Re: Boil your own kettle
by Bruce Ure at 18:02 05/11/04 (Blogs::Bruce)
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):

--8<------------------------------------------
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".
--8<------------------------------------------

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.

:bu:

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