Example Problem with Complete Solution

7B-3 : Entropy Change of an Isobaric Process 6 pts
A piston-and-cylinder device with a free-floating piston contains 2.6 kg of saturated water vapor at 150oC. The water loses heat to the surroundings until the cylinder contains saturated liquid water. 
The surroundings are at 20oC. Calculate… a.) ΔSwater, b.) ΔSsurroundings, c.) ΔSuniverse
Read : Calculating DS for the water in the cylinder is straightforward. The key to calculating DSsurr is the fact that the surroundings behave as a thermal reservoir. The temperature of the surroundings does not change and there are no irreversibilities in the surroundings that are associated with the process.  The key to calculating DSuniv is the fact that the universe is made up of the combination of the system and the surroundings. Consequently, DSuniv = DSsys + DSsurr.
Given: m 2.60 kg Find: DSsys ??? kJ/K
T1 150 oC DSsurr ??? kJ/K
x1 1 kg vap/kg DSuniv ??? kJ/K
Tsurr 20 oC
P1 = P2
x2 0 kg vap/kg
Assumptions: 1 - Changes in kinetic and potential energies are negligible.
2 - Boundary work is the only form of work that crosses the system boundary.
3 - The surroundings behave as a thermal reservoir.
Equations / Data / Solve:
Part a.) Because both states are saturated we can obtain the specific entropies directly from the Steam Tables or the NIST Webbook.
At T1 = 150oC : S1 6.8371 kJ/kg-K
S2 1.8418 kJ/kg-K
Therefore :
Eqn 1
DSsys -4.9953 kJ/kg-K
-12.988 kJ/K
Part b.) The surroundings behave as a thermal reservoir.
We can calculate DSsurr from:
Eqn 2
We can determine Qsys by applying the 1st Law using the water within the cylinder as the system.
Eqn 3
Eqn 4
Because the process is isobaric, the boundary work is :
Eqn 5
Now, substitute Eqn 5 into Eqn 4 to get :
Eqn 6
We can look up enthalpy values for states 1 & 2 in the Saturated Steam Table or in the NIST Webbook.
At T1 = 150oC : H1 2745.9 kJ/kg
H2 632.18 kJ/kg
Next plug H1 and H2 into Eqn 6.  Qsurr is equal in magnitude, but opposite in sign to Qsys because the heat leaving the system enters the surroundings.
Qsys -2113.7 kJ/kg Qsurr 2113.7 kJ/kg
Now, we can plug numbers into Eqn 2 to calculate DSsurr. DSsurr 7.2105 kJ/kg-K
18.747 kJ/K
Part c.) The universe is made up of the combination of the system and the surroundings.  Therefore :
Eqn 7
So, all we need to do is plug values into Eqn 7 that we determined in parts (a) and (b).
DSuniv 5.7595 kJ/K
DSuniv > 0 because the heat transfer to the surroundings was not reversible.
Verify: None of the assumptions made in this problem solution can be verified.
Answers : DSsys -12.988 kJ/K
DSsurr 18.747 kJ/K
DSuniv 5.7595 kJ/K