In a Carnot heat engine, the temperature of source and sink are 500K and 375K. If the engine consumes 25×105J per cycle, then work done by engine per cycle is
A
6.25×105J
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B
100×105J
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C
2.5×105J
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D
0.625×105J
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Solution
The correct option is A6.25×105J Temperature of source, T1=500K
Temperature of sink, T2=375K
Heat absorbed, Q1=25×105J
Efficiency of a Carnot's heat engine, η=WQ1=1−T2T1 ⇒ Workdone per cycle is, W=Q1(1−T2T1) W=25×105(1−375500) ∴W=25×105×0.25=6.25×105J