posted by Lauren on .
A quantity of ice at 0.0 degrees C was added to 33.6 of water at 21.0 degree C to give water at 0.0 degrees C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol and the specific heat is 4.18 J/(g * degrees C)
q = mass x specific heat x delta T?
q = mass x heat fusion?
I don't understand what to plug in
Heat removed from water in going from 21.0 to zero C is
mass x specific heat x delta T.
mass = 33.6 g
specific heat is 4.184 J/g*K
delta T is 21.0 - 0.
heat added to ice to melt it is
mass x heat fusion.
mass is the unknown.
heat fusion is given.
Note: you need to work in the same units. The heat of fusion is quoted in the problem in kJ/mol but I quoted the specific heat of water in J/gram. One needs to be changed. I would suggest the 4.184 be changed.
ok so i did 33.6g (4.184 J/g)(21.0-0)
So how do i get J to kJ/mol
The specific heat of water in J/mol = 75.3. You can change 33.6 g water to mols by 33.6/18 = ??
Then (33.6/18)*75.3*21 = ?? Joules.
You can change to kJ by ??/1000 = xx.