During an integral number of complete cycles a reversible carnot engine (shown by a cycle) absors 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 thether 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

Heat absorbed by working substance
`Q_1=1200J`
work done by engine, `W=200J`
Heat rejected to sinks
`Q_2+Q_3=Q_1-W=1200-200=1000`
According to property of reversible carnot's cycle
`(sum(Q)/(T))_(sources)=(sum(Q)/(T))_(si nks)`
`(Q_1)/(T_1)=(Q_2)/(T_2)+(Q_3)/(T_3)`
`(1200)/(400)=(Q_2)/(300)+(Q_3)/(200)`
`(Q_2)/(3)+(Q_2)/(2)=300`
`2Q_2+3Q_3=1800` .(ii)
Solving (i) and (ii) `Q_2=1200J,Q_3=-200J`
As `Q_3` is negative i.e., working substance takes heat `Q_3` from the reservoir of temperature `T_3`.
Heat rejected to reservoir at temperature `T_2=1200J`
Heat absorbed from reservoir at temperature `T_3=200J`
(b) Change in entropy:
First reservoir `DeltaS_1=(Q_1)/(T_1)=(-1200)/(400)=-3(J)/(K)`
Second reservoir `DeltaS_2=(Q_2)/(T_2)=(1200)/(300)=4(J)/(K)`
Third reservoir `DeltaS_3=(Q_3)/(T_3)=(-200)/(200)=-1(J)/(K)`
(c) Change in entropy of universe
`DeltaS=sum(Q)/(T)=DeltaS_1+DeltaS_2+DeltaS_3=-3+4-1=0`