Example Problem with Complete Solution

7B-3 : Entropy Change of an Isobaric Process 6 pts
Consider a process in which 1.00 kg of saturated water vapor at 100oC is condensed to a saturated liquid in an isobaric process by heat transfer to the surrounding air, which is at 25oC.  What is the change in entropy of the water ?  What is the change in entropy of the surroundings ?  What is the change in the entropy of the universe ?
Read : Calculating ΔS for the water in the cylinder is straightforward.  The key to calculating ΔSsurr 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, ΔSuniv = ΔSsys + ΔSsurr.
Given : m 1.00 kg Find :
T1 100 oC ΔSsys ??? kJ/K
x1 1 ΔSsurr ??? kJ/K
Tsurr 25 oC ΔSuniv ??? kJ/K
P1 = P2
x2 0
Diagram : Figure 1
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 T = 100oC : S1 7.3541 kJ/kg-K
S2 1.3072 kJ/kg-K
Therefore : Equation 1
Eqn 1
ΔSsys -6.0469 kJ/kg-K
-6.0469 kJ/K
Part b.) The surroundings behave as a thermal reservoir.
We can calculate DSsurr from:
Equation 2
Eqn 2
We can determine Qsys by applying the 1st Law using the gas within the cylinder as the system.
Equation 3
Eqn 3
Equation 4 Eqn 4
Because the process is isobaric, the boundary work is :
Equation 5
Eqn 5
Now, substitute Eqn 5 into Eqn 4 to get :
Equation 6
Eqn 6
We can look up enthalpy values for states 1 & 2 in the Saturated Steam table or in the NIST Webbook.
At T = 100oC : H1 2675.6 kJ/kg
H2 419.17 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 -2256.43 kJ/kg Qsurr 2256.43 kJ/kg
Now, we can plug numbers into Eqn 2 to calculate ΔSsurr.
ΔSsurr 7.5681 kJ/kg-K
7.5681 kJ/K
Part c.) The universe is made up of the combiation of the system and the surroundings.  Therefore :
Equation 7
Eqn 7
So, all we need to do is plug values into Eqn 7 that we determined in parts (a) and (b).
ΔSuniv 1.5212 kJ/K
ΔSuniv > 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: ΔSsys -6.0469 kJ/K
ΔSsurr 7.5681 kJ/K
ΔSuniv 1.5212 kJ/K