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

B

Explanation

`eta=1-(T_2)/(T_1) implies 1/2=1-(500)/(T_1) implies 1/2` ..(i)
`60/100=1-(T'_2)/(T_1)implies (T'_2)/(T_1)=2/5` ..(ii)
Dividing equation (i) by (ii)
`500/(T'_2) = 5/4 implies T'_(2) = 400K`