Efficiency of a carnot engine is `0.4`, when temp. of sink is `300K`. What is temp. of source? If temp. of source is kept same and that of sink is lowered by `50^(@)C`, what would be the efficiency?

Answer

`500K ; 0.5`

Explanation

Here, `eta=0.4, T_(2)= 300K, T_(1)=?`
`eta= 1-(T_(2))/(T_(1)) , (T_(2))/(T_(1))= 1-eta=1-0.4= 0.6`,
`T_(1)=(T_(2))/(0.6)= (300)/(0.6)= 500K`
`eta'= 1-(T'_(2))/(T_(1))=1 - ((300-50))/(500)= 0.5`