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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2025 · Paper 2 · Physics · Thermodynamics

The efficiency of a Carnot engine operating with a hot reservoir...

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:

Options

  1. AWork extracted from the Carnot engine in one cycle is 60 J .
  2. BTemperature of the cold reservoir of the Carnot engine is 600 K .
  3. CTemperature of the cold reservoir of the heat pump is 270 K .
  4. DHeat supplied to the hot reservoir of the heat pump in one cycle is 540 J .

Correct answer

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

Step-by-step solution

η=0.4(COP)HP=10W=η×Q1=0.4×150=60JCOP= Dissered Effect  Work input 10=Q3WQ3=600J Also η=0.4=1−T2T1T2=600KCOP=T3T3−T4=10\begin{aligned} & \eta=0.4 \\ & (\mathrm{COP})_{\mathrm{HP}}=10 \\ & \mathrm{ W}=\eta \times \mathrm{Q}_1=0.4 \times 150=60 \mathrm{ J} \\ & \mathrm{COP}=\frac{\text{ Dissered Effect }}{\text{ Work input }} \\ & 10=\frac{\mathrm{Q}_3}{\mathrm{ W}} \\ & \mathrm{Q}_3=600 \mathrm{ J} \\ & \text{ Also } \\ & \eta=0.4=1-\frac{\mathrm{T}_2}{\mathrm{ T}_1} \\ & \mathrm{ T}_2=600 \mathrm{ K} \\ & \mathrm{COP}=\frac{\mathrm{T}_3}{\mathrm{ T}_3-\mathrm{T}_4}=10\end{aligned}=300300−T4=10T4=270K\begin{aligned} & =\frac{300}{300-\mathrm{T}_4}=10 \\ & \mathrm{ T}_4=270 \mathrm{ K}\end{aligned}$
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