If hydrogen fusion is ever developed as an energy source, approximately how much energy is available in a cubic foot of seawater?

What I've read is that, as a rough approximation, "a bucket of seawater [close enough to a cubic foot?] contains enough deuterium to power the City of Los Angeles for a year."

Unfortunately I haven't actually seen the calculation. I think approximately one out of every 10,000 hydrogen atoms is deuterium, which is to say it's quite common.

Compared to our present energy needs this is practically an unlimited supply of energy, and clean too! Now if we could just solve the problem of ignition and containment...

No telling exactly which "bucket" you are using, but there are 7.48 gallons to a cubic foot. A bucket holding much more than 2 or 3 gallons is difficult to carry very far! A cubic foot is more likely 3 or 4 "buckets." Not very scientific, but maybe it will lead us somewhere.

According to this site, one liter of water can provide about 2.5 x 10⁶ kilocalories. One cubic of water is a little over 28.3 liters. The available power in a cubic foot of water, therefore, would be around 70,000,000 kilocalories.

[This message was edited by coldfuse on 06-27-02 at 08:13 PM.]

Thanks, Coldfuse. According to your numbers (which appear to be quite reliable, from your link to an article by Isaac Asimov) I have been severely misled by over-hype!

It also says that one liter of seawater yields the equivalent of burning 300 liters of gasoline.

Since 1 kcal = 1.16x10^-3 KW-H, the 7x10^7 kcal contained in a cubic foot of seawater is equivalent to about 85 Megawatt-Hours, which is just shy of 10 Kilowatt-years.

Can anyone complete the calculation to determine how many cubic feet of seawater are required to "power a city" (by complete nuclear fusion) for a year?

According to this article from my local paper, 57,000 megawatts are required to power 57 million homes. That comes to one killowatt per home.

Certainly, our communities include businesses, schools, industrial plants, etc. that consume a great deal of power. But keeping the concept of one killowatt per home is useful.

Professor's 85 megawatt-hours translates into keeping 3,542 homes going for 24 hours at this rate of consumption.