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Question:

A Carnot engine operates with sources at 127°C and sink at 27°C. If the source supplies 40 kJ of heat energy, the work done by the engine is?

30 kJ

10 kJ

4 kJ

1 kJ

Solution:

Efficiency, η=1−T₂/T₁
∴η=1−(273+27)/(273+127)=1−300/400=1/4
Also η=Work done by engine/Heat supplied by source=W/40 kJ
∴W=40η=40×1/4=10 kJ.