The efficiency of carnot engine is 50% and temperature of sink is 500K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of the sink will be : -

Options

A. 600 K
B. 500 K
C. 400 K
D. 300 K

Answer

C

Explanation

For the heat engine, `eta=1-T_(2)/T_(1)`
`therefore1-500/T_(2)=1/2" "therefore500/T_(1)=1/2" "thereforeT_(1)=1000K`
In the second case, `eta=60/100=1-(T'_(2))/(1000)`
`therefore(T'_(2))/1000=1-0.6=0.4" "thereforeT'_(2)=1000xx0.4=400K`