Question

The molar heat of fusion for water is $6.01kJ{\text{ }}mo{l^{ - 1}}$. How much energy is released when $36.8g$ of water freezes at its freezing point?

Answer 1

Hint: As we know that the heat of fusion changes enthalpy take place when one mole of solid is converted into liquid at its melting point. It tells us about how much heat is removed to freeze one mole of any substance at its freezing point.

Complete step-by-step answer:
As we are given that the heat of fusion of water is $6.01kJ{\text{ }}mo{l^{ - 1}}$ and we are also given the condition that energy is released when water freezes. We know that enthalpy of fusion is basically the change in the state of a matter like a solid is converted to its liquid form or a liquid is converted into vapour form.

So, the heat of fusion released while freezing would be given as:
$\Delta {H_{fusion}} = - 6.01kJ{\text{ }}mo{l^{ - 1}}$
Now, we are given that $36.8g$ of water freezes and we also know that molecular mass of water is $18g$. So we will calculate the number of moles using these two quantities and we will get:
$moles = \dfrac{{mass}}{{molecular{\text{ }}mass}}$
$moles = \dfrac{{36.8}}{{18}} = 2.043mol$

Now we know that the heat is related to the molar heat of fusion by the below given relation:
$q = n\Delta H$
Where, $n$ is the number of moles of the given substance, $q$ is the amount of heat energy released or absorbed and $\Delta H$ is the heat or enthalpy of fusion.
After putting all the given and calculated values in the above formula we will get:
$q = 2.043 \times ( - 6.01)$
$q = - 12.3kJ$

Therefore, from the above explanation we can say the correct answer is $ - 12.3kJ$ of energy is released.

Note: Always remember that the negative sign symbolises the heat energy released in the course of a reaction and if there is positive sign before the amount of energy then it symbolises the energy absorbed during the reaction.