(a) In an experiment to measure the specific latent heat of vapourisation of water, a student places a heater in a beaker containing water. The beaker stand...

(a)(i) Completed table

The mass evaporated is the decrease in balance reading:

\[m=203.22-\text{balance reading}\]

(ii) The energy supplied is \(E=Pt\), where \(P=38\text{ J s}^{-1}\). Thus, for example, at \(100\text{ s}\), \(E=38\times100=3800\text{ J}\).

(iii) Graph of energy supplied against mass evaporated

(iv) Slope of the graph

Using two well-separated points on the line of best fit, \((0.8\text{ g},2000\text{ J})\) and \((6.1\text{ g},14000\text{ J})\):

\[\text{Slope}=\frac{\Delta E}{\Delta m}=\frac{(14000-2000)\text{ J}}{(6.1-0.8)\text{ g}}=\frac{12000}{5.3}=2.26\times10^3\text{ J g}^{-1}\]

\[\therefore\ \text{Slope}=2.26\times10^6\text{ J kg}^{-1}.\]

(v) The slope represents the specific latent heat of vapourisation of water. It is the heat energy required to change unit mass of water at its boiling point into vapour at the same temperature.

Thus, approximately \(2.26\times10^3\text{ J}\) is required to evaporate \(1\text{ g}\) of water, or \(2.26\times10^6\text{ J}\) is required for \(1\text{ kg}\) of water.