Entropy as three steps

[alicechien_4F]

For 4.17, it asks us "Calculate the standard entropy of vaporization of water at 85 degrees C, given that its standard entropy of vaporization at 100 degrees C is 109.0 J/Kmol and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J/Kmol and 33.6 J/Kmol, respectively, in this range."

In the solutions, entropy is broken up into three steps: heating, vaporizing, and then cooling. Can someone explain why you have to cool the product back down at 85C if you are vaporizing it?

[AArmellini_1I]

You have to cool it because in order to vaporize you must be at 100C. So first you have to increase the temperature to 100, then vaporize, then reduce the temp back down to 85degrees in order to satisfy the question

[Sebastian Lee 1L]

I also had a question on this problem. I get how to do the problem using the entropy equations when temperature changes. But why is it that you can cool the water vapor back down to 85 Centigrade and still have it be in the vapor state? I assume it's in the vapor state since the molar heat capacity for the vapor is used to calculate the entropy.