Suppose a bag full of ice (450 g) at 0.0 °C sits on the counter and begins to melt to liquid water. How much energy must be absorbed by the ice if 2/3 of it melted?

Dec 23, 2015

$q = 1.0 \cdot {10}^{2} \text{kJ}$

Explanation:

All you have to do here is figure out how much heat is required to convert ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, i.e. how much heat is required for that much ice to undergo a solid $\to$ liquid phase change.

As you know, phase changes take place at constant temperature and depend on the given substance's enthalpy of fusion, which in water's case is equal to about

$\Delta {H}_{f} = 334 \text{J"/"g}$

http://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html

What that means is that in order to melt $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, you need to provide it with $\text{334 J}$ of heat.

In your case, you need to know how much heat is required to melt $\frac{2}{3}$ of that $\text{450-g}$ sample of ice. Well, if $\text{1 g}$ requires $\text{334 J}$ of heat, it follows that you have

2/3 * 450 color(red)(cancel(color(black)("g"))) * "334 J"/(1color(red)(cancel(color(black)("g")))) = "100,200 J"

Rounded to two sig figs and expressed in kilojoules, the answer will be

$q = \textcolor{g r e e n}{1.0 \cdot {10}^{2} \text{kJ}}$