How many grams of ice at -14.0 degree Centigrade can be completely converted to liquid at 13.7 degree Centigrade if the available heat for this process is 5.91×10^3 kJ?

Question

For ice, use a specific heat of 2.01J/(g.degree Centigrade}) and H2 = 6.01 kJ/mol.

Express your answer numerically in grams.

Answer 1

5.91 x 10^6J = [mass ice x specific heat ice x 14]+[(mass ice x heat fusion)]+[mass water x specific heat water x 13.7].
Change your 6.01 kJ/mol for heat fusion (I assume H2 stands for heat fusion) to J/g, substitute into the above and solve for grams ice. I get something like 14 kg.