A carnot engine operates between `227^(@)C and 127^(@)C`. It absorbs 80 kilocalories of heat from the source. What is the work done in joule ?

Options

A. `4.5xx10^(4)J`
B. `5.3xx10^(4)J`
C. `6.72xx10^(4)J`
D. `7.32xx10^(4)J`

Answer

C

Explanation

Heat converted into work (W) = `Q_(1)-Q_(2)`
But `Q_(1)/Q_(2)=T_(2)/T_(1)=400/500=4/5`
`thereforeQ_(2)=4/5,Q_(1)=4/5xx80=64`Kilocalorie
`thereforeW=80-64=16`Kilocalorie (in Heat units)
`W=JH=4200J/("Kilocalorie")xx16"Kilocalories"`
`=67200J=6.72xx10^(4)J`