A Carnot engine whose sinl is at `300K` has an efficiency of `40%`. By how much should the temperature of source be increased so as to increase its efficiency by `50%` of original efficiency.

Options

A. `380K`
B. `275K`
C. `325K`
D. `250K`

Answer

D

Explanation

Here, `T_(2)= 300K, eta_(1)= 40%`
`eta-(1)= 1-(T_(2))/(T_(1))`
`(40)/(100)= 10(300)/(T_(1)) :. T_(1)= 500K`
`eta_(2)= 40%+(50)/(100)xx40%= 60%`
`T_(2)= 300K, T_(1)=?`
As `eta_(2)= 1-(T'_(2))/(T'_(1))`
`:. (60)/(100)= 1-(300)/(T'_(1)) or T'_(1)=750K`.
Increase in temp. of source
`=T'_(1)-T_(1)= 750-500= 250K`