# Question 04705

Dec 29, 2015

$\Delta {H}_{\text{fus" = -"6.005 kJ/mol}}$

#### Explanation:

As you know, the molar enthalpy change of fusion, $\Delta {H}_{\text{fus}}$, is defined as the change in enthalpy when one mole of a substance undergoes a liquid to solid or a solid to liquid phase change at its melting / freezing point.

More specifically, the molar enthalpy change of fusion will tell you

• how much heat must be added to one mole of a substance at its melting point in order for a solid $\to$ liquid phase change to take place

• how much heat must be given off by one mole of a substance at its freezing point in order for a liquid $\to$ solid phase change to take place

In your case, you are interested in finding out the molar enthalpy change of fusion for the freezing of $\text{30.00 g}$ of water at ${0}^{\circ} \text{C}$. This means that the molar enthalpy change of fusion must carry a negative sign, since it represents heat lost.

The equation you'll use here looks like this

color(blue)(q = n * DeltaH_"fus")" ", where

$q$ - heat lost / absorbed
$n$ - the number of moles of the substance
$\Delta {H}_{\text{fus}}$ - the molar enthalpy change of fusion

Use water's molar mass to determine how many moles you have in that $\text{30.00-g}$ sample

30.00 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.6653 moles H"_2"O"

No,w you know that $\text{10.00 kJ}$ of heat are given off when this sample of water goes from liquid water at ${0}^{\circ} \text{C}$ to ice at ${0}^{\circ} \text{C}$.

This means that the value of $q$ will be

$q = - \text{10.00 kJ}$

Again, the negative sign is used to symbolize heat lost.

This means that you have

$\Delta {H}_{\text{fus}} = \frac{q}{n}$

DeltaH_"fus" = (-"10.00 kJ")/"1.6653 moles" = color(green)(-"6.005 kJ/mol")#

The answer is rounded to four sig figs.

So, you can say that the molar enthalpy change of fusion for water will be equal to

• $\Delta {H}_{\text{fus" = +"6.005 kJ/mol}} \to$ when ice at ${0}^{\circ} \text{C}$ melts to liquid water at ${0}^{\circ} \text{C}$
• $\Delta {H}_{\text{fus" = -"6.005 kJ/mol}} \to$ when liquid water at ${0}^{\circ} \text{C}$ freezes to ice at ${0}^{\circ} \text{C}$