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Physics · Thermodynamics

JEE Advanced 2025 — Paper 2 — Question 8

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:

  1. Option A:

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

    Correct
  2. Option B:

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

    Correct
  3. Option C:

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

    Correct
  4. Option D:

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

Answer: A, B, C

Step-by-step solution

η=0.4\eta=0.4

(COP)HP=10(\mathrm{COP})_{\mathrm{HP}}=10

W=η×Q1=0.4×150=60 J\mathrm{W}=\eta \times \mathrm{Q}_{1}=0.4 \times 150=60 \mathrm{~J}

COP = Dissered Effect  Work input =\frac{\text { Dissered Effect }}{\text { Work input }}

10=Q3W10=\frac{Q_{3}}{W}

Q3=600 J\mathrm{Q}_{3}=600 \mathrm{~J}

Also η=0.4=1−T2 T1\eta=0.4=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}

T2=600 K\mathrm{T}_{2}=600 \mathrm{~K}

COP⁡=T3T3−T4=10\operatorname{COP}=\frac{T_{3}}{T_{3}-T_{4}}=10

=300300−T4=10=\frac{300}{300-\mathrm{T}_{4}}=10

T4=270 K\mathrm{T}_{4}=270 \mathrm{~K}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics
Topic
Entropy, Carnot's Engines and Refrigerators

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