A reversible engine converts one fifth of heat which it absorbs from source into work. When the temp. of sink is reduced by `70^(@)C`, its efficiency is doubled. Calculate the temperature of the source and sink.

Answer

`350K ; 280K`

Explanation

Here, `W=1/5Q_(1)`,
`Q_(2)=Q_(1)-W=Q_(1)-1/5Q_(1)=4/5Q_(1)`
`:. (Q_(2))/(Q_(1))=4/5`
As `(T_(2))/(T_(1))=(Q_(2))/(Q_(1)) :. (T_(2))/(T_(1))=4/5`
`eta= 1-(T_(2))/(T_(1))= 1 -4/5= 1/5`
On reducing the temp. of sink by `70^(@)C`, efficiency becomes double.
`:. 2eta= 1-((T_(2)-70)/(T_(1))) or 2xx1/5=1-(T_(2))/(T_(1))+(70)/(T_(1))`
`=1-4/5+(70)/(T_(1))=1/5+(70)/(T_(1))`
`(70)/(T_(1))=2/5-1/5=1/5`
`T_(1)= 350K`
As `(T_(2))/(T_(1))=4/5, T_(2)=4/5T_(1)=4/5xx350=280K`