# How do you calculate the heat in joules needed to freeze 295 g of water at 0*C?

Jul 4, 2016

$\text{98.4 kJ}$

#### Explanation:

The first important thing to mention here is that you don't need heat to freeze water at its freezing point!

In fact, heat is being given off when water goes from liquid at ${0}^{\circ} \text{C}$ to solid at ${0}^{\circ} \text{C}$, i.e. when it undergoes a solid $\to$ liquid phase change.

In other words, freezing is an exothermic process because the system is giving off heat to its surroundings.

Now, the enthalpy of fusion, $\Delta {H}_{f}$, tells you how much heat is needed in order to convert ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$. For water, the enthalpy of fusion is given as

$\Delta {H}_{f} = {\text{333.55 J g}}^{- 1}$

The trick here is to realize that the amount of heat needed to melt ice will be equal to the amount of heat given off when liquid water freezes.

You can thus say that when $\text{1 g}$ of water freezes at ${0}^{\circ} \text{C}$, $\text{333.55 J}$ of heat are being given off to the surroundings. Since your sample has a mass of $\text{295 g}$, it follows that it will release

295 color(red)(cancel(color(black)("g"))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = "98,397 J"

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

"heat given off" = color(green)(|bar(ul(color(white)(a/a)color(black)("98.4 kJ")color(white)(a/a)|)))