Question

We have of ice at -11.9°C and it is ultimately completely converted to liquid at 9.9°C. For ice, use a specific heat (Cs) of 2.01 J/gK, for liquid water use a specific heat of 4.18 J/gK, and ΔHfusion = 6.01kJ/mole, Molar mass H20 = 18.01g/mol. If 4.19x10^3 kJ is transferred as heat to the ice for the above process is, how many grams of ice are melted and warmed to 9.9°C?

Answer 1

Do this steps.
q1 = energy to raise T from zero for liquid H2O to 9.9 liquid water.
q1 = (mass H2O x specific heat H2o x (Tfinal-Tinitial). You have all of these numbers but mass H2O.

q2 = heat to convert solid ice to liquid water at zero C is
q2 = mass ice x heat fusion. You have heat fusion.

q3 is heat to raise T of solid ice from -11.9 C to solid ice at zero C.
q3 = mass ice x specific heat ice x (Tfinal-Tinitial). You have all but mass ice.
All of these must add up to heat you have and that is 4.19E3 kJ which I would convert to J.
4.19E6J = q1 + q2 + q3. Substitute the equations above for q1, q2, and q3 and solve for mass ice. I estimated about 10,000 grams ice.