Question

What is the final temperature and state of water when 8.50 kJ of energy is added to 10.0 g of ice at 0 degree Celsius.

Helpful information:

Cs (ice)= 2.1 J/g•C

Heat of vaporization (water)= 40.7kJ/mol

Ca (water)= 4.184 J/g•C

Heat of fusion (water)= 6.02 kJ/mol

Cs (steam) = 2.0 J/g• C

Answer 1

Energy used to melt ice at zero C to liquid water at zero C.
dq = mass ice x heat fusion = (10/18) x 6.02 kJ/mol = 3.34 kJ.
Energy remaining to be used is 8.50 kJ - 3.34 = 5.15 kJ.

Energy used to raise T of water at zero C to 100 C.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = (10) x 4.184 x (100-0) = about 4184 J or 4.184 kJ.
How much is left now? That's 5.15 kJ - 4.18 kJ = about 0.98 kJ.

That can be used to vaporize some of the water at 100 C.
q = mass H2O x heat vap
0.97 kJ = mols H2O x 40.7 kJ/mol and solve for mols H2O, then convert to grams.
You need to go through all of these calculations again but you see you will have some H2O liquid and some water steam. Check all of these numbers.