MadSci Network: Earth Sciences |

Posted By:

Area of science:

ID:

Hi Robert:

If there's one thing an oceanographer loves, it's a back of the envelope calculuations! I'll use scientific notation because the numbers are going to be pretty large.

Lets's take your 10 by 10 kilometer square patch of ocean, so a hundred
square kilometers. Next, lets say that we want to decrease the temperature in
the top metre of water by 3 degrees celcius. The amount of water in that 10 km
by 10 km by one metre is 10000 metres times 10000 metres times 1, or
10^{8} cubic metres. A cubic metre of water weighs 1000 kilograms
(we'll ignore salt in this example), so that volume of water will weigh
10^{8}m^{3} times 1000 kg/m^{3}, which equals
10^{11} kg. That's a lot of water, it's equivalent to the mass of
almost 50 000 statues of liberty.

The amount of energy required to heat or cool a parcel of water is calculated
by multiplying the specific
heat by the mass of water, and the temperature change. The specific heat
of water is 4.186 kJ/kg ^{o}C (in other words, 4.186 kilojoules
must be removed to cool one kilogram of water by one degree).

So in our example, the amount of energy required is 4.186 kJ/kg
^{o}C times 10^{11} kg times 3 ^{o}C, which works out to
1.3x10^{12} kJ. That is a lot of energy - Little Boy, the atomic bomb
dropped on Hiroshima, released about 1.8x10^{10} kJ, so we are talking
about an amount of energy roughly equivalent to about 70 Little Boys.

Moving on, the heat of solution of ammonium nitrate is 26.2 kJ/mole (i.e.,
26.2 kJ is removed for every mole
of ammonium nitrate that is dissolved). So we will need 1.3x10^{12}
÷ 26.2 kJ/mole, which equals 4.8x10^{10} moles. The molar mass of
ammonium nitrate is 80.04 grams per mole (equivalent to 0.08004 kilograms per
mole), so that works out to a grand total of 4.8x10^{10} moles times
0.08004 kilograms/mole, which equals 3.8x10^{9} kilograms of ammonium
nitrate required, or a little under 4 million metric tons.

I think you'll agree that that is a lot of ammonium nitrate! I can see a few problems with this plan:

- How many ships would it take to carry that much? I couldn't find any decent
estimates of what the cargo capacity of your average bulk carrier is, so we'll
make a conservative guess - some reasonable dimensions for a Panamax freighter
(one that can just fit through the Panama Canal) are something like 300 by 30
metres. We'll guess that their cargo holds are also something like 30 metres
high, which gives us a cargo-carrying volume of 270000 cubic metres (which will
probably be high, because we're ignoring things like fuel tanks, engine rooms,
and the fact that ships are not rectangles). The density
of ammonium nitrate is 1700 kg per cubic metre, so each ship will be able to
carry a maximum of about 4.6x10
^{8}kilograms, so we'd need at least 9 ships to carry that much. - What would that cost? The price per ton varies quite a lot, between about 80 and 220 dollars in the last four years for fertilizer grade ammonium nitrate (which is far from pure). Even if we take the lowest price, we'd be looking at a bill of about 800 million dollars.
- Ammonium nitrate is explosive, and has been involved in numerous large explosions, with significant loss of life (and with far fewer amounts that we're talking about here).
- We'd probably have to do this to a
*lot*more that a 10x10 km patch of ocean. For instance, the recent hurricane Katrina lost some energy travelling over a considerable portion of the tip of Florida, but once back over warm waters it spun up again to a Category 5 Hurricane.

So, in short, I think we're just going to have to put up with the weather for the forseeable future.

I hope that helped!

Rob Campbell, MAD Scientist

Try the links in the MadSci Library for more information on Earth Sciences.

MadSci Network, webadmin@madsci.org

© 1995-2005. All rights reserved.