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A heat engine operating between $227^{0} C$ and $77^{0} C$ absorb 10 Kcal of heat from $227^{0} C$ reservoir reversibly per cycle, calculate the total work done (in Kcal) in two cycles.

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answer is 6.

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Detailed Solution

$η = \frac{T_{2} - T_{1}}{T_{2}} = \frac{| - w_{T o t a l} |}{q_{2}} \frac{500 - 350}{500} = \frac{| - w_{t o t a l} |}{10}$

Work done in one cycle =3, in two cycles = $2 \times 3 = 6$

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