Respuesta :
Answer: The change in entropy of the given process is 1324.8 J/K
Explanation:
The processes involved in the given problem are:
[tex]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)[/tex]
Pressure is taken as constant.
To calculate the entropy change for same phase at different temperature, we use the equation:
[tex]\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})[/tex] .......(1)
where,
[tex]\Delta S[/tex] = Entropy change
[tex]C_{p,m}[/tex] = specific heat capacity of medium
m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)
[tex]T_2[/tex] = final temperature
[tex]T_1[/tex] = initial temperature
To calculate the entropy change for different phase at same temperature, we use the equation:
[tex]\Delta S=m\times \frac{\Delta H_{f,v}}{T}[/tex] .......(2)
where,
[tex]\Delta S[/tex] = Entropy change
m = mass of ice
[tex]\Delta H_{f,v}[/tex] = enthalpy of fusion of vaporization
T = temperature of the system
Calculating the entropy change for each process:
- For process 1:
We are given:
[tex]m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K[/tex]
Putting values in equation 1, we get:
[tex]\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K[/tex]
- For process 2:
We are given:
[tex]m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K[/tex]
Putting values in equation 2, we get:
[tex]\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K[/tex]
- For process 3:
We are given:
[tex]m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K[/tex]
Putting values in equation 1, we get:
[tex]\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K[/tex]
- For process 4:
We are given:
[tex]m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K[/tex]
Putting values in equation 2, we get:
[tex]\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K[/tex]
- For process 5:
We are given:
[tex]m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K[/tex]
Putting values in equation 1, we get:
[tex]\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K[/tex]
Total entropy change for the process = [tex]\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5[/tex]
Total entropy change for the process = [tex][21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K[/tex]
Hence, the change in entropy of the given process is 1324.8 J/K
