During an integral number of complete cycles, a reversible engine (shown by a circle) absorbs $1200$ joule from reservoir at $400 K$ and performs $200$ joule of mechanical work.
(a) Find the quantities of heat exchanged with the second reservoir.

(b) Find the change of entropy in second reservoir (in $J / K$ ).
(c) What is the change in entropy of the universe?

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Solution

(a)The engine absorbs the heat from the first resevoir at $T_{1} = 400 K$ .
Let us assume that the other two resevoirs are acting as sink.(if our assumption is wrong, then will will get thier values as negative)
Using conservation of energy,
$Q_{s o u r c e} = W + Q_{s i n k}$
$Q_{1} = W + Q_{2} + Q_{3}$
$Q_{2} + Q_{3} = 1200 - 200 = 1000 J$ ....(i)

Since it is a reversible engine the net change in entropy is zero.
$Δ S_{1} + Δ S_{2} + Δ S_{2} = 0$
$\frac{Q_{1}}{T_{1}} + \frac{Q_{2}}{T_{2}} + \frac{Q_{3}}{T_{3}} = 0$
$\frac{Q_{2}}{300} + \frac{Q_{3}}{200} = \frac{1200}{400} = 3$
$2 Q_{2} + 3 Q_{3} = 1800$ ...(ii)

Solving (i) and (ii), we get
$Q_{2} = 1200 J$ and $Q_{3} = - 200 J$
The negative sign of the third reservoir indicates that it is acting as a source and not as sink.

(b) Change in entropy, $Δ S = \frac{Δ Q}{T}$
$Δ S_{1} = \frac{Δ Q_{1}}{T_{1}} = \frac{- 1200}{400} = - 3 J / K$
$Δ S_{2} = \frac{Δ Q_{2}}{T_{2}} = \frac{1200}{300} = 4 J / K$
$Δ S_{3} = \frac{Δ Q_{3}}{T_{3}} = \frac{- 200}{200} = - 1 J / K$

(c)The net change in the entropy of Universe (any reversible process) $= 0$

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