Experiment:

Take a beaker and place it over a stand. Put small pieces of ice in the beaker and suspend a thermometer in the beaker to measure the temperature. Place a burner under the beaker. The ice will start melting. The temperature of the mixture containing ice and water will not increase above 0°C until all the ice melts. Note the time which the ice takes to melt completely into the water at 0°C

Continue heating the water at 0°C in the beaker. Its temperature will begin to increase. Note the time which the water in the beaker takes to reach its boiling point at 100°C from 0°C.

Draw a temperature-time graph as shown in figure 8.11. Calculate the latent heat of fusion of ice from the data as follows:

Let, the mass of ice = m

Finding the time from the graph:

Time taken by ice to melt completely at 0°C =$ t_{f} = t_{2-} t_{1} $ =  3.6 min.
Time taken by water to heat from 0°C to 100°C =$ t_{0} = t_{3-} t_{2}$ =  4.6 min.
Specific heat of water c = $4200 J〖kg〗^{-1} K^{-1}$.
Increase in the temperature of water = $\triangle T$ = 100°C =100 K
Heat required by water from 0°C to 100°C =$ \triangle Q$ = $ mc \triangle T$
=$m \times 4200 Jkg^{-1} K^{-1} \times 100K$
= $m\times 420 000 Jkg^{-1}$
= $m \times 4.2 \times 10^{5} Jkg^{-1}$

Heat $ \triangle Q$ is supplied to water in time $t_{0}$ to raise its temperature from 0°C to 100°C. Hence, the rate of absorbing heat by the water in the beaker is given by:

Rate of absorbing heat =  $ \triangle Q/(t_{0})$
∴ Heat absorbed in time  $t_{f} = Δ Q = (ΔQ× t_{f})/(t_{0})$ =$\frac{\triangle Q \times ( t_{f})}{(t_{0})}$ 
Since, $\triangle Q_{f}$ = $ m \times H_{f}$                  (from eq. 8.7)

Putting the values, we get:

$m \times H_{f}$  =  $m \times 4.2\times 10^{} Jkg^{-1}\times ( t_{f})/(t_{0})$
or, $H_{f}$= $\frac{4.2 \times 10^{5} Jkg^{-1} \times ( t_{f})}{(t_{0})}$ 
The values of $t_{f}$ and$ t_{0}$ can be found from the graph. Put the values in the above equation to get
$H_{f}$ = $4.2 \times 10^{5} Jkg^{-1} \times ( 3.6min)/(4.6min )$
= $3.29 \times 10^{5}  Jkg^{-1}$

     
The latent heat of fusion of ice found by the above experiment is $3.29 \times 10^{5}  Jkg^{-1}$   while its actual value is $3.36 \times 10^{5} Jkg^{-1}$.