# Buoyancy problem and density of ice

1. Jun 22, 2009

### ash4741

1. The problem statement, all variables and given/known data
The density of ice is 920kg/m^3, and the density of sea water is 1030kg/m^3. What is the fraction of the total volume of an iceberg that is exposed?

given:
density sea water = 1030kg/m^3
density ice = 920kg/m^3

2. Relevant equations

density water/density ice

3. The attempt at a solution
I am really not sure of how to start this problem.

I divided 1030 by 920 and got 1.101kg/m^3

how do I start this problem?

2. Jun 22, 2009

### mgb_phys

First step in this sort of problem is to estimate (guess) what answer you expect.
You know icebergs float, with most of the ice below the water - if you are a native English speaker you know 'tip of the iceberg'

The next step is that you know your answer can't have any units - it's just a ratio = a number.
So if you have two numbers with units and you want an answer with no units - what do you have to do with the numbers?

Last edited: Jun 22, 2009
3. Jun 22, 2009

### tiny-tim

Hi ash4741!

Do it logically …

how much volume of water is displaced by the ice?

4. Jun 22, 2009

### bucher

Start with the fundamental concept of buoyancy. The mass of the ice is equal to the mass of the displaced water. The volume of the displaced water is equal to the volume of the submerged iceberg.

This should get you on the right track.

5. Jun 22, 2009

### ash4741

So the 1.101 is the mass of the iceberg? and the mass of the water displaced? But I thought that was the volume of the iceberg that is under the water.

6. Jun 22, 2009

### LowlyPion

One thing that might help is that 1030/920 is not 1.101. At least calculate it more precisely.

7. Jun 22, 2009

### Naty1

What do you think this answer MEANS?
Also write down the units in your calculation inputs...because You have the wrong units in your answer....

no.... to calculate the MASS you would need to know the volume of the iceberg in question....but you don't need to know that....

What fraction of the iceberg will be underwater?.....something less than the total iceberg,right,because it floats and some stick up above water but most of it is underwater.....

8. Jun 22, 2009

### ash4741

I though that the answer to this division (which is actually 1.119) was the volume of the iceberg that is under water. Sorry, I do not quite understand.

9. Jun 22, 2009

### ash4741

So the answer is 0.931 for the ratio that is exposed?
I am just guessing because what I thought was the portion under water is 1.119.

10. Jun 22, 2009

### tiny-tim

Why do you keep guessing?

use Archimedes' principle (the one about displacement of water), and work it out!

11. Jun 22, 2009

### LowlyPion

Maybe try a less abstract approach? If you have a 1 m3 block of ice how much does that weigh?

What percentage of that volume if filled with the heavier sea water would weigh the same? That's all the sea water that needs to be displaced for it to float.

So ... how much is left over? That's the percentage above the waterline "exposed" isn't it?

12. Jun 22, 2009

### bucher

Alright, if you start with the approach that the mass of the iceberg must equal the mass of the water displaced you will get to the answer. Use your densities to help you out here.

You should get

Fraction Exposed = 1 - (density of ice)/(density of sea water)

13. Jun 22, 2009

### ash4741

So if the mass of water displaced= mass of ice then:
density of water * volume of water displace = density of ice * volume of ice
So
(Volume of water displaced)/(volume of ice) =1 and you would subtract 1 from density of ice/ density of water

Is that right?

14. Jun 22, 2009

### LowlyPion

The first part is correct. Determine first then the volume of water needed to be displaced. And then it is the ratio of the volumes that is a % that you subtract from 1, not as you expressed.

15. Jun 22, 2009

### ash4741

Thanks everyone