Required practical: thermal insulation

The question is how well a material slows the transfer of energy from hot water to the room. You can change one of three things: the material, the number of layers, or the thickness.

1. investigation: which material is the best thermal insulator?
2. independent: the type of material
3. dependent: the temperature drop of the water in a set time
4. control: the volume of water, the starting temperature, the beaker, the lid, the time

A beaker with no insulation gives you something to compare with.

1. Wrap each beaker in its material (the same thickness each time) and hold it on with elastic bands.
2. Pour the same volume of hot water from a kettle into each, measured with a measuring cylinder.
3. Put on a lid, or leave the lid off every one, and record the starting temperature.
4. Read the temperature every minute (or let a temperature probe and data logger do it) for the same time.
5. Repeat each material and work out a mean temperature drop.

Safety: hot water from a kettle can scald, so pour it into beakers that are standing on a heatproof mat on the bench, never into one you are holding. Keep the kettle cable tidy at the back of the bench and wipe up spills straight away.

Plot temperature against time for each material. The better insulator gives the curve that falls more slowly. Every curve gets less steep with time, because the water gets closer to room temperature and loses energy more slowly.

If the starting temperatures were not quite the same, compare the drops, not the end temperatures.

rate of cooling = temperature drop ÷ time

You can also find the energy the water lost, using its mass (1 cm³ of water has a mass of 1 g):

1. 200 cm³ of water (0.2 kg) cools by 12 °C in 10 minutes; c = 4200 J/kg °C
2. ΔE = m c Δθ = 0.2 × 4200 × 12 = 10 080 J
3. power = 10 080 ÷ 600 = 16.8 W

Adding layers helps less and less: the drop gets smaller with each layer, and the graph of drop against layers levels off.