Problem with phase changes.?

Question

Can someone show me step by step how to work this problem:
The heat of fusion of water is 80 cal/g, the heat of vaporization of water is 540 cal/g, the specific heat of ice is .48 cal/deg/g, the specific heat of steam is .48 cal/deg/g, and the specific heat of liquid water is 1.0 cal/deg/g. How much heat would be needed to convert 12.37 g of ice at -35 degrees C to steam at 146 degrees C.

Answer 1

It's been awhile since i've done this but i'll give it my best

First off, the first energy you need to calculate is the energy needed to raise the temperature of ice to 0 degrees C. so...

(12.37 g)(.48 cal/deg/g)(35 deg) = +207.816 cal

When the ice changes to water you have a positive heat of fusion. So...

(12.37 g)(80 cal/g) = 989.6 cal

Next you have to raise the water temperature from 0 degrees C to 100 degrees C. So...

(12.37 g)(1.0 cal/deg/g)(100 deg) = 1237 cal

Next you have the positive heat of vaporization.

(12.37 g)(540 cal/g) = 6679.9 cal

Next once it is steam, you must raise the temperature from 100 degrees C to 146 degrees C. So...

(12.37g)(.48 cal/deg/g)(46 deg) = 273.1296 cal.

Now that you have the energy for each step add them together.

207.816 + 989.6 + 1237 + 6679.9 + 273.1296 = 9387.45 cal

I'm pretty confident in my breakdown but it defintely wouldn't hurt to have someone else check it

Answer 2

First Ice at -35 to Ice at 0

12.37x.48x35

Ice at 0 to water at 0

12.37x80

Water at 0 to water at 100

12.37x1x100

Water at 100 to Steam at 100

12.37x540

Steam at 100 to Steam at 146

12.37x.48x146

Add them all up.

Answer 3

just simply add up the calories required for the temperature change within each phase and add the calories required for the two phase changes.

You do the rest of your homework.