A student determined the specific latent heat of vaporisation of water - AQA - GCSE Physics - Question 5 - 2022 - Paper 1

The power output of the heater is a control variable because it is kept constant throughout the experiment to ensure that it does not affect the results.

The beaker of boiling water could be moved safely using tongs or gloves to avoid direct contact with the hot beaker.

To calculate the specific latent heat of vaporisation, use the formula:

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

where:

• $Q = 25,200 \, J$
• $m = 0.0090 \, kg$

Substituting the values:

$L = \frac{25,200}{0.0090} = 2,800,000 \, J/kg$

Thus, the specific latent heat of vaporisation of water is $2.8 \times 10^6 \, J/kg$. The unit is J/kg.

A source of error in the student's experiment was the transfer of thermal energy from the water to the surroundings, which could lead to inaccuracies in the mass measurement and the calculated specific latent heat.