To use standard enthalpies of formation to calculate the standard change in enthalpy for the melting of ice. kJ/mol
∆Hf° = standard heat of formation  can be use to claculate the std enthalpy change as follows;
∆Hf°reaction = ∆Hf° p - ∆Hf° r
if the reaction is Hâ‚‚O (s) ---- Hâ‚‚O(l)
But
∆Hf°reaction = ∆Hf° H₂O(l)  - ∆Hf° H₂O (s), which were -285kJ/mol and -291.8kJ/mol (obtained from the back of chemical process analysis text)
∆Hf°reaction = -285.8 kJ/mole - (-291.8 kJ/mole) = 6.0 kJ/mole = 6000 J/mole
heat gained by ice melting = heat lost by cooling water
(mass x ∆Hf°) ice = (mass x Cp x dT) liquid water
mass ice = (mass x Cp x dT) liquid water / (∆Hf°) ice
mass ice = 480 mL x 1.00 g/mL x 4.18 J/gC x (25-0 C) / 6000 J/mole
= 5.66 mole ice = 5.66 mole x 18.0 g /mole = 150 g  of ice required to cool the 480mL of water.
