The temperature of sink of Carnot engine is `27^(@)C`. Efficiency of engine is `25%`. Then temeperature of source is

Options

A. `127^(@)C`
B. `227^(@)C`
C. `327^(@)C`
D. `27^(@)C`

Answer

A

Explanation

`eta=1-T_(2)/T_(1)=(T_(1)-T_(2))/T_(1)`
`therefore25/100=1/4=(T_(1)-T_(2))/T_(1)`
`therefore4T_(1)-4T_(2)=T_(1)`
`therefore3T_(1)=4T_(2)=4xx(273+27)=1200`
`thereforeT_(1)=400K=127^(@)C`