How heavy is a cloud? An activity for using maths in science lessons
This is another blog entry about using maths in science teaching. This one is a calculation exercise to bring home the reality of something children (and adults) probably don't spend much time thinking about, but is absolutely fascinating - the mass of a cloud.
Ask a child to draw a cloud and they will probably draw something 'fluffy'. Ask them where clouds can be found and they will (hopefully) tell you, "In the sky." All very routine and ordinary. The next question is the real clincher however. Asking them if clouds are heavy or light will possibly produce some thought, probably followed by an assertion that they are light, because they are floating in the sky. This is faulty reasoning, though understandable because we traditionally think of heavy things falling to the ground. As the following exercise demonstrates, clouds have an enormous mass, despite the fact that they float in the sky. (What they do have is a very low density, and it is this which enables them to float above the denser, drier air below).
Assumptions and simplifications
Real clouds have somewhat irregular shapes. What's more there are different types of cloud, and they don't have fixed sizes. Hence for the calculation I've chosen a particular type of cloud, and assigned it a reasonable size for its type. I've chosen a cumulonimbus, or storm cloud. These are typically seen in Britain after hot summer days and give warning of probable rain, with thunder and lightning. They are tall and usually described as anvil-shaped, which refers to the way they get wider at the top. For simplicity in our calculation I'm going to assume the cloud is actually a simple cylinder.
There is some data I've looked up which is needed for our calculation:
Nominal diameter of a cumulonimbus cloud = 10km, so the radius = 5km (this is a conservative estimate - ie. on the small side).
Nominal height of a cumulonimbus cloud = 10km (again, on the smallish end of the scale).
(The data above is from cloudatlas.wmo.int/en/observation-of-clouds-from-aircraft-descriptions-cumulonimbus.html and en.wikipedia.org/wiki/Cumulonimbus_cloud)
Nominal density of a cloud = 0.5g/m3
(this is actually the given density of a cumulus cloud, which is a bit less dense than a cumulonimbus cloud, but I couldn't find any other suitable data so we'll go with it. The final answers will all be somewhat lower than true values therefore, but this doesn't matter too much as the point is more about the magnitude than the actual numerical values).
The calculation
Volume of our cylindrical cloud = πr2h where r = radius of the circular base and h = height of the cylinder.
= π x 52 x 10
= 785.40 km3
We need to convert this to m3 (as this is the unit of volume in our density figure)
1 km3 = 109 m3
785.40 km3 = 7.854 x 1011 m3
As each m3 of the cloud has a mass of 0.5g, our cloud's mass = 7.854 x 1011 x 0.5
= 3.927 x 1011 g
This is a big mass, and we can put it into more appropriate units:
1 kg = 1000 g and 1 tonne = 1000 kg
Hence, 3.927 x 1011 g = 3.927 x 108 kg = 3.927 x 105 tonnes
That last figure is 392,700 tonnes; not far off half a million tonnes. That's heavy! That it floats above the ground should not really be surprising, but there is a general tendency (until we really think about it that is) to view gases, vapours and clouds as having either no mass or very little mass. The reason for the deception it that their density is low - in other words their mass is very spread out.
Extending the calculation
The mass calculated above is for water in a cloud in which it is in the form of liquid droplets but these are tiny and widely dispersed (hence the low density of the bulk cloud). If the liquid water were to be brought together into one place its volume would be 3.927 x 105 m3 (because the density of liquid water is 1 tonne/m3).
Suppose the cylindrical cloud were to drop all this water over an area corresponding to its 'footprint' - ie. its circular base. What would the depth of water dropped be?
We can do this by imagining a cylindrical container into which the cloud drops its water. We know the volume of the water (just done, above); we can work out the area of the cylinder's base - it's the same as the area of the cloud's base and we've already worked it out above when we calculated the cloud's volume. It's 78.54 km2.
Using the volume formula for a cylinder, Volume = πr2h, we actually want to know h (that's the depth of the rainfall).
Rearranging gives h = Volume / πr2 which is the same as volume/area.
Hence h = 3.927 x 105 m3 / 7.854 x 107 m2 = 0.005m
This is half a millimetre! A seemingly tiny depth of water. And yet, if we scooped up all this water, (which, remember, fell over an area of 78.54 km2) it would be over 300,000 tonnes, as our earlier calculation showed. The tiny depth and the huge mass seem hard to reconcile, don't you think?
What's the point?
There are some useful learning points here. One is the surprising mass of clouds, and the fact that they can float above drier, less dense air. This can be connected to the fact that the air itself has mass, something children will not intuitively grasp. This could have relevance when teaching about floating and sinking, although as we normally do this in the context of objects floating on water that may be a step too far.
The other learning points are about the surprising results we can get when dealing with very big or very small numbers, and the way that the squaring of length to get area, or cubing to get volume, can magnify numbers in an often unexpected way. So for instance, the towering cumulonimbus cloud, with its mass of almost half a million tonnes, and the volume of water within it, still only results in a seemingly meagre half a millimetre depth of rainfall. It perhaps shouldn't be surprising, but it is.
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