Physics, asked by sahilgod6072, 1 year ago

A heat engine operates between a source at 550◦c and sink at 25◦c. if the heat is supplied to the heat engine at the rate of 1200 kj/min, determine maximum power output of this heat engine.

Answers

Answered by Simran1796

2

Efficiency (₩) =work done/heat=W/Q
Also ₩=1- (T2/T1)
Therefore W/Q=1 -(T2/T1)
W/1200=525/550
OR W=1,145.45

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