“Pasta Quick! – thinking through the science (part 1)”
Extending the experiment for slightly older scientists
Many people boil water in an electric kettle before pouring into the pan to cook pasta. So is this a really good idea for energy use?
Suitable for Key Stage 3 (ages 11 to 13) and Key Stage 4 (ages 14-16) if continuing with part 2.
Skills: Thinking about a Method, Measurements, Comparisons, Arithmetic, using Units and using Formulae.
Here we explain the scientific method and this time we are wanting to compare cooking pasta with a pan with a lid on to boiling water in an electric kettle before we pour it into a pan to continue cooking with a lid. So here the experiment is about using an electric kettle to speed up the first stage of cooking.
Explain hypothesis. Scientists want to test a question.
An experiment is made up of a hypothesis or question for testing, then thinking about how to test it (method) and carrying it out taking observations (results). Finally from the results we find out what the answer to the question is (conclusion).
Here we will form the hypothesis as something like “boiling the water first in an electric kettle saves time and saves energy”. We will work out whether that is actually right at the end.
First get them to have a think about whether it will make a difference and they can guess which is better.
Now we want to formulate a method. How can we test that? What can we do?
Think about keeping the test fair. We want to boil exactly the same amount of water, add the same amount of salt and cook the same amount of pasta. We want to start off with the pan at the same temperature.
How will we measure it? We can measure the speed using a timer or a clock (which measures seconds too). We can measure the quantity of water using a jug or often even using the kettle itself. We can measure the salt with a teaspoon and the pasta with a mug or weigh it on kitchen scales.
Can we make sure we are the only users of gas and electricity in the household for the duration of the experiment? If we can we can use the gas and electricity meters. Maybe we can use the smart meter too if we have one? If we can’t switch off other electrical appliances, can we estimate the energy use from the time and the kettle power rating?
Try and write a step-by-step method like the one in the first experiment. Remember to keep things fair.
We should note down our readings and observations in our logbook (on paper).
What was our start time and end time. What was our start meter reading and end meter reading. How long was the kettle on for.
As with the previous experiment we are looking for the quicker one which will be a simple subtraction of times.
Here we are trying also to compare energy use. This may need some guidance.
We should try and compare using kiloWatt-hours, kWh (you could also compare using Joules, J if you want to, but we do need to be consistent).
Gas units are measured in cubic metres (m^3). The conversion to kWh is in your gas bill (if you don’t have access to the bill then use the example here).
Units (m^3) x Volume Correction x Calorific Value / 3.6 = Usage
Units x 1.02264 x 39.4 / 3.6 = Usage in kWh
Note that 1W = 1J/s (one Watt is equal to one Joule per second). So 1kW = 1000J/s (one kiloWatt is equal to one thousand Joules per second). So 1kWh = 1000J/s x number of seconds in an hour = 1000J/s x 60 x 60s = 1000J/s x 3600 = 3 600 000 J = 3600 kJ = 3.6 MJ (3.6 MegaJoules). That’s where the 3.6 in the formula above comes from.
The Volume Correction is applied to adjust for temperature and pressure and is an industry accepted correction.
The Calorific Value is the energy contained in one cubic meter of gas (in MJ/m^3). The value on your bill is the one to use as there are very slight differences around the world due to differences in the composition of natural gas.
Your electric kettle should have a rating plate with the power in W. To estimate the energy used we will multiply the power in kW by the time taken to boil the kettle in hours.
Example: say it says 2200 W, and it took us 6 minutes to boil the kettle
2200 W = 2.2 kW
6 min = 6/60 hours = 0.1 hour
Energy = 2.2 kW * 0.1 hour = 0.22 kWh
So which method is quicker?
And which method used less energy?