What is the volume of water in a cylindrical cumulus cloud of given dimensions?

  • Thread starter scmejla
  • Start date
In summary, the conversation revolves around the calculation of the amount of water in a cylindrical cumulus cloud of height 3.6 km and radius 1.1 km. The lower value and higher value for the amount of water, given the range of 50 to 500 water drops with a typical radius of 10 µm, are 0.286614 m^3 and 2.86614 m^3 respectively. The conversation also includes a discussion about the accuracy of the calculations and the estimation of the amount of water in the cloud.
  • #1
scmejla
4
0
Does the solution to this problem look right?

A cubic centimeter in a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 µm. For that range, give the lower value and the higher value, respectively, for the following.
(a) How many cubic meters of water are in a cylindrical cumulus cloud of height 3.6 km and radius 1.1 km?

Vdrop = 4/3 * pi * (10^-6 m)^3 = 4.18879(10^-15) m^3

Vdrop*50 = 2.0944(10^-13) m^3/(10^-2) m^3 = 2.0944(10^-11)
Vdrop*500 = 2.0944(10^-12) m^3/(10^-2) m^3 = 2.0944(10^-10)

Vcloud = pi * 3.6(10^3) m * (1.1(10^3) m)^2 = 1.36848(10^10) m^3

Lower value = 2.0944(10^-11) * 1.36848(10^10) m^3 = 0.286614 m^3
Higher value = 2.0944(10^-10) * 1.36848(10^10) m^3 = 2.86614 m^3
 
Physics news on Phys.org
  • #2
scmejla said:
Does the solution to this problem look right?

A cubic centimeter in a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 µm. For that range, give the lower value and the higher value, respectively, for the following.
(a) How many cubic meters of water are in a cylindrical cumulus cloud of height 3.6 km and radius 1.1 km?

Vdrop = 4/3 * pi * (10^-6 m)^3 = 4.18879(10^-15) m^3

Vdrop*50 = 2.0944(10^-13) m^3/(10^-2) m^3 = 2.0944(10^-11)
Vdrop*500 = 2.0944(10^-12) m^3/(10^-2) m^3 = 2.0944(10^-10)

Vcloud = pi * 3.6(10^3) m * (1.1(10^3) m)^2 = 1.36848(10^10) m^3

Lower value = 2.0944(10^-11) * 1.36848(10^10) m^3 = 0.286614 m^3
Higher value = 2.0944(10^-10) * 1.36848(10^10) m^3 = 2.86614 m^3

Hello Scmejla. Interesting problem. You know that's a big cloud but you're saying it has only .2 m^3 of water. That's only a few buckets full. Surely it has more than that, few truckloads at least. I checked your work. You calculated volume of water in a drop (in cubic meters). That's 4.18*10^-15. For the lower limit, there are 50 of those drops in every centimeter and 1 million cubic centimeters in a cubic meter. Thus, the low limit of the amount of water in 1 cubic meter is how much?

Edit: Alright, maybe 10 or 12 you know what I mean. So how many bucket-fulls of water are in .28 m^3? You know, without calculating it, I'd guess now around 15. I say we figure how many bucket-fulls of water are in that cloud anyway. :smile:
 
Last edited:
  • #3
Thanx for setting me straight, I should have realized that was an awful small amount for such a big cloud.
 

1. What is the purpose of asking "Does this problem look right?"

The purpose of asking this question is to ensure that the problem is accurately represented and that all necessary information is provided to solve it. It also allows for double-checking of calculations and identifying any potential errors.

2. How can I determine if a problem looks right?

To determine if a problem looks right, you should first read through the problem carefully to understand what is being asked. Then, you can check if all the given information is relevant and if the problem can be solved using the given information. You can also compare your solution to similar problems or ask for a second opinion from a colleague or mentor.

3. What should I do if the problem does not look right?

If the problem does not look right, you should first check for any missing or incorrect information and make sure your calculations are accurate. You can also try approaching the problem from a different angle or using different methods. If you are still unsure, it is always helpful to seek guidance from someone with more experience in the subject area.

4. How can I avoid making mistakes when solving problems?

To avoid making mistakes when solving problems, it is important to pay attention to details, double-check your work, and use multiple methods to solve the problem. It is also helpful to take breaks and come back to the problem with a fresh perspective. Additionally, seeking feedback from others and practicing problem-solving regularly can help improve accuracy.

5. Are there any common mistakes to watch out for when checking if a problem looks right?

Some common mistakes to watch out for when checking if a problem looks right include misinterpreting information, using incorrect formulas or equations, and calculation errors. It is also important to make sure all units are consistent throughout the problem and to consider any assumptions that may have been made.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
11K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Back
Top