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Q. A Carnot engine takes heat from a reservoir at $ 627{}^\circ C $ and rejects heat to a sink at $ 27{}^\circ C $ . Its efficiency will be

JamiaJamia 2008

Solution:

$ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}} $ $ \therefore $ $ \eta =1-\frac{(27+273)}{(273+627)} $ $ =1-\frac{300}{900}=\frac{600}{900}=\frac{2}{3} $