Change in Mass of Hurricane Harvey's Rain

Game Maker
Messages
4
Reaction score
4
Homework Statement
Energy plus mass is always conserved. Hence molecules of vaporized water will have a greater mass than molecules of liquid water, due to the heat of vaporization. Similarly, potential energy will increase the mass of a molecule at high altitude compared to when the molecule is close to the earth’s surface.

In 2017, hurricane Harvey stalled over Texas dumping unprecedented amounts of rain over a vast area. The heaviest rain totals averaged 25 inches and covered an area 150 miles in diameter. Assuming the rain fell from a height of 25,000 ft, how much mass was given up by the water vapor as it condensed and fell to earth.

Assume Texas is flat and at sea level.
Use a water density of 1gm/cc
Ignore the energy required to change the temp of liquid water.
Relevant Equations
e=m⋅C^2, e=Δh⋅a⋅m , e=∆Hvap⋅m
e = energy, m = mass, C= speed of light, Δh=change in altitude (height), a= acceleration of gravity
∆Hvap= latent heat of vaporization
First I have to explain that this is not actually a homework problem. It is a problem I created to use in a sort of game. If I made a mistake, it tends to ball up the game. So I am looking for someone to double check my answer, and to point out any errors in my statement about physics. I prefer not to provide the full details of my solution at this time, because I have found that it is sometimes possible to lead others down a wrong path without errors being caught. If there is a better forum for this, please let me know. Thank you in advance for your help.

My answer: 748kg
 
Physics news on Phys.org
Game Maker said:
I prefer not to provide the full details of my solution at this time,
Then why should we provide ours? That's not how this forum works. If you want us to check you work you need to show it. If you want us to just solve your problem for you, then fagedaboudit.
 
Why should you help me at all; even if I provide my solution?
But if that the way this forum works, then I don't mind posting my solution, even if it does provide a less rigorous check.

150mi diameter ⋅ 1603.9m/mi = 240585m in diameter
Area of rain = 3.14 ⋅ 240585^2/4 = 4.55E10m^2
Depth of rain = 25 in ⋅ .0254 m/in = .635m
Volume of rain = 4.55E10m^2 ⋅ .635m = 2.89E10m^3
Mass of rain = 2.89E10m^3 ⋅ 1g/cc ⋅ 1E6cc/m^3 .001kg/g =2.89E13kg

Energy for vaporization = 2.89E13kg ⋅ 2.257E6J/kg = 6.52E19J
Potential energy = 25000ft ⋅ 0.3048m/ft ⋅ 9.8m/s^2 ⋅2.89E13kg = 2.16E18J
Total energy = 6.52E19J + 2.16E18J = 6.736E19J
Mass lost = 6.736E19J/(3E8m/s)^2 = 748kg
 
  • Like
Likes DEvens
Yes, near enough.

Heh heh. A "gentle rain" is about 600 times the total ordinance dropped during WWII.
 
  • Like
Likes Game Maker
Game Maker said:
Similarly, potential energy will increase the mass of a molecule at high altitude compared to when the molecule is close to the earth’s surface.

Wait, that potential energy (2E18 J) wasn't "lost." It was used to make the velocity of the raindrops. And once they reached terminal velocity, it was used to heat up and stir up the surrounding air.
 
gmax137 said:
Wait, that potential energy (2E18 J) wasn't "lost." It was used to make the velocity of the raindrops. And once they reached terminal velocity, it was used to heat up and stir up the surrounding air.

We are in agreement that the potential energy was not lost. It was transferred from the water to the surroundings. My understanding is that in doing so, it lost mass and the surroundings gained an equivalent mass. This would be similar to the concept that a battery or capacitor has a greater mass when it is charged than when it is discharged. I would appreciate feedback on whether or not my understanding is correct.
 
Game Maker said:
We are in agreement that the potential energy was not lost. It was transferred from the water to the surroundings. My understanding is that in doing so, it lost mass and the surroundings gained an equivalent mass.
No, I don't think this is the right way to look at it. The liquid at 25000 feet lost altitude; the air between the ground and 25000 feet gained velocity.

A battery is different because the energy gained when it is charged is "stored" in the chemical bonds (as opposed to kinetic energy), and the binding energy has a E=mc^2 mass.

You may want to read this
https://en.wikipedia.org/wiki/Energy–momentum_relation
 
It's a gnarly thing this mass we are talking about here. The question is, how might we measure it?

You couldn't, for example, grab some water molecules and measure their mass. You'd have to do that "locally." You wind up getting the same old mass every time. Proper time and local coordinates and all that. For example, if you were to put the water molecules through a mass spectrometer, it's all local. Even supposing the instrument was accurate enough.

This is an inference. We are saying that, with the water molecules as vapor, and sitting in clouds at altitude, the complete earth-air-system has a mass. Then later, when that energy has gone away, presumably as photons that escaped to space, the new configuration of the earth-air-system has a new lower mass. And since it's 700-something kg difference, it's unreasonably small to ever consider measuring it.

So, basically the inference is, what is the mass equivalent of the energy released by the storm? If you were to provide the energy by converting mass to energy (say by half matter and half anti-matter), and if you were able to do it at 100% efficiency (of course you can't) then how much mass would you need?

I suspect that the assignment was motivational, to get the class to do the arithmetic for an accessible natural occurrence they can understand. This will prep them for things like the mass anomaly for nucleons. So you compare the mass of two Helium atoms with the mass of a Beryllium atom, or three He's with a Carbon, and so on. Then you predict the energy release from various nuclear interactions, and then go look up what is measured.
 
  • #10
As I mentioned in the original statement, this isn't actually a homework assignment. I came up with the problem myself, for an unusual application.

I am pretty comfortable with the change in mass during condensation, and I will simply use that for my application (a sort of game). However GMAX137 has me seriously reconsidering my position on mass change due to potential energy. If I moved this problem to deep space, and replaced gravity with a spring mechanism, then cocking the spring (adding potential energy), should increase the mass of the system, but it would increase the mass of the spring, not the water. I suspect that in the earthbound case, the potential energy created when the water gains altitude actually gets stored in the gravitational field and not in the water, hence the water mass does not change due to potential energy. Dealing with the energy of the gravitational field is way beyond me.

I would like to thank everyone for their help, this was fun.
 
Back
Top