A carnot engine absorbs `2000J` of heat from the source of heat engine at `227^(@)C` and rejects `1200J` of heat to the sink during each cycle. Calculate
(i) Temp.of sink (ii) efficiency of engine
(iii) amount of work done during each cycle.

Answer

`27^(@)C ; 40% ; 800J`

Explanation

Here, `Q_(1)= 2000J, T_(1)=227^(@)C=(227+273)K= 500K`
`Q_(2)= 1200J, T_(2)=? Eta=? W=?`
As `(Q_(2))/(Q_(1))=(T_(2))/(T_(1)) :. T_(2)=(Q_(2))/(Q_(1))xxT_(1)`
`=(1200)/(2000)xx500= 300K`
`=(300-273)^(@)C`
`27^(@)C`
(ii) `eta=1 -(Q_(2))/(Q_(1))= 1- (1200)/(2000)`
`=2/5=2/5xx100%= 40%`
(iii) `W=Q_(1)-Q_(2)`
`=2000-1200= 800J`