A heat engine is haivng a source at temperature `527^(@)C` and sink at temperature `127^(@)C` . If the useful work is required to be done by the engine at the rate of 750 watt, then the amount of heat absorbed per second from the source in calories and the efficiency of heat engine are:

Options

A. `482.2cals^(-1),50%`
B. `482.2cals^(-1),25%`
C. `357.14cals^(-1),50%`
D. `357.14cals^(-1),25%`

Answer

C

Explanation

Given `T_(1)=800K` and `T_(2)=400K`
`W=Q_(1)-Q_(2)=750W`
Amount of heat absorbed is given by
`W/(Q_(1))=(T_(1)-T_(2))/(T_(1))=(800-400)/800=1/2`
`implies750/(Q_(1))=1/2`
`impliesQ_(1)=1500`Watt. OrJ `sec^(-1)`
`Q_(1)=1500/4.2=357.14calsec^(-1)`
Efficience `eta=(T_(1)-T_(2))/T=(800+40)/000=1/2` or 50%