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A Carnot engine of efficiency 40% , take...

A Carnot engine of efficiency 40% , takes heat from a source maintained at a temperature of 500 K . If is desired to have an engine of efficiency 60% . Then , the source temperature for the same sink temperature must be

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`eta=1-(T_(2))/(T_(1)),0.4=1-(T_(2))/(500)rArr T_(2)=300K`
`0.6=1-(T_(2))/(T_(1)^(1))=1-(300)/(T_(1)^(1))rArr T_(1)^(1)=(300)/(0.4)=750K`
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