Question

A Carnot engine whose
temperature of the source is 400
K takes 200 calories of
heat at this
temperature and it rejects 150
calories of heat to the sink?​

Answer 1

Heat of source, QH= 200 cal and temperature TH= 400 K, heat reject into sink QL= 150 cal and temperature is TL

Work done by the engine will be, W= QH - QL = 200 - 150= 50 cal

Efficiency of engine, Eff = W*100/QH= 50*100/200 = 25%

In terms of temperature Carnot engine efficiency we can estimate..

Eff = 1 - (TL/TH) after rearranging for TL we get..

TL = (1 - Eff)*TH = (1 - 0.25)*400 = 0.75*400 = 300K