I would work this in two stages.
{mass ice x specific heat ice x [0-(-30)]} + [(mass warm water x specific heat water x (x-87)] = 0
That will take care of moving the ice T to zero C. x is the temperature of the mixture. The next step actually is three steps in one.
heat to melt ice at zero C + heat added to melted ice to final T + heat lost by water at higher T to Tfinal.
[mass ice x heat fusion] + [mass melted ice x speicif heat ice water x (Tfinal-Tinitial)] + [mass warm water at x temperature x specific heat water x (Tfinal-Tinitial)] = 0
Tinitial for the warm water will be x value determined in the first step.
Post your work if you get stuck.
A 14.0-g sample of ice at -18.0°C is mixed with 122.0 g of water at 87.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J/g · °C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
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