How many grams of ice at -26.5\, ^\circ C can be completely converted to liquid at 15.0\, ^\circ C if the available heat for this process is 4.50×103 kJ?

Question

For ice, use a specific heat of 2.01J/(g\cdot{ ^\circ C}) and H2 = 6.01 kJ/mol.

Answer 1

I word to the wise should be sufficient; i.e., you would do well not to try to type in the special symbols. The ones displayed on my screen look like Sanskrit and they are not readable.

For your problem, assuming I have translated the Sanskrit correctly, is
4.50E6 J = mass ice x heat fusion + [mass water x specific heat water x (Tfnal-Tinitial)]
You have one unknown, mass ice and mass water (same unknown).