How much ice (at 0°C) must be added to 1.15 kg of water at 81 °C so as to end up with all liquid at 26 °C? (ci = 2000 J/(kg.°C), cw = 4186 J/(kg.°C), Lf=3.35×10^5 J/kg, Lv= 2.26×10^6 J/kg)

Answer 1

The sum of the heats gained will be zero (some will be negative gained).

heaticemeleting+heat waterwarmcooling+heaticewaterwarming=0
M*3.35e5+1.15*4186*(26-81)+M*4186(26-0)=0
M(3.35e5+4186*26)=1.15*4186(55)
M= ... ? I get about 600 grams (.6kg)