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?

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Answer 1 · 2014-05-30T00:27:00

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.

$q_{1} + q_{2}$ $q_1 + q_2$ = 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

$m =26600/(333.55"g⁻¹")$ = 79.7 g