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

Answer

C

Explanation

`eta=50%=(50)/(100)=(1)/(2)`
`eta_1=1-(T_2)/(T_1)`
`(1)/(2)=1-(500)/(T_1)impliesT_1=1000K`
`eta_2=60%=(60)/(100)=0.6`
`eta_2=1-(T_2')/(T_1)`
`0.6=1-(T_2')/(1000)impliesT_1'=600K`