Calculate the heat released when 42.0 g of water at 35.0 degrees Celsius is converted to ice at -5.0 degrees Celsius. The specific heat of ice is 2.03 J/(g C), the molar heat of fusion of ice is 6010 J/mol, and the specific heat of water is 4.18 J/(g C)

Question

(Hint: Be sure to consider the sign of the heat of fusion vs. heat of freezing.)

I obviously don't know how to do this since my two out three attempts were wrong. can someone please explain so I can figure out what i'm doing wrong?

Answer 1

Why didn't you show your work so we could figures out what you were doing wrong?
You want to add the heat removed in changing T from 35 to zero, to the heat removed @ zero C to convert to ice, to the heat removed in changing T from zero C to -5.

Answer 2

q=mass*Specific heat of liquid*change in temp
q=42*4.18*35=6144.6

q=mass* specific heat of ice*change in temp
q=42*2.03*5=426.3

q=heat of fusion of ice*moles
q=6010*2.3=14023

Then add all three together and get 20593.9 Joules

Answer 3

Your numbers look ok. I suspect the problem is that you are reporting too many significant figures.
For example, that first one of 5144.6 is allowed only 3 places. Also, shouldn't you have a - sign for the total. Each of the three are exothermic; i.e.,, the first one is mass x sp. h. x (Tfinal-Tinitial) = Tf is 0 and Ti is 35 so that gives you a negative sign. Frankly, I think when the problem says realeased that you need not consider the - sign because you already know it's released. However, it appears obvious from the problem that the author expect the sign to be included.

Answer 4

You were right the answer was -20600.