Answer:
from the steam table
we get to know the values
[tex]p_{1} =200Kpa\\x_{1} =0.25[/tex]
[tex]therefore\\v_{i} =v_{f} +x_{1} v_{fg} \\v_{i} =0.001061+(0.25)(0.88578-0.001061)\\v_{i} =0.22224m^3/kg[/tex]
we have
[tex]s_{i} =s_{f} +x_{1} s_{fg} \\s_{i} =1.5302+(0.250)(5.5968)=2.9294 Kj/kg.K[/tex]
to find s 2
[tex]v_{2} =v_{1} =0.22224m^3/kg\\x_{2} =1\\s_{2} =6.6335Kj/kg.K\\[/tex]
so the entropy change is
[tex]S=m(s_{2}- s_{1} )[/tex]
put values
[tex]S=(3kg)(6.6335-2.9294)Kj/kg.K\\S=11.1123Kj/K[/tex]
