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Question

A heat engine receives 50 kcal of heat from the source per cycle, and operates with an efficiency of 20%. Find the heat rejected to the sink per cycle.

A
50 kcal
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B
10 kcal
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C
42 kcal
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D
40 kcal
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Solution

The correct option is D 40 kcal
Heat supplied, Q1=50 kcal
Efficiency, η=20%
Using formula for efficiency,
η=WQ1
W=η×Q1
i.e., work obtained per cycle W=20 %×50 kcal=10 kcal
Or, W=10 kcal
From energy conservation,
Q1=Q2+W
Or, Q2=Q1W ...(i)
From Eq.(i) heat rejected to the sink per cycle is given by,
Heat rejected, Q2=50 kcal10 kcal
Q2=40 kcal

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