(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...
Question 1 Report
(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 stands on an electronic balance so that the mass of the beaker and water could be measured. The heater is switched on and readings were taken every 100s when the water starts boiling.
The table below shows the readings.
| Time/s | 0 | 100 | 200 | 300 | 400 |
| Reading on balance/g | 203.22 | 201.62 | 199.79 | 198.26 | 196.50 |
| Mass of water evaporated/g | 0 | ||||
| Energy supplied by heated/J | 0 |
(I) Fill in the mass of water evaporated.
(ii) Given that the heater supplies energy at the rate of 38J/s, fill in the values of the energy supplied by the heater in 100s, 200s, 300s, and 400s.
(ii) Plot a graph of energy supplied on the vertical axis and mass of water evaporated on the horizontal axis, starting both axes from the origin(0,0).
(iv) Determine the slope of the graph.
(v) what does the value of the slope mean?
Answer Details
(a)(i) Completed table
The mass evaporated is the decrease in balance reading:
\[m=203.22-\text{balance reading}\]
| Time / s | 0 | 100 | 200 | 300 | 400 |
|---|---|---|---|---|---|
| Balance reading / g | 203.22 | 201.62 | 199.79 | 198.26 | 196.50 |
| Mass of water evaporated / g | 0.00 | 1.60 | 3.43 | 4.96 | 6.72 |
| Energy supplied / J | 0 | 3800 | 7600 | 11400 | 15200 |
(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.
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