Question

If you add 163.11 g of ice at 0.0 °C to 0.087 L of water at 39.9 °C and allow the water to cool to 0.0 °C, how many grams of ice will remain?

Answer 1

`124.16g`

Explanation:

Conservation of energy - Energy lost by hot water equals energy gained by ice which goes towards melting the ice.

Hence : `(mcDeltaT)_(water)=(mL_f)_(ice)`

`therefore (1000kg//m^3xx0.000087m^3)(4180J//kg.K)(39.9^@C)=(m)(334000J//kg)`

Solving we get `m=0.03895kg=38.95g`

Thus grams of ice remaining unmelted `=163.11-38.95=124.16g`