A Carnot engine whose sink is at 300 K 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. 380 K
B. 275 K
C. 325 K
D. 250 K

Answer

D

Explanation

`eta =1-(T_(2))/(T_(1)) rArr (40)/(100)=1-(300)/(T_(1)) rArr T_(1)=500 K`
New efficiency.
`eta. =eta +(50)/(100) cdot eta =40% +(50)/(100)xx40% =60%`
`therefore eta.=1-(T_(2))/(T_(1)) rArr (60)/(100)=1-(300)/(T_(1))w`
`rArr T_(1) =750 K`
`therefore` Increase in temperature of source
`=T_(1).-T_(1)=750-500=250 K`.
`therefore` Correct choice is (d).