A carnot engine operating between temperature `T_(1) andT_(2)` has efficiency `(1)/(6)`. When `T_(2)` is lowered by 62 K, its efficiency increases to `(1)/(3)`. Then find the values of `T_(1) and T_(2)`.

Explanation

`"Efficiency "eta=1-(T_(2))/(T_(1))`
`"In the first case : "1-(T_(2))/(T_(1))=(1)/(6) rArr (T_(2))/(T_(1))=(5)/(6)`
`T_(1)=(6)/(5)T_(2)`
`"In the second case :"1-((T_(2)-62))/(T_(1))=(1)/(3),(T_(2)-62)/(T_(1))=(2)/(3)`
`(T_(2)-62)/(((6)/(5))T_(2))=(2)/(3)`
`T_(2)-62=(2)/(3)xx(6)/(5)T_(2)=(4)/(5)T_(2)rArr T_(2)=310K`
`T_(1)=(6)/(5)xx310=372K`
`T_(1)=372K and T_(2)=210K.`