How much water is contained in a cumulus cloud and what is its mass?

[AnkhUNC]

[SOLVED] Mass of water in a cloud

Homework Statement

One cubic centimeter of a cumulus contains 220 water drops, which have a typical radius of 10 μm. (a) How many cubic meters of water are in a cylindrical cumulus cloud of height 3.0 km and radius 1.0 km? (b) How many 1-liter pop bottles would that water fill? (c) Water has a density of 1000 kg/m^3. How much mass does the water in the cloud have?

Homework Equations

The Attempt at a Solution

OK here is what I have so far I know I'm almost there I'm just making some simple mistake somewhere along the line.

First 1 cm^3 of a cloud = (220 drops/cm^3) which equals 220*10^6 drops/m^3

10 microns = 10*10^-6m = radius of a drop. So the density of a drop = 4.188790205E-5m

Volume of the cloud = (Pi)r^2*h = (Pi)(1000m)^2(3000m) = 942477961m (? messing up here?) But this isn't the answer to (A)?

(B) is just a conversion of (A) so..

(C) Well i guess I only got to (A)


Any help of what I'm doing wrong or how I should be going about this would be appreciated. Thanks,

 

[Kurdt]

You've calculated the volume of the cloud for a) but what they want is the volume of water in the cloud in cubic meters. What you were doing before calculating the clouds volume was along the right lines to help get the final answer.

 

[AnkhUNC]

When I cube the answer for (A) it still comes up wrong

 

[AnkhUNC]

To continue: The #drops = ((Whatever A should be)(220*10^(6) drop/m^3) = ?

So mass of cloud = (# drops)(4/3 (Pi)(10*10^(-6)m^3)) right? Don't know where I'm going wrong with (A).

 

[Kurdt]

You've calculated the volume of the cloud. Now you need to know the volume of water in the cloud. Since you know the volume of each drop (because you have the radius and assuming they're spheres) and you know how many drops are in a cubic centimeter you can work out the volume of water in the whole cloud.

 

[AnkhUNC]

I added it together and still ended up getting it wrong so I'm not sure what I'm doing wrong :( Its frustrating though.

 

[hage567]

The volume of the cloud and the volume of the water are not the same. I think that's where you are getting confused.

To continue: The #drops = ((Whatever A should be)(220*10^(6) drop/m^3) = ?

I think you are getting ahead of yourself here.
The number of drops is part of determining the answer to (a). First, find the volume of the cloud. You've done this in your first post, but I think you're answer is off by a factor of ten. Double check it. Once you have this, you can find the total number of drops in the cloud. The number of drops will be the volume of the cloud * the number of drops/m^3.

So mass of cloud = (# drops)(4/3 (Pi)(10*10^(-6)m^3)) right? Don't know where I'm going wrong with (A).

This equation does not give you mass, check the units. This gives you the volume of water in the cloud. Which is what you are trying to calculate for (a).

Don't worry about mass and density yet, that's for part (c).

 

[AnkhUNC]

Well I figured it out finally,

(a)8.685E3m^3

(b)8685000bottles

(c)8685000kg

Thanks for the help!

 

[hage567]

Good work!