The efficiency of Carnot's enegine is `50%`. The temperature of its sink is `7^(@)C`. To increase its efficiency to `70%`. What is the increase in temperature of the source?

Explanation

Efficiency in first state
`eta = 50% =1//2`
`T_(2) =273 + 7= 280 K`
Formula
`eta =1 - (T_(2))/(T _(1))`
`1/2 =1 - (280)/(T_(1))implies (280)/(T_(1)) =1/2`
or `T_(1) =560^(@)K` (temperatue of source)