A Carnot engine whose sink 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

The efficiency of carnot engine is defined as the ratio of work done to the heat supplied i.e.,
`eta = ("Work done")/("Heat Supplied") = (W)/(Q_1)=(Q_(1)-Q_(2))/(Q_1)`
`=1-(Q_2)/(Q_1)=1-(T_2)/(T_1)`
Here, `T_(1)` is the temperature of source and `T_(2)` is the temperatrue of the sink.
As given `eta = 40%=40/100=0.4`
and `T_(2) = 300K`
so, `0.4=1-(300)/(T_1)`
`T_(1)=(300)/(1-0.4)=300/0.6=500K`
Let temeperature of the source be increased by `x K`, then effeciency becomes
`eta^(')=40%+50% or eta`
`=40/100+50/100xx0.4+0.5xx0.4=0.6`
`hence, 0.6=1-(300)/(500+x)`
`(300)/(500+x)=0.4`
`500+x = 300/0.4 = 750`
`x=750-500=250K`.