Answer:
-6.326 KJ/K
Explanation:
A) the entropy change is defined as:
![delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}](https://tex.z-dn.net/?f=delta%20S_%7B12%7D%3D%5Cint%5Climits%5E2_1%20%20%5C%2C%20%5Cfrac%7BdQ%7D%7BT%7D)
In an isobaric process heat (Q) is defined as:
![Q= m*Cp*dT](https://tex.z-dn.net/?f=Q%3D%20m%2ACp%2AdT)
Replacing in the equation for entropy
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:
Solving the integral we get the expression to estimate the entropy change in the system
![delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})](https://tex.z-dn.net/?f=delta%20S_%7B12%7D%3D%20m%2ACp%20%2Aln%28%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%29)
The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is ![0.85\frac{kJ}{Kg*K}](https://tex.z-dn.net/?f=0.85%5Cfrac%7BkJ%7D%7BKg%2AK%7D)
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.
![Q= m*Cp*dT](https://tex.z-dn.net/?f=Q%3D%20m%2ACp%2AdT)
With
clearing for T2 we get:
![T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K](https://tex.z-dn.net/?f=T_%7B2%7D%3D%5Cfrac%7BQ%7D%7Bm%2ACp%7D%2BT1%3D%20%5Cfrac%7B-976.71kJ%7D%7B5.25Kg%2A0.85%5Cfrac%7BkJ%7D%7BKg%2AK%7D%7D%2B288.86%20K%20%3D69.98%20K)
Now we can estimate the entropy change in the system
![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}](https://tex.z-dn.net/?f=delta%20S_%7B12%7D%3D%20m%2ACp%2Aln%28%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%29%3D%205.25Kg%2A0.85%5Cfrac%7BkJ%7D%7BKg%2AK%7D%2Aln%28%5Cfrac%7B69.98%7D%7B288.861%7D%29%3D%20-6.326%5Cfrac%7BkJ%7D%7BK%7D)
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.