The efficiency of the carnot engine is `(1)/(6)`. On decreasing the temperature of the sink by 65 K, the efficiency increases to `(1)/(3)`. Find the temperatures of the source and the sink.

Explanation

We have,
`eta=(T_(2)-T_(1))/(T_(2))` where `T_(1)` and `T_(2)` are the temperatures of sink and source respectively
`:.eta=(T_(2)-T_(1))/(T_(2))=(1)/(6)`. ……………….(1)
Now the temperature of the sink is reduced by 65K.
`:.` temp of the sink `=(T_(1)-65)`
`:.eta=(T_(2)-(T_(1)-65))/(T_(2))=(1)/(3)`. ...............(2)
On solving Eqns. (1) and (2) we get
`T_(1)=325K`
`T_(2)=390K`