T_C}{T_H}

Given $\eta=0.4$ and $T_H=1000\,\text{K}$ :

0.4=1-\frac{T_C}{1000}

\frac{T_C}{1000}=0.6

T_C=600\,\text{K}

Hence Option B is correct.

Heat pump data

The work produced by the engine is fully used to run the heat pump, so

W_{\text{pump}}=60\,\text{J}

Coefficient of performance (COP) of heat pump is 10. For a heat pump,

\text{COP}=\frac{Q_H}{W}

where $Q_H$ is the heat delivered to the hot reservoir.

Thus,

10=\frac{Q_H}{60}

Q_H=600\,\text{J}

So the heat supplied to the hot reservoir of the heat pump per cycle is $600\,\text{J}$ , not $540\,\text{J}$ . Hence Option D is incorrect.

Cold reservoir temperature of the heat pump

For an ideal (Carnot) heat pump,

\text{COP}=\frac{T_H}{T_H-T_C}

Given:

$\text{COP}=10$
Hot reservoir temperature of pump: $T_H=300\,\text{K}$

So,

10=\frac{300}{300-T_C}

300-T_C=30

T_C=270\,\text{K}

Hence Option C is correct.

Final evaluation of options

A: Correct
B: Correct
C: Correct
D: Incorrect

Therefore, the correct options are:

\boxed{A,\ B,\ C}

Comparison with stored correct answer

Stored correct answer: $A, B, C$

Our derived answer matches the stored answer exactly.

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