Efficiency of a Carnot engine is `50%` when temperature of outlet is `500 K`. In order to increase efficiency up to `60%` keeping temperature of intake the same what is temperature of outlet?

Options

A. 200K
B. 400 K
C. 600 K
D. 800 K

Answer

B

Explanation

`eta=1-(T_2)/(T_1)`
`implies 1/2 =1-(500)/(T_1) implies T_1=1000K`
Now,
`(60)/(100)=1-(T_2)/(T_1) implies (T_2)/(1000)=1-(6)/(10)=(4)/(10)`
`implies T_2=400K`