# 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?

May 31, 2016

$124.16 g$

#### Explanation:

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

Hence : ${\left(m c \Delta T\right)}_{w a t e r} = {\left(m {L}_{f}\right)}_{i c e}$

$\therefore \left(1000 k g / {m}^{3} \times 0.000087 {m}^{3}\right) \left(4180 J / k g . K\right) \left({39.9}^{\circ} C\right) = \left(m\right) \left(334000 J / k g\right)$

Solving we get $m = 0.03895 k g = 38.95 g$

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