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Boil your own kettle
by Bruce Ure at 11:34 05/11/04 (Blogs::Bruce)
Fascinating, this.
I was discussing with a friend recently the possibilities of utilising energy generated during exercise to power common electrical items.

He got quite heated, if you'll pardon the pun, when I explained I thought it would take far too much energy to boil a kettle, for one person to be able to make a pot of tea after a swift workout.

I calculated that if you have an 80 percent efficient mechanism for converting a dropping one-tonne mass into heat, you would need to raise it almost 5 metres in order to boil half a litre of water.

Now call me an unfit slob if you will, but there is no way I'm lifting a tonne through five metres for a cup of tea, cunning effort-reducing pulley mechanisms or no.

I'd be interested to see if anyone else can get to about the same figure, and how. It's entirely possible, in fact I'd go so far as to say likely, that my maths is way out.


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Boil your own kettle Bruce Ure - 11:34 05/11/04
Re: Boil your own kettle Simon - 17:11 05/11/04
Specific heat capacity of water is 4.186 joule/gram/°C

Half a litre of water has a mass of 0.5Kg (at 4°C).

Let's assume we're going to boil water that's initially at 20°C, and let's further assume that the mass difference at this temperature is negligible.

So we want to raise the temperature of 0.5Kg of water by 80°C, which needs this much energy:

Energy required = 4.816J * 500g * 80°C Bollocks - typo (pointed out by Bruce)

Energy required = 4.186J * 500g * 80°C

= 192,640 167,440 J

Gravitational potential energy E (in Joules) of a mass = mgh, so to get enough to boil your half litre of water (and assuming a 100% efficient conversion of this PE) you need to drop your 1 tonne weight from a height of:

h = E/mg = 192,640J / (1000Kg * 9.81 m/sec2)

h = E/mg = 167,440J / (1000Kg * 9.81 m/sec2)

= 17.07m

Numerologists please note that this is the same height in metres as the number of Joules needed to boil water from 20°C, if you remove the digits '2' and '0' from 192640 J and then divide the result by 100 (the boiling point of water in °C).

Numerologists should further note that 1964 is the year I was born.

Numerologists should sod right off!

Re: Boil your own kettle Bruce Ure - 18:02 05/11/04
Absolutely correct.

Note how I deliberately took the erroneous value of "1" to be the number of joules required to raise the temperature of 1cc (gram, effectively) of water by 1 degree centigrade.

I also assumed 20 degree start point and before my 80% efficiency calculation had as near as dammit 4 metres. So if we take the correct SHC of 4.186 and multiply by my 4m we should get to your 19.64m, and what is very interesting, is that we don't.

Hang on, I'll dig out me workings (shuffle shuffle):

Distance from which you'd have to drop a 1 tonne mass in order to boil 0.5 Litres of water:

Energy required to raise temp of 1cc water by 1 degree = 1 joule [wrong!].

So energy required to raise 500ccs water by 80 degrees = 40,000 Joules

So we need 40KJ from our 1-tonne mass.

E = mgh so h = E/mg

= 40,000 / (1000 * 9.8)

= 4.0816326530612244897959183673469, otherwise known as "four".

So why did that not work, then? Ah! Your second specific heat capacity is wrong, you've typo'd it to 4.816 instead of 4.186.

If we do 4.186 * 4.08 we get (tappity tap) 17.07888 (as opposed to your 4.816 which gives us 19.64928, perilously close to your 19.64).

The final answer, then, is just over 17 metres, so I can safely say that what I was getting at, which is that if you want me to make you a cup of tea by lifting and dropping a heavy weight a long way you can forget it, still stands.

Thank fuck for that. I was beginning to think we might have to rewrite maths. And I don't know about you but I'm busy all next week.


Re: Boil your own kettle Simon - 18:23 05/11/04
Grrr, damn typos - good job I'm not responsible for anything important like planetary spacecraft :-)


Re: Boil your own kettle Steve - 04:08 06/11/04
Um, Bruce, I didn't quite understand that bit...
Re: Boil your own kettle Bruce Ure - 10:03 08/11/04
Which bit?


Re: Boil your own kettle Steve - 15:27 08/11/04
The bit after "Fascinating this"...
Re: Boil your own kettle Julian Freeman - 14:33 07/11/04
1 Calorie heats 1g water by 1 degree, as you kow.

So that would be a need for 40 kilocalories..

Less than half a bar of chocolate or more specifically this 107kcal of 2 fingers of Christmas Pudding flavour Kit Kat.

So using one finger I could boil a kettle.

Re: Boil your own kettle David Crowson - 16:16 07/11/04
"So using one finger I could boil a kettle."

Now, that I'd pay to see :)


Re: Boil your own kettle Bruce Ure - 10:05 08/11/04
Oh it's calorie! I obviously wasn't paying attention at school. I was probably flicking rubber bands at the girls.


Re: Boil your own kettle Paul Wakeford - 23:37 19/12/04
They have Christmas Pudding flavour Kit Kat??

Why did nobody tell me? Maybe it's not made it to the frozen north.