The efficiency of a Carbot's heat engine is `0.25`. If on reducing the temperature of the sink by `50^(@)C`, the efficiency increase, `0.35`, then calculate the temperature of the source and sink.

Explanation

`eta=1-(T_(2))/(T_(1)) " " i., (T_(2))/(T_(1))=1-0.25=0.75=(3)/(4)" " :. T_(2)=((3)/(4))T_(2)`
Also `0.35=1-((T_(2)-50)/(T_(1)))=(T_(1)T_(2)+50)/(T_(1))`
i.e. `0.35 T_(1)=T_(1)+50-T_(2) or T_(2)-50=0.65T_(1)=0.65((4)/(3))T_(2)`
and `T_(1)=1.333xx375.1=500K`
`T_(2)(1-0.8667)=50,`
`T_(2)=375.1K`