Question

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Answer 1

Answer: The change in entropy of the given process is 1324.8 J/K

Step-by-step explanation:

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)`

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

`\Delta S=m* C_(p,m)* \ln ((T_2)/(T_1))` .......(1)

where,

`\Delta S` = Entropy change

`C_(p,m)` = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

`T_2` = final temperature

`T_1` = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

`\Delta S=m* (\Delta H_(f,v))/(T)` .......(2)

where,

`\Delta S` = Entropy change

m = mass of ice

`\Delta H_(f,v)` = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

• For process 1:

We are given:

`m=150g\\C_(p,s)=2.06J/gK\\T_1=255K\\T_2=273K`

Putting values in equation 1, we get:

`\Delta S_1=150g* 2.06J/g.K* \ln((273K)/(255K))\\\\\Delta S_1=21.1J/K`

• For process 2:

We are given:

`m=150g\\\Delta H_(fusion)=334.16J/g\\T=273K`

Putting values in equation 2, we get:

`\Delta S_2=(150g* 334.16J/g)/(273K)\\\\\Delta S_2=183.6J/K`

• For process 3:

We are given:

`m=150g\\C_(p,l)=4.184J/gK\\T_1=273K\\T_2=373K`

Putting values in equation 1, we get:

`\Delta S_3=150g* 4.184J/g.K* \ln((373K)/(273K))\\\\\Delta S_3=195.9J/K`

• For process 4:

We are given:

`m=150g\\\Delta H_(vaporization)=2259J/g\\T=373K`

Putting values in equation 2, we get:

`\Delta S_2=(150g* 2259J/g)/(373K)\\\\\Delta S_2=908.4J/K`

• For process 5:

We are given:

`m=150g\\C_(p,g)=2.02J/gK\\T_1=373K\\T_2=393K`

Putting values in equation 1, we get:

`\Delta S_5=150g* 2.02J/g.K* \ln((393K)/(373K))\\\\\Delta S_5=15.8J/K`

Total entropy change for the process =
`\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5`

Total entropy change for the process =
`[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K`

Hence, the change in entropy of the given process is 1324.8 J/K