# Question #387cd

Sep 24, 2017

$\text{33,360 J}$ of heat must be given off.

#### Explanation:

For starters, it's very important to realize that you don't need energy to convert liquid water to ice because this process actually releases energy.

In other words, when water freezes, heat is being given off, not absorbed.

Now, the problem provides you with the enthalpy fusion of water

$\Delta {H}_{\text{fus" = "333.6 J g}}^{- 1}$

For a given substance, its enthalpy of fusion tells you how much energy is needed to convert $\text{1 g}$ of this substance from solid at its melting point to liquid at its melting point.

In your case, the enthalpy of fusion of water tells you that in order to convert $\text{1 g}$ of ice at its normal boiling point of ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, you need to provide $\text{333.6 J}$ of heat.

This means that when $\text{1 g}$ of liquid water at ${0}^{\circ} \text{C}$ is converted to ice at ${0}^{\circ} \text{C}$, $\text{333.6 J}$ of heat are being given off.

So remember, you have

• $\text{solid " -> " liquid " implies " heat absorbed}$
• $\text{liquid " -> " solid " implies " heat given off}$

This means that when $\text{100.0 g}$ of liquid water freeze at ${0}^{\circ} \text{C}$, you have

$100.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * overbrace("333.6 J"/(1color(red)(cancel(color(black)("g")))))^(color(blue)(=DeltaH_"fus")) = color(darkgreen)(ul(color(black)("33.360 J}}}}$

that are being given off. The answer is rounded to four sig figs, the number of sig figs you have for the mass of liquid water.