Carnot efficiency: $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{600} = 0.5$
$W = \eta Q_1 \Rightarrow Q_1 = \frac{W}{\eta} = \frac{150}{0.5} = 300\,\text{J}$
Back to JEE Main 2021 Thermodynamics
JEE Main 2021 — Physics Thermodynamics
easy
mcq
2021
Official previous-year question
Verified 30 May 2026.
Question
A Carnot engine operates between temperatures $600\,\text{K}$ and $300\,\text{K}$. If the engine produces $150\,\text{J}$ of work per cycle, the heat absorbed from the source per cycle is:
Options
- A
$600\,\text{J}$
- B
$450\,\text{J}$
- C
$300\,\text{J}$
- D
$150\,\text{J}$
Solution
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