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

Answer

C

Explanation

`eta = 1 -(T_(2))/T_(1)implies 1-(300)/(T_(1)) =0.4 implies T_(1) =500K`
now `eta in=0.4+0.4xx(50)/(100)=0.6`

therefore 0.6=1-`(300)/(500+DeltaT)`
implies 500 +`DeltaT=750impliesDeltaT=250K`