Apologies for the hiatus recently, it's been a busy time (when isn't it). I hope to get back to talking about experiments soon, but for now I wanted to write up a quick problem I thought up a while back. The question is this: how long does a molecule of H2O on earth remain in the liquid state, on average?

I'm going to treat this purely as an order of magnitude problem. I'm also going to have to start with one assumption that is almost certainly inaccurate, but makes things a lot easier. I'm going to assume perfect mixing of all of the water on earth.

Given that assumption, I really only need to figure out two things. The first is how much liquid water there is on earth. The second is now much liquid water leaves the liquid phase each year. Let's start with the total amount of liquid water on earth. This is relatively easy to estimate. I happen to know that about 70% of the earth's surface is covered in water. Most all of that is ocean. To an order of magnitude, the average depth of the ocean must be 1 km, as it is certainly not 100 m or 10 km [1]. For a thin spherically shell, the volume of the shell is roughly

\[ 4 \pi r_e^2 \Delta r \]

where r_e is the radius of the earth. Thus, the total volume of water on the earth is

\[.7*4 \pi r_e^2 (1 km)\]

Now, we need to figure out how much H20 leaves the liquid phase every year. To an order of magnitude, it rains 1 m everywhere on earth each year, it's not .1 m or 10 m [2]. I'm going to ignore any freezing/melting in the ice caps, assuming that is small fraction of the water that leaves the liquid phase each year. Since we have a closed system, all the water that rains must have left the liquid phase. So, on average, the total volume of water that leaves the liquid phase is

\[4 \pi r_e^2 (1 m) \]

Thus, the fraction of liquid water that changes phase per year is

\[ \frac{4 \pi r_e^2 (1m)}{.7*4\pi r_e^2 (1 km)} = .0014 \]

This means that, given my assumption of perfect mixing, in somewhere around 1/.0014 = 700 yr all of the water on earth will have cycled through the vapor phase. Since we're only operating to an order of magnitude, I'll call this 1000 years. This is the answer to our question if every molecule has been in the vapor phase once in 1000 years, then we expect a molecule to stay in the liquid phase for 1000 years

[1] According to wikipedia, this is really about 4 km, so we're underestimating a bit.

[2] According to wikipedia, this is spot on (.99 m on average).

## Sunday, July 10, 2011

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I expect the standard deviation of a random selection of molecules, interviewed and asked "the last time you were in the liquid phase, what was the duration?" would be very, very large. Further, I'm speculating the probability distribution is not normal for said randomly selected water molecule. Lognormal possibly?

ReplyDeleteI wonder how much water each year is created (via combustion, respiration, and other chemical reactions which liberate water) and destroyed (via photosynthesis, hydrolysis, etc. That could be an interesting back-of-the-envelope, order of magnitude calculation.

ReplyDeleteIn the same vein as what King of the Road said, it seems like there's a large population of water molecules diffusing around in the deep ocean which wouldn't reach the surface in quite a while (although maybe this time would be significantly shortened by ocean currents, so this is more of an upper limit). At any rate, I get a diffusion time of ~ 4 million years by assuming a mean free path based on the inverse cube root of the number density of water times a typical velocity at T=300 K (which yields a diffusion coefficient of 10^-3 cm^2/g). Dividing (4 km)^2 by this gives the 4 megayear timescale.

ReplyDeleteKing of the Road and Forbes are both quite right in that the assumption of perfect mixing is probably completely wrong. However, as Forbes says, I'd expect that the time scale is not a purely diffusive one. Still, if we use that as an upper limit and my estimate as a lower limit, I'd be inclined to take a geometric mean and come out with ~65000 years as the timescale for the average mixing.

ReplyDeleteAs for what type of probability distribution we get . . . I'd be shocked if it were normal. Lognormal is as good a guess as any, I'd imagine.

Finally, Blaise has an interesting question, about how much is created or destroyed, but I don't think I can even write down all the (common) reactions that do either, so I don't know that I can start that. Still, I'd posit that we're in a zero sum game at the moment (at least on human timescales), so we probably only need to estimate creation or destruction. If someone wants to make a list, I'd give the estimation a try!

That would be a fun problem, Blaise. My first guess would be that it's all plants and bacteria and the like. So then it's just how much water does a plant need to make the food that it eats in one day.

ReplyDeleteAnother related problem could be how much carbon dioxide is taken in (or released) by plants. Then we could compare that to how much people put out.

Hang on, I don't think you've answered the question. Question was "how long does a molecule of H2O on earth remain in the liquid state, on average?"

ReplyDeleteWhat you worked out was: how long is the average cycle of liquid to vapour to liquid again? The remaining question is: what proportion of that cycle is the liquid phase, and what proportion is the vapour phase? (Or: what proportion of the Earth's water is vapour at any given moment?)

At the levels of accuracy you're working at, the vapour phase is likely to be vanishingly small, granted. Still, it's fun to pick nits!

The mixing time of the oceans is on the order of 1000 years or less. See, for example, Matsumoto (2007).

ReplyDeleteThat's an interesting question, chinaphil. I hadn't quite thought of it in those terms. Let's see if I can quickly estimate that. Now, to a first approximation, we can find the mass of the earth's atmosphere by taking the pressure at sea level, multiplying by the earth's surface area, and dividing by g, the acceleration of gravity at the earth's surface. This has the advantage of being quick and easy, and the disadvantage of not being accurate (need to account for gravitational force varying with height, not all of earth at sea level, etc). Still, let's try it.

ReplyDelete\[m_{atm}=(1\cdot 10^5 Pa)(4\pi(6500km)^2/(10) =5.3 \cdot 10^18 kg\]

According to wikipedia, this is only 3% larger than the actual mass of the atmosphere, not bad! Next, we'll need to know how much of that is water. Again according to wikipedia, water takes up about .4% of the atmosphere. That's still a whopping 2.1*10^(16) kg!

The total mass of the water on earth is simply the amount (found in the original post) times by the density:

\[(.7)4\pi r_e^2 (1km)\rho=(.7)(4\pi)(6500km)^2(1km)(1000kg*m^{-3})=3.7\cdot10^{20} kg \]

Of course, if I had been clever to start with, I could have done a ratio immediately, and canceled some things:

\[\frac{m_{atm}}{m_{liq}}=\frac{.004*P_{atm}}{.7*1km*1000kg*m^{-3}*10m*s^{-2}}=5.7\cdot10^{-5}\]

So the fraction of mass of water in the atmosphere compared to the liquid phase is small. Very small! As you surmised, it spends, for our purposes, a vanishing amount of time in the vapor phase, ~1/20th of a year, if my original estimate for the cycle time is correct.

Hmm. Looks like my tex is a little funny in a few places, and I can't seem to edit the comments on here. The first equation should finish:

ReplyDelete\[m_{atm}=5.3\cdot10^{18}kg\]

And should have another closing parenthesis somewhere in it!

The second equation, which got truncated (at least in my browser), is

\[m_{liq}=3.7\cdot10^{20}kg\]