AP CHEM
posted by Jake on .
An ice cube tray contains enough water at 22.0 C to make 18 ice cubes that each have a mass of 30.0g. The tray is placed in a freezer that uses CF2CL2 as a refridgerant. The heat of vap of CF2CL2 is 158 J/g. What mass of CF@CL2 must be vaporized in the refridgeration cycle to convert all the water at 22.0 C to ice at 5.o C.
this is all i have so far. please help? thanks
H capacity H20 (s) = 2.08 J/g C
H capacity H20 (l) = 4.18 J/g C
ethanply of fusion of ice =6.02 KJ/mol.

You have the mass of the water to be frozen, figure the heat lost to convert it to ice at 5C.
You will have to watch units, on the heat of fusion, convert that to J/g.
Then, the heat lost has to equal the heat gained, or
Massfreon*Heatvap= heat lost 
mass of water = 30g x 18 cubes = 540 g.
heat required to lower T of water from 22 to zero is mass x 4.18 x 22 = xx
heat required to freeze 540 g water is
mass x heat of fusion water = yy
heat required to lower T of 540 g ice from zero to 5 is mass x 2.08 x 5 = zz
Total heat required to lower T of 540 g water from 22 to 5 C is xx + yy + zz.
What do you have to do this job? You have a refrigerant which soaks up 158 J for each gram of the stuff used. So how many grams does it take to equal the total heat required from above?
Check my thinking. Check my work.