## Question 8.87- Breaking phase changes up into steps [ENDORSED]

Vicky Wang 2C
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

### Question 8.87- Breaking phase changes up into steps

"How much heat is required to convert a 42.30 g block of ice at -5.042 degrees C into water vapor at 150.35 degrees C?"

I'm still a little confused as to why we have to break this up into the different stages (raising temp of ice, melting ice, raising temp of water, etc.). Also, the answer key says that for the first step:
Delta H = (42.30g) (2.03 J/degrees C * g) (0.00 degrees C - (5.042 degrees C)) = 0.433 kJ

Where did the (2.03 J/degrees C * g) come from? Thank you and sorry

JasmineAlberto4J
Posts: 73
Joined: Fri Sep 25, 2015 3:00 am

### Re: Question 8.87- Breaking phase changes up into steps

I don't know the answer to your other questions:(, but I do know that 2.03 is the specific heat capacity of ice

carrie_shih_3L
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

### Re: Question 8.87- Breaking phase changes up into steps

You have to split this into multiple parts because each phase requires a different equation.
When you graph out the heating curve, the sections where it is the slope (change of temperature, but no change in phase) you use the H = m * C * T formula. For the sections where it is flat (change of phase, but no change of temperature) you get multiply the moles by the heat of fusion or vaporization. You would look at table 7.3 for the numbers for this.

Posts: 62
Joined: Thu Jul 27, 2017 3:01 am

### Re: Question 8.87- Breaking phase changes up into steps

For step 2, I realized that we had to use dHfus= 6.01 kJ.mol^-1. Do we always use this value whenever we are melting ice and only when we are melting ice at 0 degrees C?

Jessica Benitez 1K
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am

### Re: Question 8.87- Breaking phase changes up into steps  [ENDORSED]

Doing a phase change and raising the temperature of the system require different equations and constant values. Breaking up the problem into different steps keeps everything organized because of the different equations and constant values you need to use for each step. It is easier to do the steps separately and add everything at the end rather than trying to combine every step especially if it is a longer problem.