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A cyclic heat engine does \( 50 \mathrm{kJ} \) of work per cycle. If efficiency of engine is \( 75 \% \) the heat rejected per cycle will be:

1. 60.6
2. 16.6
3. 200
4. 600

Correct answer: 60.6

Solution

For a heat engine, efficiency is η = W/Q_in. With W = 50 kJ and η = 0.75, the heat input is 66.7 kJ, so the rejected heat is Q_out = Q_in - W = 16.7 kJ. However, the provided correct option indicates the intended calculation is likely based on a different convention or a typo in the problem statement.

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