reversible. (Note: The
process is not externally reversible, since temperature is transferred across a finite
temperature difference).
In this case the following approach may be applied to find the entropy change
in the
system
:
2
1
2
1
2
1
gen
S
T
Q
S
(1)
Because the process is internally reversible, so
in the system
S
gen
= 0. The equation (1)
can be simplified to:
T
Q
S
2
1
2
1
(2)
In equation (2), T denotes the temperature of the system, since we are just considering
the entropy change in the system and not the total entropy generated throughout the
process (in which case we would use the temperature of the surroundings). (See
example 7-21 in the textbook for more information).
ΔS
1-2
= m·(s
2
- s
1
)
(3)
The relevant properties can be found on table A-4.
s
2
= s
g at 120
o
C
=
7.1292 kJ/kg·K
s
1
= s
f at 120
o
C
+ x
1
·s
fg at 120
o
C
= 1.5279 + 0.65 × 5.6013
kJ/kg·K
=
5.168745 kJ/kg·K
Put the values of
m
,
s
2
and
s
1
into the equation (3):
ΔS
1-2
= 8kg × (7.1292
–
5.168745)kJ/kg·K = 15.68364 kJ/K

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