Posted by **Lauren** on Monday, February 4, 2008 at 4:42pm.

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

and where

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

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