The specific latent heat of vaporisation of water can be determined using the apparatus shown, a) A student planned to vary the current in the heater from 0 A to 5 A - Edexcel - A-Level Physics - Question 3 - 2023 - Paper 3

The graph would show a straight line with a negative gradient, illustrating that the mass of water in the beaker decreases over time as it evaporates. The mass of water would decrease steadily at first, then may level off as the water reaches boiling point, indicating a stable mass loss until all the water has evaporated.

To calculate L, we use the formula:

$L = \frac{W}{m}$

where $W = V \cdot I \cdot t$ and $m$ is the change in mass.

1. First, calculate the total energy used, $W$:

   • $V = 12 V, I = 4.2 A, t = 6.0 \text{ minutes} = 6.0 \times 60 = 360 \text{ seconds}$
   • $W = 12 \times 4.2 \times 360 = 181.8 \text{ J}$

2. The change in mass $m = 7.5 g = 0.0075 kg$.

3. Now substitute into the latent heat formula:

   • $L = \frac{181.8}{0.0075} = 2.424 \times 10^4 \text{ J/kg}$

One significant source of error could be heat loss to the surroundings, which occurs when the beaker and water lose energy during the heating process. If heat escapes, not all the energy produced by the heater is effectively used to vaporise the water. As a result, the calculated value of L will be larger than the true value. This can lead to inaccurate estimations of the latent heat due to the energy balance being disturbed.