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

Answer

C

Explanation

The efficiency of Carnot engine is defined as th ratio of work done to the heat supplied i.e.
`eta=("Word 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 temperature of sink.
As given `eta=40%=40/100=0.4`
adn `T_(2)=300K`
so `0.4=1-300/(T_(1))`
`impliesT_(1)=300/(1-0.4)=300/0.6=500K`
Let temperature of the source be increased by x K, then efficiency becomes `eta.=40%+50%` of `eta`
`=40/100+50/100xx0.4=0.4+0.5xx0.4=0.6`
Hence `=0.6=1-300/(500+x)`
`implies300/(500+x)=0.4`
`implies500+x=300/0.4=750`
`x=750-500=250K`