A Carnot engine, whose efficiency is `40%`, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency `60%`. Then, the intake temperature for the same exhaust (sink) temperature must be:

Options

A. 1200 K
B. 750 K
C. 600 K
D. efficiency of carnot engine cannot be made larger than 50%

Answer

B

Explanation

For a carnot engine, `eta=1-T_(2)/T_(1)`
In the first case, `40/100=1-T_(2)/500`
`therefore6/10=T_(2)/500" "thereforeT_(2)=500xx6/100=300K`
In the second case, `6/10=1-300/(T'_(1))`
`therefore0.4=300/(T'_(1))" "thereforeT'_(1)=300/0.4=750K`