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:

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Use \( \eta = 1 - \frac{T_C}{T_H} = \frac{W}{Q_H} \) and find \( Q_C = Q_H - W \) for energy rejected.
Updated On: May 20, 2025
  • 800 J
  • 1200 J
  • 400 J
  • 1600 J
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The Correct Option is B

Solution and Explanation

Efficiency of a Carnot engine: \[ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{400} = \frac{1}{4} \] \[ \text{Efficiency} = \frac{W}{Q_H} \Rightarrow \frac{1}{4} = \frac{400}{Q_H} \Rightarrow Q_H = 1600\, \text{J} \] Energy exhausted (heat rejected): \[ Q_C = Q_H - W = 1600 - 400 = 1200\, \text{J} \]
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