To heat ice from 273.15 K to water at 293.15 K (0 °C to 20 °C) requires::(1) 333.55 J/g (heat of fusion of ice) = 333.55 kJ/kg = 333.55 kJ for 1 kg of ice to melt:PLUS:(2) 4.18 J/(g·K)· 20K = 4.18 kJ/(kg·K)· 20K = 83.6 kJ for 1kg of water to go up 20 K:= 417.15 kJOr to restate it in everyday terms, one part ice at 0 °C will cool almost exactly 4 parts water at 20 °C to 0 °C.