Clouds are visible suspensions of billions of tiny water droplets and/or ice crystals, and are formed when air rises and cools to its dewpoint, causing its water-vapour to condense out. You may wonder how a cloud can therefore remain suspended, given that it’s composed of water, but the reality is that each tiny particle acts independently, remaining suspended by the buoyant force of the air. But what if you were to measure the total weight of a cloud? What would it weigh? The answer is staggering, but not in the way you would expect!
Firstly, a “cloud” does not weigh anything as it is not a solid mass in itself but is a collection of independent water particles (droplets or ice crystals). Each particle (measuring less than a millimetre) has a weight and volume, which is perfectly balanced by buoyant force of the air. The particles are separated by air, so if we take a cloud to be both the water particles plus the air in between them, then we can start to think about how much mass we’re talking about.
I took the above photo of a cumulus mediocris cloud early this afternoon. Its base was at 2,500 ft (750 m) above ground level and was around 1 km wide. It looked around twice as wide as it was high, so it was around 500 m deep and we can estimate its volume by treating it as roughly a half sphere of radius 500 metres. Its volume is therefore
V = 2/3(Pi)R³ = 261.8 million m³
Cumulus clouds have been measured to have an average water-loading of around 0.6 g/m³, so the total water contained in this volume is
261,800,000 x 0.0006 kg = 157,080 kg of water
That’s 157, 080 litres of water. Not that much when you think about it, enough to fill about 1050 bathtubs. What is surprising is that the huge majority of the mass of such a cloud is actually due to the AIR in the cloud, not the water! There is around 1000 times more air than water! Crazy, but true. So how much air is there? Let’s look at its density.
Station pressure at nearby Casement Aerodrome (92 metres) at the time was 997 hPa, the temperature was 15 °C and the dewpoint was 7 °C. This makes the air density at this level 1.200 kg/m³.
This air rose and cooled at a rate of 9.8 °C/km until it saturated, so at this point (the cloud-base) it will have had cooled by about 7.5 degrees. Temperature and dewpoint were therefore around 7.5 °C, and with a pressure of around 910 hPa at this level the air density was around 1.124 kg/m³.
As the air was now saturated it would continue to rise but cool at a slower rate of around 6.5 °C/km, falling to around 4 °C at the top of the cloud. The pressure here was around 860 hPa, so the density was about 1.075 kg/m³.
The average density of the cloud through its depth was therefore (1.124 + 1.075)/2 = 1.100 kg/m³. Note that this is way, way higher than the mere 0.0006 kg/m³ water-loading. Multiplying this density by the volume of the cloud we get the total mass of the air in the cloud to be around 288 million kg, or 288,000 tonnes! That’s about the weight of around 528 fully-laden Airbus 380 aircraft or around 41,150 male African elephants!
So taking the weights of the water and air again we have:
Water: 157 tonnes + Air: 288,000 tonnes -> Total: 288, 157 tonnes
So in our small, single cumulus cloud, a staggering 99.95% of the mass of the cloud is from the air between the water droplets, with only 0.05% coming from the water itself. If you take a much higher and thinner cloud, such as cirrus, those wispy clouds composed of ice-crystals, we get an even lower ice-loading per volume ratio, but now air-density is only around 30-40% of what it is in low cumulus clouds, so the ice to air ratio will be different.
Nevertheless, it is still staggering to think that something as common and annoying to us as clouds are in fact composed mostly of thin air! Well I’ll be damned!