Lab - How to estimate the heat of fusion

 Here,  the case of the ice shall be consider. However, some simmilarities should be found for other materials.

The problem

For short, the problem may be posed as a question:

How much energy would you need to melt a solid material?

This question have practical implications since chemical reactions may be better performed if raw materials are in liquid phase. Aso, if for other reason you need to melt something, at industrial scale for example, then you must know how much energy would be needed for this process.

The answer to the question in hand would depend on the temperature changes and the mass of the material.

The experiment - For melting an ice cube

The basic idea on which the measurement of the heat of fusion comes from the working principle of the calorimeter. In other words, if you place an ice cube into a portion of water, which is of course inside a thermally insulated container, it would melt and a corresponding temperature gradient would be set. This suggests that energy, thermal in this case, was transferred from one material to the other along with change of phase.

Fig. 1 An experimental sketch to measure the heat of fusion

Understanding the heat flow

The following facts can be useful to understand what happens in the sketch of Fig. 1:

• If the vessel in Fig. 1 is thermally insulated then no energy flows to the surroundings nor viceversa. In other words, the energy of the water, before pouring the ice cube, remains constant as $Q_{H2O}$.

• The ice cube also has its own amount of energy. Let us say: $Q_{ice}$.

• Once the ice cube is poured into the water, the vessel is kept closed and the energy is added from or given to the surroundings. In other words:

$Q_{ice} + Q_{H2O}=0$

$Q_{ice}=-Q_{H2O}$        Eq. (01)

• The energy of ice cube $Q_{ice}$, in the water, is balanced by two temperature gradients (that are in fact thermal energy contributions):

• the first temperature gradient is due to the temperature difference between the ice cube wall and its center. This temperature gradient produces $Q_{ice-melt}$, or the energy driving the melt of the ice cube,

• the second temperature gradient is due to the temperature difference between th ice cube wall and the surrounding warm water. This temperature gradient produces $Q_{w-surro}$, or the energy coming from the warm water far from the ice cube,

in other words:

$Q_{ice}=Q_{ice-melt}+Q_{w-surro}$        Eq. (02)
• The previous statments also mean that:

$-Q_{H2O}=Q_{ice-melt}+Q_{w-surro}$        Eq. (03)

• In other words, the energy required to melt the ice cube in the warm water in the insulated vessel is equal to:

$Q_{ice-melt}=-Q_{w-surro}-Q_{H2O}$        Eq. (03)

From the point of view of the experiment, Eq. (03) should be enough to estimate the heat of fusion of the ice cube. Next, if the thermal energy is in way general way written as,

$Q=C_p m \Delta T$        Eq. (04)

then Eq. (03) can be rewritten as:

$Q_{ice-melt}=-C_{H2O}m_{ice+H2O}\left( T_{H2O-final} - T_{ice} \right) \\-C_{H2O}m_{H2O}\left( T_{H2O-final} - T_{H2O-ini} \right) $        Eq. (05)

Measuring the stuff

Care has to be paid to the experimental procedure when measuring temperature. 

First, the ice temperature is 0 °C as you may know. This is, $T_{ice}=0$.

You choose the initial temperature of the water $T_{H2O-ini}$ since you heat the water to a desired temperature.

Measuring $T_{H2O-final}$ can be tricky since the final stage involves time. The temperature falls from $T_{H2O-ini}$ to $T_{H2O-final}$ in a given amount of time that may be different for different experiments and conditions. You should be aware of a temperature behavior as follows:

Fig. 2 Temperature gradient evolution for the water inside the calorimeter

On the other hand,

$Q_{ice-melt}= c_{ice}m_{ice}\left( T_{H2O-final} - T_{ice} \right)$        Eq. (06)

However, since $c_{ice}$ is unknown, then $Q_{ice-melt}$ can not be determined from Eq. (06). In other words, $Q_{ice-melt}$ must be estimated from Eq. (05). 

Finally, the heat of fusion is defined as,

Ice heat of fusion [J/g] $=c_{ice}\left( T_{H2O-final} - T_{ice} \right)\\
= \dfrac{Q_{ice-melt}}{m_{ice}}\\
\approx 333.6 J/g$

Any question? Write in the comments and I shall try to help.

Other stuff of interest

• Some basic concepts on thermodynamics
• Who was Clausius?
• Processes at constant volume and pressure
• Composition variables in mixtures
• The first law of thermodynamics and some concepts
• Partial and total derivatives (for thermodynamics)
• First law of thermodynamics for chemical reactions
• Solving the Colebrook equation

==========
Ildebrando.