JEE Advanced 2025 — Paper 2 — PHY-II solution

JEE Advanced2025Paper 2PHY-II

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:

A
Work extracted from the Carnot engine in one cycle is 60 J.
B
Temperature of the cold reservoir of the Carnot engine is 600 K.
C
Temperature of the cold reservoir of the heat pump is 270 K.
D
Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J.

Solution

For the Carnot engine: $W = η Q_{1} = 0.4 \times 150 = 60$ J. (A) is correct.

From $η = 1 - T_{2} / T_{1}$ with $T_{1} = 1000$ K, $T_{2} = 600$ K. (B) is correct.

For the heat pump (driven by $W = 60$ J), the COP definition for a pump is $COP = Q_{h} / W$ , where $Q_{h}$ is heat delivered to the hot reservoir. The problem states the hot reservoir is at 300 K and COP $= 10$ , and the COP for a Carnot-type heat pump is $COP = \frac{T _{h}}{T _{h} - T _{c}}$ :

$10 = \frac{300}{300 - T _{c}} \Rightarrow 300 - T_{c} = 30 \Rightarrow T_{c} = 270$ K. (C) is correct.

Heat delivered to the hot reservoir: $Q_{h} = COP \times W = 10 \times 60 = 600$ J $\neq = 540$ J. (D) is incorrect.

$∴$ Correct options are (A), (B), (C).

← Back to full paper

Practice more PHY-II

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →