An ideal engine operates by taking in steam from a boiler at `327^(@)C` and rejecting heat to a sink at `27^(@)C`. The engine runs at `500 rpm` and heat taken is `600 k cal` in each revolution. Calculate(i) efficiency of engine (ii) work done in each cycle (iii) heat rejected in each revolution and (iv) power output of engine.

Answer

`50% ; 1.26xx10^(6)J ; 300 K cal; 1.05xx10^(4)kW`

Explanation

Here, `327^(@)C= (327+273)K= 600K`,
`T_(2)=27^(@)C= (27+273)= 300K`
`n=500 rpm= (500)/(60)rps, Q_(1)= 600kcal`
(i) `eta= 1-(T_(2))/(T_(1))= 1-(300)/(600)=0.5= 50%`
(ii) `eta=W/(Q_(1))`
`:. =etaQ_(1)= 0.5xx600= 300Kcal`
`= 300xx4.2xx10^(3)J=1.26xx10^(6)J`
(iii) `Q_(2)=Q_(1)-W=600 kcal-300kcal`
`=300kcal`,
(iv) `"Power output" = ("work")/("time")= (500)/(60)xx1.26xx10^(6)J//s`
`=1.05xx10^(4)KW`