Naked Science Forum
General Science => General Science => Topic started by: sciencepm on 11/02/2021 08:05:33
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Hey guys, I was looking into the sky yesterday and was asking myself: Is ist possible to measure the weight of clouds? If so, how is it done?
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Yes, but!
The density of dry air at 5000 ft is about 1000 gram/cubic meter. The densest cumulonimbus clouds contain an additional 3 gram/m3 of water, so you can estimate the mass of a cloud as being pretty much the same as the mass of dry air occupying the same volume.
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Hey!
just read your question, never thought about the weight of clouds before but was intrigued and happy to see someone already had an answer to this topic. So I do not have an answer but wanted to say that it is a great question to ask. ;D
Yes, but!
The density of dry air at 5000 ft is about 1000 gram/cubic meter. The densest cumulonimbus clouds contain an additional 3 gram/m3 of water, so you can estimate the mass of a cloud as being pretty much the same as the mass of dry air occupying the same volume.
That is so interesting, thank you for sharing your knowledge!!
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Hey Guys,
I think it might be difficult to calculate the weight of a cloud exactly. In itself, the weight of a cloud depends on its size, temperature and density. However, with these factors you can estimate the weight.
Best regards :)
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you can estimate the mass of a cloud as being pretty much the same as the mass of dry air occupying the same volume.
Mass is not weight. Most clouds are weightless, buoyed by the air just as a person is weightless while in water, buoyed by said water.
Weight is a force, and if it required a net positive force to hold them up, they'd fall to the ground due to lack of that force. Some clouds do make it to the ground and are presumably prevented from going further down by minimal force exerted by the ground, and hence these clouds are not entirely weightless.
Edit: My analogy is not entirely appropriate since a human, while made largely of water, is not just water in a certain state, while a cloud is not a discreet object like a human, but rather a region of atmosphere that is above 100% humidity. This latter state can come and go quickly with any change in temperature or pressure, hence the cloud being able to precipitate 'out of thin air', quite unlike the human in the water.
Perhaps the question is asking the weight of the portion of the atmosphere that meets this humidity requirement, but the atmosphere (as a whole) exerts force in all directions on Earth with no significant net force, so it effectively is weightless unless you piled it all on one side. You can multiply pressure by the area of Earth, but it would seem to require a flat-Earth to make a coherent force vector out of that scalar pressure value. Weight is force which is a vector, but pressure is not a vector quantity.
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Weight is the gravitational force acting on a body. Gravity acts on every element of a cloud.
Weight is what a spring balance measures. If I stand on a spring balance I am "buoyed up" by the springs but,sadly, they indicate that I am far from weightless.
If gravity did not act on the molecules of air and water, we would have no atmosphere.
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Weight is what a spring balance measures. If I stand on a spring balance I am "buoyed up" by the springs but,sadly, they indicate that I am far from weightless.
Try putting your bathroom scales at the bottom of a swimming pool.
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What a spring balance measures, plus any buoyancy correction, of course. Something to do with Archimedes. My weight hasn't changed because I'm still in the earth's gravitational field, but I've displaced a few liters of water that used to be held up by the rest of the pool.
W = mg and neither m nor g is altered by the surrounding water.
If you don't believe me, weigh the swimming pool with and without me.
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My weight hasn't changed because I'm still in the earth's gravitational field,
...OK so, your weight doesn't change when you are in free fall then?
It's a matter of definition(s).
An object's "weight in air" is a well known concept.
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It's an interesting subject for philosophical discussion!
On the one had we have the definition w = mg so weight defined as the force vector acting on a body due to gravitation. But Einstein says (and who are we to disagree?) that any other acceleration is equivalent to gravitation. If you put your spring balance in the seat of a plane or a car you can measure the force vector under the sum of gravitation and an accelerating force that you won't allow me to call centrifugal. Now fly a loop, parabola or orbit and note how your maps, pencils and lunch are no longer held to the table with a net force. But m hasn't changed and you are still fairly close to the planet, so neither has gg. What has happened is that ga = - gg so gtot = 0
In the case of a cloud, we have a vaguely definable mass that is not accelerating (much) so ga = 0 and w = mgg. If there were no other air around it, we could indeed measure w with a spring balance but since it is surrounded by a medium of pretty much the same density (see reply #1) there is an upward displacement force of almost equal magnitude.
Watch this space for a discussion of orbiting spherical cows in a vacuum. Bullshit in free fall.
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Interesting. Thanks for the answers so far.
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I don't believe there is a standard way to measure the weight of a cloud. But if you can determine the volume of the white sky(cloud) in liters.......and if you assume the density is constant thru-out......Then filter the water out of one liter of it.......you can multiply that water weight by the volume of cloud in liters.
That will be the weight of the cloud.
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@sciencepm
Hello!
😊
Very Interesting Question.
👍
Hoping this might clear out a lil smog & fog & might just help You to finally see the silver lining of the Clouds.
☁️🌥️☁️
https://www.usgs.gov/special-topic/water-science-school/science/how-much-does-a-cloud-weigh?qt-science_center_objects=0#qt-science_center_objects
Thanks & Credits - USGS
🙏
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Yes, but!
The density of dry air at 5000 ft is about 1000 gram/cubic meter. The densest cumulonimbus clouds contain an additional 3 gram/m3 of water, so you can estimate the mass of a cloud as being pretty much the same as the mass of dry air occupying the same volume.
There's one more interesting wrinkle though, often the weight of the cloud can be negative!
The reason is that water vapour is H2 + O, whereas air is mostly nitrogen which is 80% N2 and 20% O2.
Very, very roughly N and O atoms weigh the same (7 and 8 units), and H is much lighter (1 unit), so H2O vapour is a bit more (weighs 10 units) than half the weight of N2/O2 gas (about 14 units).
Because the mass of gases is proportional to the weight of the molecules that make it up, those grams of water vapour are lighter than air! It's only water droplets that add to the weight, and they are often very small and light.
While clouds can form from upward winds, often they get up there by simply being lighter than air, either from the extra water vapor or from being warmer than the surrounding air.
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The distinction is between humid air (effectively transparent with mostly mono and dimer H2O) and a cloud (reflective and absorbent, with mesoscopic droplets) is pretty obvious. Even the most apparently static clouds are actually dynamic, with falling droplets or ice particles being replaced by rising and condensing vapor: the shape of a "cloud" is actually the shape of the condensation isotherm.
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There is no hard distinction though, there's a continuum. Any cloud is a mixture of dry air, water vapour and fog, rain or snow etc. If clouds were always systematically much heavier than air like you've been implying they would be a lot rarer.
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There is no hard distinction though, there's a continuum.
No.
There's a phase change.
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Oh really? So all the water vapor suddenly condenses out at a particular temperature does it? It really doesn't- there's a shifting equilibrium.
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Oh really? So all the water vapor suddenly condenses out at a particular temperature does it? It really doesn't- there's a shifting equilibrium.
Straw man.
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Condensation is fairly rapid at or below the dewpoint. This is a function of local pressure (altitude) and relative humidity. If you look at "active" (developing) clouds you will see that they have flat or concave bottoms and well-defined edges. The classic cumulus cloud drifts downwind, first forming and growing as the moist air mass condenses, then merging or dispersing.
A truly dynamic equilibrium cloud is called a lenticular, formed at the top of a standing wave in the lee of a mountain. Although the shape remains fairly constant, the air inside is usually moving at 30 - 100 knots: the cloud is continuously forming at the leading edge (as rising air passes the dewpoint) and dispersing at the trailing edge where the moist air is descending.
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One estimate of cumulus cloud density is as a density of about 0.5 gram per cubic meter. A 1 cubic kilometer (km3) cloud contains 1 billion cubic meters. Doing the math: 1,000,000,000 x 0.5 = 500,000,000 grams of water droplets in our cloud.
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Assuming you mean kg/m3, that's still a very high cloud - about 20,000 ft. You might encounter cumulus at that height in Texas but it's about the top of a very big cumulonimbus in the UK!