# What amount of heat is required to completely melt a 29.95-gram sample of H_2O(s) at 0° C?

Dec 20, 2015

$1.000 \cdot {10}^{4} \text{J}$

#### Explanation:

When a sample of water melts from ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, it undergoes a phase change.

As you know, phase changes take place at constant temperature. All the heat added to the sample goes into disrupting the strong hydrogen bonds that keep the water molecules locked in place in the solid state.

This means that you can't use water or ice's specific heat, since the heat added does not change the sample's temperature.

Instead, you will use water's enthalpy of fusion, $\Delta {H}_{f}$, which tells you what the change in enthalpy is when heating a substance at its melting point to make it undergo a solid $\to$ liquid phase change.

Water's enthalpy of fusion is approximately equal to

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

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

This tells you that in order to convert $\text{1 g}$ 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, the sample is said to have a mass of $\text{29.95 g}$, which means that you'll need

29.95 color(red)(cancel(color(black)("g"))) * "334 J"/(1color(red)(cancel(color(black)("g")))) = "10,003.3 J"

Rounded to four sig figs, the answer will be

$q = \textcolor{g r e e n}{1.000 \cdot {10}^{4} \text{J}}$