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Numerical · Q22

Q.A heat engine absorbs 2000 J2000\ \text{J} of heat from its source and rejects 1500 J1500\ \text{J} of heat to its sink in each cycle. Calculate the work done by the engine per cycle and its efficiency.

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Given: heat absorbed from the source Q1=2000 JQ_1 = 2000\ \text{J}, heat rejected to the sink Q2=1500 JQ_2 = 1500\ \text{J}, per cycle.

Work done per cycle:

W=Q1−Q2=2000−1500=500 JW = Q_1 - Q_2 = 2000 - 1500 = 500\ \text{J}

Efficiency:

η=WQ1=5002000=0.25\eta = \frac{W}{Q_1} = \frac{500}{2000} = 0.25 …

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