<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K
<u>Explanation:</u>
The processes involved in the given problem are:
![1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)](https://tex.z-dn.net/?f=1.%29H_2O%28s%29%28-18%5EoC%2C255K%29%5Crightarrow%20H_2O%28s%29%280%5EoC%2C273K%29%5C%5C2.%29H_2O%28s%29%280%5EoC%2C273K%29%5Crightarrow%20H_2O%28l%29%280%5EoC%2C273K%29%5C%5C3.%29H_2O%28l%29%280%5EoC%2C273K%29%5Crightarrow%20H_2O%28l%29%28100%5EoC%2C373K%29%5C%5C4.%29H_2O%28l%29%28100%5EoC%2C373K%29%5Crightarrow%20H_2O%28g%29%28100%5EoC%2C373K%29%5C%5C5.%29H_2O%28g%29%28100%5EoC%2C373K%29%5Crightarrow%20H_2O%28g%29%28120%5EoC%2C393K%29)
Pressure is taken as constant.
To calculate the entropy change for same phase at different temperature, we use the equation:
.......(1)
where,
= Entropy change
= specific heat capacity of medium
m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)
= final temperature
= initial temperature
To calculate the entropy change for different phase at same temperature, we use the equation:
.......(2)
where,
= Entropy change
m = mass of ice
= enthalpy of fusion of vaporization
T = temperature of the system
Calculating the entropy change for each process:
We are given:
![m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K](https://tex.z-dn.net/?f=m%3D150g%5C%5CC_%7Bp%2Cs%7D%3D2.06J%2FgK%5C%5CT_1%3D255K%5C%5CT_2%3D273K)
Putting values in equation 1, we get:
![\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K](https://tex.z-dn.net/?f=%5CDelta%20S_1%3D150g%5Ctimes%202.06J%2Fg.K%5Ctimes%20%5Cln%28%5Cfrac%7B273K%7D%7B255K%7D%29%5C%5C%5C%5C%5CDelta%20S_1%3D21.1J%2FK)
We are given:
![m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K](https://tex.z-dn.net/?f=m%3D150g%5C%5C%5CDelta%20H_%7Bfusion%7D%3D334.16J%2Fg%5C%5CT%3D273K)
Putting values in equation 2, we get:
![\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K](https://tex.z-dn.net/?f=%5CDelta%20S_2%3D%5Cfrac%7B150g%5Ctimes%20334.16J%2Fg%7D%7B273K%7D%5C%5C%5C%5C%5CDelta%20S_2%3D183.6J%2FK)
We are given:
![m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K](https://tex.z-dn.net/?f=m%3D150g%5C%5CC_%7Bp%2Cl%7D%3D4.184J%2FgK%5C%5CT_1%3D273K%5C%5CT_2%3D373K)
Putting values in equation 1, we get:
![\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K](https://tex.z-dn.net/?f=%5CDelta%20S_3%3D150g%5Ctimes%204.184J%2Fg.K%5Ctimes%20%5Cln%28%5Cfrac%7B373K%7D%7B273K%7D%29%5C%5C%5C%5C%5CDelta%20S_3%3D195.9J%2FK)
We are given:
![m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K](https://tex.z-dn.net/?f=m%3D150g%5C%5C%5CDelta%20H_%7Bvaporization%7D%3D2259J%2Fg%5C%5CT%3D373K)
Putting values in equation 2, we get:
![\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S_2%3D%5Cfrac%7B150g%5Ctimes%202259J%2Fg%7D%7B373K%7D%5C%5C%5C%5C%5CDelta%20S_2%3D908.4J%2FK)
We are given:
![m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K](https://tex.z-dn.net/?f=m%3D150g%5C%5CC_%7Bp%2Cg%7D%3D2.02J%2FgK%5C%5CT_1%3D373K%5C%5CT_2%3D393K)
Putting values in equation 1, we get:
![\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K](https://tex.z-dn.net/?f=%5CDelta%20S_5%3D150g%5Ctimes%202.02J%2Fg.K%5Ctimes%20%5Cln%28%5Cfrac%7B393K%7D%7B373K%7D%29%5C%5C%5C%5C%5CDelta%20S_5%3D15.8J%2FK)
Total entropy change for the process = ![\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5](https://tex.z-dn.net/?f=%5CDelta%20S_1%2B%5CDelta%20S_2%2B%5CDelta%20S_3%2B%5CDelta%20S_4%2B%5CDelta%20S_5)
Total entropy change for the process = ![[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K](https://tex.z-dn.net/?f=%5B21.1%2B183.6%2B195.9%2B908.4%2B15.8%5DJ%2FK%3D1324.8J%2FK)
Hence, the change in entropy of the given process is 1324.8 J/K