How many kg of ice needs to be added to 1.31 kg of water at 64.9°C to cool the water to 14.9°C when the Latent heat of ice = 80 kcal/kg?

Answer 1

m₁cΔt₁=λm₂+m₂cΔt₂
m₂=m₁cΔt₁/(λ+cΔt₂) =
=1.31•4218•(64.9-14.9)/(336000+4218•14.9)=
=0.69 kg
(Latent heat of ice = 80 kcal/kg=336000 J/kg)