Physics · Thermodynamics

NEET (UG) 2023 — Question 177

A Carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is

Option A:
15°C
Option B:
100°C
Option C:
200°C
Option D:
27°C
Correct

Answer: D

Step-by-step solution

Efficiency of Carnot engine is $\eta = 1 - \frac{T_2}{T_1}$ . Given $\eta = 50\% = 0.5$ , source temperature $T_1 = 327^\circ\text{C} = 327 + 273 = 600\,\text{K}$ . Substitute: $0.5 = 1 - \frac{T_2}{600}$ . Thus $\frac{T_2}{600} = 0.5$ , so $T_2 = 300\,\text{K}$ . Convert to Celsius: $T_2 = 300 - 273 = 27^\circ\text{C}$ . Hence the sink temperature is $27^\circ\text{C}$ , which corresponds to option D.

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Exam: NEET (UG) 2023
Paper: 2023 paper
Subject: Physics
Chapter: Thermodynamics
Topic: Entropy, Carnot's Engines and Refrigerators