# 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?

May 30, 2014

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.

${q}_{1} + {q}_{2}$ = 0

mΔH_(fus) + mcΔT = 0

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

333.55 $m$ g⁻¹- 26 600 = 0

$m = \frac{26600}{333.55 \text{g⁻¹}}$ = 79.7 g