What amount of ice must be added to 540.0 g of water at 25.0 °C to cool the water to 0.0 °C and have no ice?

1 Answer

You must add 79.7 g of ice.

There are two heats involved: the heat to melt the ice and the heat to cool the water.

Heat to melt the ice + Heat to cool the water = 0.

q_1 + q_2q1+q2 = 0

mΔH_(fus) + mcΔTmΔHfus+mcΔT = 0

mm × 333.55 J·g⁻¹ + 254 g × 4.184 J·g⁻¹°C⁻¹ × (-25.0 °C) = 0

333.55 mm g⁻¹- 26 600 = 0

m =26600/(333.55"g⁻¹") = 79.7 g