Re: Boil your own kettle
by Simon at 17:11 05/11/04 (Blogs::Bruce)
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!
--
simon

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