Sunday, October 17, 2010

Making numbers stick - desalination, melting, and boiling

I'm always looking for new ways to make physical numbers memorable.
  • One method is to use a picture (eg nuclear waste, per person, per year) [page 170, SEWTHA]
  • Another general rule is to choose units such that the answer to be remembered comes out between "1 unit" and "200 units", because smallish numbers are easier to remember.
  • Another idea is to reexpress the quantity in completely different units, which may be more familiar and more memorable, as illustrated in this earlier post where I converted an incomprehensible 20 x 1022 J into a hopefully more human-friendly ocean temperature rise of 0.2 degrees C.

I'd like to give a few more examples of this trick, all converting unmemorable numbers in awkward units into temperature rises.
Example 1: the cost of desalinating sea water. [This method of making it stick came from Jim Gill, Chancellor of Curtin University, via Sam Wylie.] In SEWTHA (p 93), I report that desalination has an energy cost of 8 kWh per m3. A nice way to make this number more meaningful is to work out what temperature rise you would get if the same energy were put directly into heat in the same volume of water. The answer is ((8 kWh) / (1000 litres)) / (4.2 ((kJ / C) / litre)) = 7 degrees C.
This result brings home that if the desalinated water is going to be used for a shower or for cooking, the energy cost of the desalination is fairly tiny compared to the energy that will be used later in the water's lifecycle.

Example 2: melting ice. The latent heat of melting of ice is 6 kJ/mol, or 333 kJ per kg, a quantity I have never been able to memorise... until now! Using the same trick as above, we can convert this into an equivalent temperature rise, by dividing by the heat capacity. The answer is "the latent heat of melting of ice 'is' 80 degrees C".
I don't think I'll forget that number! It really brings home why mountaineers spend so much time melting snow. The energy to melt the snow is roughly the same as the energy to bring the melted snow up to boiling point!
Example 3: vaporizing water. We can apply the same trick to the heat required to vaporize water (2258 kJ/kg). The answer is (2258 kJ/kg) / (4.2 kJ/kg/C) in C = 538 C. This number violates the "should be between 1 and 200" rule, so it is not super-memorable, but it is quite striking, isn't it - whereas near-boiling water is 373 degrees above absolute zero, the energy required to actually boil it is equivalent to another 538 degrees of temperature rise! Maybe the best way to obey the "1-200" rule is to reexpress this heat once more, comparing it to the energy required to bring the water from 0 to 100 C. It is bigger by a factor of 5.4. So "the time for the kettle to boil itself dry is about 5 times the time taken to bring it to the boil".
Here ends the lesson.


David B. Benson said...

The mountaineer, if high enough up, won't be able to have 80 degree Celcius water before it is boiling. I've had so-called boiling hot water tea several times in the mountain tent; distinctly tepid and most welcome.

Crno Srce said...

This is an important question when trying to communicate complex ideas that involve numbers, especially to the average numerophobic :-) I like the idea of relating the quantities to more day-to-day experiences.

I wonder if the comparison of desal cost with water heating cost is a little misleading. Even if your water is very cold (near zero) and you're aiming at reaching 65 degrees for your hot water, 7 degrees of temperature rise is roughly 1/8th of that, or 12% - this is not insignificant in terms of efficiency gains! It's like using 12% more hot water.

From my point of view in Australia, the desal plants are generally closer to the water users than the dams, so perhaps a comparison in terms of energy cost for pumping water X kilometres or whatever, would be more interesting :-)

However, thinking about it now - I have to admit that I think this number is going to stick with me!