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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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Answer» Solution :`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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