A carnot engine absorbs `1000J` of heat energy from a reservoir at `127^(@)C` and rejecs `600J` of heat energy during each cycle. Calculate (i) efficiency of the engine, (ii) temperature of sink, (iii) amount of useful work done per cycle.

Explanation

Here, `Q_(1)= 1000J, Q_(2)=600J`
`T_(1)= 127^(@)C= 127+273= 400K`,
`eta=? T_(2)=? W=?`
From `(Q_(2))/(Q_(1))= (T_(2))/(T_(1)), T_(2)= (Q_(2))/(Q_(1))xxT_(1)`
`T_(2)= (600)/(1000)xx400= 240K= 240-273= -33^(@)C`
`eta=1 -(T_(2))/(T_(1))=1 -(240)/(400)= 0.4= 40%`
Also, `W=Q_(1)-Q_(2)= 1000-600= 400J`