A reversible engine takes in heat from a reservoir of heat at `527^(@)C`, and gives out to the sink at `127^(@)C` How many calories per second must it take from the reservoir in order to produce useful mechanical work at the rate of `750 wat t`? `1 cal. = 4.2J`.

Answer

`357.1 cal//s`

Explanation

Here, `T_(1)= 527^(@)C= 800K, T_(2)= 127^(@)C= 400K`
`Q_(1)=?, W=750 "wa tt"= 750 "joule"//sec`
Efficiency, `eta=1 - (T_(2))/(T_(1))=1 - (400)/(800)= 1/2`
As `eta=W/(Q_(1)) :. Q_(1)=W/(eta)=(750//4.2)/(1//2)=(1500)/(4.2)`
`=357.1 cal//s`