An ideal Carnot's engine whose efficiency 40% receives heat of 500K. If the efficiency is to be 50% then the temperature of sink will be

Options

A. 900 K
B. 800 K
C. 700 K
D. 600 K

Answer

D

Explanation

Given : `eta=0.4andT_(1)=500K`
For the Carnot engine, `eta=40%=0.4`
`becauseeta=1-T_(2)/T_(1)" "thereforeT_(2)/T_(1)=1-eta=1-0.4=0.6`
`thereforeT_(2)/500=0.6" "thereforeT_(2)=300K`
In the second case, `T_(2)/T_(1)=1-0.5=0.5`
`thereforeT_(1)=T_(2)/0.5=300/0.5=600K`