# The molar heat of fusion for water is 6.01 kJ/mol. How much energy is released when 36.8 g of water freezes at its freezing point?

Jan 25, 2016

$\text{12.3 kJ}$

#### Explanation:

For a given substance, the molar heat of fusion basically tells you one thing from two perspectives

• how much heat is needed in order to melt one mole of that substance at its melting point
• how much heat must be removed in order to freeze one mole of that substance at its freezing point

It is very important to realize that the molar enthalpy of fusion will carry a positive sign when you're dealing with melting and a negative sign when you're dealing with freezing.

That is the case because heat released carries a negative sign, while heat absorbed carries a positive sign. So, for water, you can say that

$\Delta {H}_{\text{fus" = +"6.01 kJ/mol}} \to$ heat needed for melting

$\Delta {H}_{\text{fus" = -"6.01 kJ/mol}} \to$ heat released when freezing

You're interested in finding out how much heat is released when $\text{36.8 g}$ of water freeze at water's freezing point. The first thing to do here is use water's molar mass to calculate how many moles you have in that sample

36.8 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "2.043 moles H"_2"O"

The heat released can be calculated using the equation

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

$q$ - heat released
$n$ - the number of moles of the substance
$\Delta {H}_{\text{fus}}$ - the molar enthalpy of fusion for that substance

Since you're dealing with freezing, you will have

q = 2.043 color(red)(cancel(color(black)("moles"))) * (-6.01"kJ"/color(red)(cancel(color(black)("mol")))) = -"12.3 kJ"

What this means is that when $\text{36.8 g}$ of water freeze at water's freezing point, $\text{12.3 kJ}$ of heat are being released to the surroundings.

Remember, the negative sign symbolizes heat released.

Having

$q = - \text{12.3 kJ}$

is equivalent to saying that $\text{12.3 kJ}$ of heat are being released.