chem help needed!?

Question

From the following data, calculate the total heat needed to convert 0.882 mol ice at -7.98°C to liquid water at 1.18°C.
Melting point at 1 atm 0°C

Csolid 2.09J/g°C
Cliquid 4.21J/g°C
DeltaHfus 6.02 kJ/mol

Answer 1

Mol. wt. H2O = 18.
0.882molH2O x 18gH2O/1molH2O = 15.9gH2O

First, raise the ice from -7.98C to 0C:

15.9gH2O(s) x 2.09J/gH2O(s)-degC x 7.98degC = 265J

Next melt the ice:

0.882molH2O(s) x 6020J/molH2O(s) = 5310J

Finally, raise the water to 1.18C:

15.9gH2O x 4.21J/gH2O(s)-degC x 1.18degC = 79.0J

265 + 5310 + 79 = 5654J