# Question 1112a

Jun 9, 2015

You'd need 97.9 kJ of heat to melt that much ice at ${0}^{\circ} \text{C}$.

#### Explanation:

The first thing you need to notice is that you need to melt the ice at its melting temperature. This means that you're dealing with a phase change and all the supplied heat will go into converting solid ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$.

The equation that you'll use looks like this

$q = n \cdot \Delta {H}_{\text{fus}}$, where

$n$ - the number of moles of ice;
$\Delta {H}_{\text{fus}}$ - the molar enthalpy of fusion.

Since the molar heat of fusion expresses the amount of heat required to convert 1 mole of solid ice to liquid water at ${0}^{\circ} \text{C}$, you're going to have to determine how many moles of ice are present in your sample.

You can write the number of moles of a substance as the ratio between mass and molar mass

$n = \frac{m}{M} _ M$

In order to get the mass of ice, you can use its density and volume

$\rho = \frac{m}{V} \implies m = \rho \cdot V$

320cancel("cm"^3) * "0.917 g"/(1cancel("cm"^3)) = "293.44 g"

Water's molar mass is equal to 18.02 g/mol, which means that you get

q = (293.44cancel("g"))/(18.02cancel("g")/cancel("mol")) * 6.009"kJ"/cancel("mol") = "97.85 kJ"#

Rounded to three sig figs, the answer will be

$q = \textcolor{g r e e n}{\text{97.9 kJ}}$

You need to supply 97.9 kJ of heat in order to convert 293.44 g of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$.