During an integral number of complete cycles a reversible carnot engine (shown by a cycle) absorbs 1200 joule from reservoir at 400 K and performs 200 Joule of mechanical work.
(a) Find the quantities of heat exchanged with the other two reservoirs. State whether the reservoirs absorb or lose heat.
(b) Find the change in entropy of each reservoir.
(c) What is the change in entropy of the universe?

Explanation

(a) By conservation of energy
`Q_(1)= W+Q_(2)+Q_(3)`
`i.e. Q_(2)+Q_(3)=Q_(1)-W=1200-200=1000`
And as change in entropy in a reversible-process is zero
`Q_(1)/T_(1)+Q_(2)/T_(2)+Q_(3)/T_(3)=0 i.e.-200/400+Q_(2)/300+Q_(3)/200=0`
i.e., 2Q+3Q=1800
Solving equation (i) and (ii) for `Q_(2)`and `Q_(3)` , we get
`Q_(2)+1200J and Q_(3)=-200J`
i.e. the reservoir at tempertature `T_(2)` absorbs 1200 J of heat while the reservoir at temperature `T_(3)` lose 200 J of heat
(b) Now as change in entropy at constant temperature is given by `DeltaS=(DeltaQ//T)`
So change in entropy of reservoir at temperatures `T_(1),T_(2)` and `T_(3)` will be respectively
`-1200/400=-3J/K,1200/300=4J/Kand (-200)/200=-1J/K`