Re: Could we use endothermic reactions to reduce hurricane strength?

Date: Wed Oct 5 03:15:33 2005
Posted By: Rob Campbell, Postdoctoral researcher, Biological Oceanography
Area of science: Earth Sciences
ID: 1127625804.Es
Message:

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 108 cubic metres. A cubic metre of water weighs 1000 kilograms (we'll ignore salt in this example), so that volume of water will weigh 108m3 times 1000 kg/m3, which equals 1011 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 oC (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 oC times 1011 kg times 3 oC, which works out to 1.3x1012 kJ. That is a lot of energy - Little Boy, the atomic bomb dropped on Hiroshima, released about 1.8x1010 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 1.3x1012 ÷ 26.2 kJ/mole, which equals 4.8x1010 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 4.8x1010 moles times 0.08004 kilograms/mole, which equals 3.8x109 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 4.6x108 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!