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Second Law of Thermodynamics

A heat engine...

Question

A heat engine is operating between $500 K$ to $300 K$ and it absorbs $10 k c a l$ of heat from $500 K$ reservoir reversibly per cycle. Calculate the work done (in kcal) per cycle.

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Solution

Heat rejected at the higher temperature is -
$Q_{2} = \frac{T_{2}}{T_{1}} \times Q_{1} = \frac{300}{500} \times 10 = 6 k c a l$

So, work done $= Q_{1} - Q_{2} = 10 - 6 = 4 k c a l$ .

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