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Question

In a Carnot heat engine, the temperature of source and sink are 827∘C and 27∘C. If the engine consumes 33 kJ per cycle, then the heat rejected to the sink per cycle is

A
11 kJ
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B
9 kJ
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C
24 kJ
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D
10 kJ
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Solution

The correct option is B 9 kJTemperature of source, T1=827∘C=1100 K Temperature of sink, (T2)=27∘C=300 K Heat absorbed, (Q1)=33 kJ Efficiency of the heat engine will be, η=1−T2T1=1−3001100 ∴η=811 From the fundamental definition of efficiency for any heat engine, η=WQ1 ⇒W=ηQ1 W=811×33×103 J=24 kJ Heat rejected to the sink, ⇒Q2=Q1−W ∴Q2=33 kJ−24 kJ=9 kJ ALTERNATIVE: For a Carnot heat engine, Q1Q2=T1T2 ⇒Q2=Q1×T2T1 ∴Q2=33×103×311=9 kJ

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