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.
AP CHEM - bobpursley, Wednesday, January 2, 2008 at 9:34pm
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
AP CHEM - DrBob222, Wednesday, January 2, 2008 at 9:34pm
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.