## Kettles, cars, and energy

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 × 46×106 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 ΔT in a mass m of a substance whose specific heat capacity is C is

E = m C ΔT

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%.

## How high am I ?

Last week I was on holiday in Wales. It wasn’t the driest of weeks, and while I was inside not climbing mountains, I finally got round to doing some mountain-related geometry that I’ve been putting off for the last 30 years or so. It’s about knowing how high you are.

If you’re climbing or descending a mountain, you sometimes want to know roughly how far (vertically) there is still to go. One way to get an idea of your altitude is to use nearby peaks or other points of known altitude as reference points. But how do you judge whether you are above or below another point? It’s not always obvious, and without some sort of rule, the worry is that you’ll make optimistic judgements, leading to disappointment in the long run.

My friend Malcolm once told me that, as a rule of thumb, you should look at the reference point relative to the distant skyline. If the point appears to be above the skyline, you are lower than it, and if it appears to be below the skyline, you are higher than it. I’ve used this skyline-rule ever since, but I’ve never checked how accurate it is.

The fact that there’s any doubt about the rule is because the Earth is not flat. If it was flat, then your line of sight to the (infinitely distant) sea-level horizon would be exactly horizontal, and the rule would work perfectly. If the skyline was made up of mountains, the rule would work perfectly as long as they were as high as your reference point.

But the Earth isn’t flat: it’s a big ball. How does this affect the accuracy of the rule? I used a wet Welsh Wednesday afternoon to find out.

It turns out that the rule is good enough for general hillwalking purposes as long as the reference point is no more than two or three kilometres away (as it usually will be). The errors are smaller if the skyline is distant mountains rather than the horizon at sea level. The rule consistently underestimates your altitude, which, in ascent at least, is probably better than the alternative. Continue reading How high am I ?

## On the fate of my food

Recently I spent a few days walking and wild camping among the mountains of the English Lake District. I was moving on every day and carrying everything I needed for the trip on my back. My rucksack got lighter and lighter as I worked my way through my food supply, and as there was no evidence that my body was getting any heavier, I started wondering where all that mass goes.

The food was largely very dry, so I’m not going to concern myself with water (though some water will be generated as the food is metabolised). Of the dry mass, I will excrete some as faeces, and some in my urine. I’ll secrete some skin oils and will also shed some skin. Then there’s hair, toenails and fingernails, and we mustn’t forget the odd bit of snot and earwax. Quite a trail of debris, really.

But what I started to wonder as I walked was: how much of this dry mass do I breathe out? I must lose some mass with each breath, because outbreaths are poorer in oxygen and richer in carbon dioxide than inbreaths, and carbon dioxide is heavier than oxygen. Can we put some numbers to this loss of mass?

It turns out that we can, and the result surprised me: by a big margin, the most important exit route for carbon appears not to be my bottom, but my lungs.

#### The sums

The volume of gas exchanged on each breath in ordinary breathing is about 0.5 litres. On the way in, the concentration of carbon dioxide is practically nothing (about 0.04%), but on the way out, it’s about 5%. So every 20 breaths results in me breathing out the equivalent of half a litre of pure CO2, which means 40 breaths to breathe out a litre.

Assuming that for every molecule of carbon dioxide breathed out, there is one molecule of oxygen breathed in, then the extra (non-water) mass breathed out is just the mass of the carbon in the carbon dioxide; we can ignore the oxygen.

How much carbon is there in one litre of CO2? One mole of carbon is 12 grams, and one mole of a perfect gas occupies roughly 24 litres in everyday conditions. So one litre of carbon dioxide contains one-24th of a mole of carbon, or about 0.5 grams.

So every 40 breaths, I breathe out 0.5 grams of carbon that originally entered my body as food. At rest, I breathe about once every 4 seconds, so it takes 2 × 40 × 4 = 320 seconds to breathe out a gram of carbon. There are 86400 seconds in a day, so over the course of a day, I’ll breathe out 86400/320 = 270 grams of carbon.

I’ve made all sorts of approximations, so let’s say that I breathe out 200-300 grams of carbon daily.

#### What about the other exit routes for carbon?

This is a much bigger number than I expected. How does it compare to the other output routes for carbon, and do the numbers stack up when we consider how much carbon I ingest?

From various internet sources, fat is something like 75% carbon, carbohydrate is about 40% carbon, and protein is about 50% carbon. On this basis, I’m going to estimate that the dry matter of faeces is in the region of 50% carbon. A typical person produces about 130 grams of faeces a day, of which 75% is water. So the amount of carbon excreted daily in faeces is about 130 × 0.25 × 0.5 = 16 grams.

Wikipedia tells me that we typically produce about 1.4 litres of urine a day, with 6.9 grams per litre of carbon, which gives about 10 grams of carbon excreted by this route per day.

Several internet sources (which may be equally wrong) suggest that we shed about 10 grams of dead skin every day. I don’t know what the moisture content of dead skin is, but it looks like we’re not going to lose more than 5g of carbon by this route every day.

Neglecting the other minor carbon loss processes, this gives a total of about 30 grams of carbon leaving the body daily by routes other than the lungs, compared to 200-300 grams via the lungs.

If these sums are right, I must ingest an equal amount of carbon in my food. This isn’t easy to estimate at all precisely. My diet is largely a mixture of fat, protein, and carbohydrate, but I don’t know what the balance is, and it’ll vary a lot between people. So let’s go to an extreme and suppose that I meet a daily energy requirement of 2500 kilocalories by eating fat only. At 9 kilocalories per gram I’d need to eat 280 grams of grease a day. At 75% carbon, this would be 210 grams of carbon. Doing a similar calculation assuming a diet of undiluted protein or carbohydrate gives a daily carbon intake of 310 grams or 250 grams respectively. So maybe I eat about 250 grams of carbon per day, which tallies reasonably well with the total figure for carbon output that I calculated earlier.

#### Not the Lake District

The picture at the top of the post wasn’t actually taken in the Lake District. It was taken in Glen Feshie, in the Cairngorms. To keep the weight down, I didn’t take my camera on the trip to the Lakes.