Experiment:

At the end of experiment 8.1, the beaker contains boiling water. Continue heating water till all the water changes into steam. Note the time which the water in the beaker takes to change entirely into steam at its boiling point of 100°C

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

 Let, Mass of ice = m

Time $ t_{0}$ taken to heat water from 0°C to 100°C (melt) = t_{0} = $t_{3-} t_{2}$ =  4.6 min.
Time taken by water at 100°C to change it into steam = $t_{4}= t_{4-} t_{3}$ =  24.4 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$
= $m \times 420 000 Jkg^{-1}$
= $m \times 4.2 \times 10^{5} Jkg^{-1}$

As burner supplies heat $\triangle$ Q to water in time t_{0}  to raise its temperature from 0°C to 100°C. Hence, the rate at which heat is absorbed by the beaker is given by
Rate of absorbing heat  =   $\triangle Q/(t_{0})$
∴ Heat absorbed in time  $t_{v}$ = $ \triangle Q_{v}$   =  $\frac{(\triangle Q\times t_{v})}{(t_{0})}$ 

=$ \frac{ \triangle Q \times(t_{v})}{(t_{0})}$
Since, $\triangle Q_v$ = $m \times H_{v}$                  (from eq. 8.8)

Putting the values, we get

$ m \times H_{v}$  = $\frac{m \times 4.2 \times 10^{5} Jkg^{-1}\times ( t_{v})}{(t_{0})}$ 
or, $H_{v}$  = $ \frac{4.2 \times 10^{5} Jkg^{-1} \times ( t_{v})}{(t_{0})}$
Putting the values of $ t_{v}$  and $ t_{0} $ from the graph, we get

$H_{v}$  = $\frac{4.2 \times 10^{5} Jkg^{-1}\times ( 24.4min)}{(4.6min )}$ 
= $2.23 \times 10^{6}  Jkg^{-1}$       

The latent heat of vaporization of water found by the above experiment is $2.23 \times 10^{6} Jkg^{-1}$ while its actual value is $2.23 \times 10^{6}  Jkg^{-1}$.