Be careful about how much water you put in your kettle and ‘do your bit’ to save energy!

That’s what we’re often told, and boiling only as much water as you need for your cup of tea can only be a good thing. But how much impact does it *really* have on your overall energy consumption?

Consider this: if you’re driving along in a car at 50 mph, you’re using enough energy to make a cup of tea *every few seconds*.

1, 2, 3, tea, 1, 2, 3, tea, 1, 2, 3, tea…

As most people don’t think twice about driving their cars for an extra few minutes, never mind an extra few seconds, it’s clear that the energy savings made by being frugal with your kettle are very small compared to your overall energy use.

Does this mean that we shouldn’t be careful with our kettles after all? No. It all matters. But having boiled exactly one cup’s worth of water for your refreshing cuppa, you shouldn’t put your feet up and think that your energy consumption issues are sorted. You’ve *much* bigger fish to fry.

#### The calculation

Imagine that you’re travelling in a car that’s doing 50 miles per hour, with a rate of fuel consumption of 50 miles to the gallon (the units are customary in a UK motoring context). This is a realistic situation; if anything it’s optimistic in terms of energy consumption.

In this scenario, your car will burn a gallon of petrol, that’s 4.5 litres approximately, every hour.

With 3600 seconds in an hour, that’s 4.5L/3600s = 0.00125 litres per second. The density of petrol is about 0.8 kg per litre, so the rate of petrol consumption comes to 0.00125 L s^{-1} × 0.8 kg L^{-1} = 0.001 kilograms per second.

The energy density of hydrocarbon fuels is about 46 megajoules per kilogram, so the rate of use of energy by your car is 0.001 kg s^{-1} × 46×10^{6} J kg^{-1} = 46000 joules per second (watts).

Now let’s think about the tea. The energy *E *required to produce a temperature change of Δ*T *in a mass *m* of a substance whose specific heat capacity is *C* is

*E* = *m C* Δ*T *

A typical mug of tea contains about 250 ml, or 0.25 kg, of water. For water, *C *= 4200 J kg^{-1} K^{-1}. The temperature change is about 90 K, if the water comes out of the tap at about 10°C and is raised to boiling point. So the amount of energy required to heat the water for a cup of tea is 0.25 kg × 4200 J kg^{-1} K^{-1}× 90 K = 94500 joules.

As the car is using 46000 joules per second, it follows that the car is using enough energy to heat the water for a cup of tea every two seconds.

Now to be fair to the car, we need to recognise that in real life there is more energy used to heat the water for the tea than actually goes into the water. There are heat losses in energy generation and transmission, and from the kettle itself.

Things get very complicated here. A thermal electricity generating plant (coal, oil, gas, or nuclear) converts energy in the fuel to electrical energy with about 35-45% efficiency. But in the UK, about 20% of our electricity is generated from renewables, where the concept of efficiency is harder to pin down. For example, I could use some oil to heat my water directly instead of doing it indirectly by generating electricity, but I couldn’t do the same with the wind. I’m going to assume that 80% of the electricity that I used was generated in thermal plants at 40% efficiency, and 20% of it was generated at 100% efficiency, giving an overall efficiency of 52%. But be aware that this is a *very* rough figure, and that there is no right answer.

According to this document, about 93% of the energy in generated electricity makes it unscathed through the transmission and distribution systems to the end user.

I did an experiment and estimated the efficiency of my kettle to be about 88%. I give details later.

This gives an overall efficiency of boiling water as 52% × 93% × 88% = 43%. So instead of using 94500 joules to boil water for a cup of tea, we’re actually using 94500J/0.43 = 220000 joules, which means that the car’s rate of energy use is equivalent to a cup of tea roughly every 5 seconds. Because of all the uncertainties involved (not least the size of a cup of tea), we should treat this figure as very approximate.

#### Efficiency of my kettle

My kettle took 180 s to raise 1 litre of water from 17°C to boiling.

The kettle is rated at 1850-2200 W for supply voltages in the range 220-240 V. When the kettle was running, I measured the supply voltage to be 242 V, so I assumed that the kettle was operating at the top of its power range, ie 2200 W.

The heat supplied by the kettle element was therefore 2200 W × 180 s = 396000 J.

Using the same equation as earlier, the energy required to raise the temperature of 1 kg of water by 83 K is 1 kg × 4200 J kg^{-1} K^{-1} × 83 K = 348600 J.

The kettle was therefore heating the water with efficiency 348600/396000 = 0.88 or 88%.