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Q&A

Can we actually "blow up" the sea?

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Reading questions of How much TNT needed to blow up Mount Everest and the Moon - and especially the comments below it, I have to ask it:

Can we blow up the sea?

Detonating atomic bomb undersea may look cool but for any sea-level attacks it is not proving anything extra. And It does not "blow up the sea"

So, what is the best way to tactically remove big body of water?

  • It has to happen fast (in matter of hours, one day is maximum)
  • It should be accompanied by some nice effect (boom!)

And, how much would it cost?

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This post was sourced from https://worldbuilding.stackexchange.com/q/8987. It is licensed under CC BY-SA 3.0.

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tl;dr: Not without killing everything.

Let's do some maths and actually figure this out.

The specific heat capacity of water is $4.186 \text{ kJkg}^{-1}$. That means it takes 4186 joules of energy to heat 1 kilogram of water up by one degree.

The average temperature of the surface of the sea is 17oC. It gets a lot colder as you go deeper, so the average temperature overall is more like 0.

The seas contain a volume of 1.3 billion cubic kilometres of water. 1 litre of water = 1 kg. 1 litre of water also = 1 dm3, so there are 1000 litres in a cubic metre and thus a cubic metre of water weighs a ton. (This is assuming freshwater to keep the numbers reasonably nice - salt water is heavier.) Then, there are $1000^3 = 1,000,000,000$ cubic metres in one cubic kilometre. That means 1.3 billion billion or 1.3 quadrillion cubic metres of water and the same number of tons, which in turn is $1.3\times 10^{21} \text{ kg}$ or 1.3 quintillion kilograms.

Now let's heat all that up by one degree.

$$ (1.3 \times 10^{21} \text{ kg}) \times 4186 \text{ Jkg}^{-1} = 5.4418 \times 10^{24} \text{ J}$$

Multiply by 100 so we can heat the water to boiling:

$$ = 5.4418 \times 10^{26} \text{ J}$$

Finally, you need around 6x the energy to actually boil it:

$$ = 3.2651 \times 10^{27} \text{ J}$$

Now while that's not quite on the order of blowing up the Earth, that's a hell of a lot of energy. You're in the perfect region for an asteroid impact. We can work out how big and fast it needs to be:

$$ \text{KE} = \frac{1}{2} mv^{2} $$ $$ 2\text{KE} = mv^{2} $$ $$ 6.5301 \times 10^{27} = mv^2 $$

We can play around with mass and velocity. Let's say this asteroid is a perfect 10km cube with 5000kg/m3 density, thus giving it a weight of $5 \times 10^{12} \text{ kg}$. That means its velocity has to be:

$$ v = \sqrt{\frac{6.5301 \times 10^{27}}{5 \times 10^{12}}} $$ $$ v = 36138898.71 \text{ ms}^{-1} $$

or around $0.12c$. That speed isn't insignificant, and while an impact from an asteroid of this size and speed wouldn't destroy Earth, it would most likely make a massive crater, not vaporise the oceans because the energy isn't distributed easily, and kill all life on Earth.

And that's before we start on the water cycle dropping all that steam straight back where it came from.

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