A pistonandcylinder device with a freefloating piston contains 2.6 kg of saturated water vapor at 150^{o}C. The water loses heat
to the surroundings
until the cylinder
contains saturated liquid
water. 
The surroundings are at 20^{o}C. Calculate… a.) ΔS_{water}, b.) ΔS_{surroundings}, c.) ΔS_{universe} 











Read : 
Calculating DS for
the water in the cylinder is straightforward. The
key to calculating DS_{surr} 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 DS_{univ} is the fact that the universe is made up of the combination of the system and the surroundings. Consequently, DS_{univ} = DS_{sys} + DS_{surr}. 










Given: 
m 
2.60 
kg 


Find: 
DS_{sys} 
??? 
kJ/K 

T_{1} 
150 
^{o}C 



DS_{surr} 
??? 
kJ/K 

x_{1} 
1 
kg vap/kg 



DS_{univ} 
??? 
kJ/K 

T_{surr} 
20 
^{o}C 







P_{1 }=
P_{2} 

^{} 







x_{2} 
0 
kg vap/kg 
















Diagram: 











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_{1} = 150^{o}C : 




S_{1} 
6.8371 
kJ/kgK 







S_{2} 
1.8418 
kJ/kgK 











Therefore : 





Eqn 1 

















DS_{sys} 
4.9953 
kJ/kgK 








12.988 
kJ/K 










Part b.) 
The surroundings
behave as a thermal reservoir. 

We can calculate DS_{surr} from: 





















Eqn 2 











We can determine Q_{sys} 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 T_{1} = 150^{o}C : 




H_{1} 
2745.9 
kJ/kg 







H_{2} 
632.18 
kJ/kg 











Next plug H_{1} and H_{2} into Eqn 6. Q_{surr} is equal
in magnitude, but opposite in sign to Q_{sys} because the heat leaving the system enters
the surroundings. 











Q_{sys} 
2113.7 
kJ/kg 



Q_{surr} 
2113.7 
kJ/kg 











Now, we can plug
numbers into Eqn 2 to
calculate DS_{surr}. 
DS_{surr} 
7.2105 
kJ/kgK 








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). 

















DS_{univ} 
5.7595 
kJ/K 











DS_{univ} > 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 : 
DS_{sys} 
12.988 
kJ/K 







DS_{surr} 
18.747 
kJ/K 







DS_{univ} 
5.7595 
kJ/K 















