<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:

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:

Putting values in equation 1, we get:

We are given:

Putting values in equation 2, we get:

We are given:

Putting values in equation 1, we get:

We are given:

Putting values in equation 2, we get:

We are given:

Putting values in equation 1, we get:

Total entropy change for the process = 
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