Question:
The work done by a Carnot engine operating between \(300\,\text{K}\) and \(400\,\text{K}\) is \(400\,\text{J}\). The energy exhausted by the engine is
The work done by a Carnot engine operating between \(300\,\text{K}\) and \(400\,\text{K}\) is \(400\,\text{J}\). The energy exhausted by the engine is
Show Hint
For a Carnot engine,
\[
\eta=1-\frac{T_C}{T_H}
\]
and
\[
Q_H=W+Q_C
\]
Always calculate efficiency first.
Updated On: Jun 22, 2026
- \(800\,\text{J}\)
- \(1200\,\text{J}\)
- \(400\,\text{J}\)
- \(1600\,\text{J}\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the efficiency formula of Carnot engine.
Efficiency of a Carnot engine is \[ \eta=1-\frac{T_C}{T_H} \] Given, \[ T_H=400\,\text{K} \] \[ T_C=300\,\text{K} \] Thus, \[ \eta=1-\frac{300}{400} \] \[ \eta=1-\frac{3}{4} \] \[ \eta=\frac{1}{4} \]
Step 2: Use the relation between efficiency and work done.
Efficiency is also given by \[ \eta=\frac{W}{Q_H} \] Given, \[ W=400\,\text{J} \] Therefore, \[ \frac{1}{4}=\frac{400}{Q_H} \] \[ Q_H=1600\,\text{J} \]
Step 3: Find the heat rejected.
Heat exhausted by the engine is \[ Q_C=Q_H-W \] \[ Q_C=1600-400 \] \[ Q_C=1200\,\text{J} \]
Step 4: Final conclusion.
Hence, the energy exhausted by the engine is \[ \boxed{1200\,\text{J}} \]
Efficiency of a Carnot engine is \[ \eta=1-\frac{T_C}{T_H} \] Given, \[ T_H=400\,\text{K} \] \[ T_C=300\,\text{K} \] Thus, \[ \eta=1-\frac{300}{400} \] \[ \eta=1-\frac{3}{4} \] \[ \eta=\frac{1}{4} \]
Step 2: Use the relation between efficiency and work done.
Efficiency is also given by \[ \eta=\frac{W}{Q_H} \] Given, \[ W=400\,\text{J} \] Therefore, \[ \frac{1}{4}=\frac{400}{Q_H} \] \[ Q_H=1600\,\text{J} \]
Step 3: Find the heat rejected.
Heat exhausted by the engine is \[ Q_C=Q_H-W \] \[ Q_C=1600-400 \] \[ Q_C=1200\,\text{J} \]
Step 4: Final conclusion.
Hence, the energy exhausted by the engine is \[ \boxed{1200\,\text{J}} \]
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