Energy can not be produced or destroyed: it must go somewhere (1st law of thermodynamics). Often when we build a machine to do work, some of the work is useful (e.g moving the car forward), but some of it is not useful (e.g. making grumbling noises or getting hot).
When we talk about efficiency, we’re talking about the ratio of the useful work performed by a machine or process to the total energy expended.
Energy efficiency = useful work performed (J) / total energy expenditure (J)
Contents
Useful work to heat water
How efficient is my kettle in boiling 500mL of water?
Firstly, we work out the energy required to raise temperature of the water from 20 degrees C to 100 degrees C. For this we need the amount of water (assumed to be 0.5 kg since the density of water is very close to 1000 kg/ m3).
We also need the specific heat capacity of water (4200 J/(kg.K) ). Note that we can use degrees C for the temperature difference (instead of Kelvin), since we’re only looking at the difference in temperature, and these scales are linearly offset from one another (i.e. an 80 degree C change is the same as an 80 Kelvin change).
So, our formula to calculate the amount of energy required to heat the 500 mL of water up from 20-100 degrees C, is
Energy = mass (kg) x Specific heat capacity (J/(kg.deg C)) x change in temperature (deg C)
or
Energy = m . Cp . Δ T = m . Cp . (T1-T2)
=0.5 kg x 4200 J/(kg.deg C) x (100-20) deg C
= 0.5 x 4200 x 80
= 168,000 J, or 168 kJ
So the energy required to heat the water is 168 kJ.
Efficiency of my Morphy Richards kettle
From an earlier post, I worked out that the kettle actually used 197 kJ to boil the kettle.
Remember that:
Energy efficiency = useful work performed (J) / total energy expenditure (J)
So in this case:
Energy efficiency = 168 kJ / 197 kJ
= 0.853, or 85.3 %
So, according to this quite simple calculation, this particular kettle is 85.3 % efficient in boiling my 500mL of water.
Some notes…
It took 74 seconds until I noticed that the water had just boiled, but the switch didn’t trip until 90 seconds…how efficient does this impact on the expected efficiency if I didn’t stop it early?
Also, what are the assumptions I’ve made in this? well first of all, I’ve assumed that the power consumption was constant throughout the heating time – and from observation of the power meter, it clearly was not constant. Ideally I should plot the power drawn over the heating time, then integrated this to get a truer value for the actual energy consumed.
I should note also that my measuring of the 500 mL of water and of the time taken was all rather crude, and to my knowledge the power meter is not particularly accurate!
We can calculate the energy to boil water, and the energy consumed by a kettle, but what about making toast? Again, we can work out how much energy is used, but how much energy is needed to toast bread? This is somewhat more complex…in these cases we need to take other measurements, of temperature for instance to work out what might be relevant for that particular product case. Here’s an example of a study looking into the heat transfer in hair dryers and irons, and here’s an article about a study that analysed the power/temperature conditions used to make improve toast making.
The Chris Hoy challenge
How many AA batteries would it take to have more energy than Chris Hoy in a sprint burst?
Let’s say we have a AA Duracel battery which is 1.5 V rated at 2500 mA.h (this is a good AA battery!).
To understand how much energy it has stored in it, we multiply 2500 mAh by 3600 (to convert the hours to seconds), which gives us 9,000,000 mA . seconds (or 9,000 Amp.seconds).
If we multiply this by 1.5 V, we get 13,500 Amp.Volt.seconds (or Joules), that’s 13.5kJ.
Remember from an earlier post that Sir Chris can churn through 92kJ before he is spent (in a full sprint)…
…that’s the equivalent of nearly 7 x AA batteries…hmmm, that doesn’t seem like much…but there’s a catch…
The AA batteries are only really rated at 50 mA…so there’s no way they could compete with that power output! They would get very hot (and may explode!).
If we drew 50 mA from each of the batteries, they would last 50 hours (50 mA x 50 hours = 2500 mA.h)…they would come in a distant second in the race if Chris Hoy finished in only 40 seconds!
Some notes…
clearly there are many assumptions here, the points really are that a) batteries have a limited current capacity that is worth being aware of right from the start, and b) Chris Hoy is an incredibly powerful individual!
