Answer:
-6.326 KJ/K
Explanation:
A) the entropy change is defined as:
[tex]delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}[/tex]
In an isobaric process heat (Q) is defined as:
[tex]Q= m*Cp*dT[/tex]
Replacing in the equation for entropy
[tex]delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}[/tex]
m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:
[tex]delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}[/tex]
Solving the integral we get the expression to estimate the entropy change in the system
[tex]delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})[/tex]
The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is [tex]0.85\frac{kJ}{Kg*K}[/tex]
We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K
The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.
[tex]Q= m*Cp*dT[/tex]
With [tex]dt=T_{2}-T_{1}[/tex] clearing for T2 we get:
[tex]T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K[/tex]
Now we can estimate the entropy change in the system
[tex]delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}[/tex]
The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.
b) see picture.
