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A certain engine which operates in a Carnot cycle absorbs 3.347 kJ at 400°C, how much work is done by the engine per cycle? The temperature of sink is 100°C.

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1.49 kJ

2.51 kJ

1.85 kJ

2.98 kJ

answer is A.

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Detailed Solution

$\begin{array}{l} η = \frac{T_{2} - T_{1}}{T_{2}} = 1 - \frac{T_{1}}{T_{2}} = 1 - \frac{373}{673} = 0.446 \\ η = \frac{Δ w}{q_{1}} or Δ W = η \times q_{1} = 0.446 \times 3.347 = 1.49 kJ \end{array}$

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