How Does Iceberg Buoyancy Relate to Sea Water Density?

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Homework Help Overview

The discussion revolves around the buoyancy of an iceberg floating in sea water, specifically focusing on the relationship between the density of the sea water and the density of ice. The original poster presents a problem where only 10% of the iceberg's volume is above water, prompting questions about the calculation of sea water density in relation to ice density.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the ratio of the visible portion of the iceberg and its implications for density calculations. There are attempts to isolate variables in the equations relating to buoyancy. Some participants question the correctness of the initial assumptions regarding density comparisons between sea water and ice.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationships between the densities and the implications of their calculations. Some guidance has been offered regarding the mass of the iceberg and the displaced water, but no consensus has been reached on the final interpretation of the results.

Contextual Notes

Participants note issues with applying subscripts in equations and the constraints of not being able to look up answers due to the nature of the homework problem. There is also a recognition of potential confusion regarding the density relationships based on everyday experiences, such as ice floating in water.

Agent M27
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Homework Statement


An iceberg is floating on sea water such that only 10% of its volume is above the water level. What is the density of sea water expressed in terms of the density of ice.


Homework Equations



[tex]\frac{Vf}{Vi}[/tex] = [tex]\frac{\rho}{\rho}[/tex]

[tex]\rho[/tex]f = density of sea water
[tex]\rho[/tex]i = density of ice

The Attempt at a Solution


I know this is a rather simple problem but where I am getting hung up is the ratio of the visible portion of the iceberg. I hate to admit but the simple maths trip me up. I began by isolating the eq so that:

[tex]\rho[/tex]f = [tex]\rho[/tex]i [tex]\frac{Vi}{Vf}[/tex]

from which I get [tex]\frac{9}{10}[/tex] [tex]\rho[/tex]i


By the way this is an even number problem so I cannot look up the answer, so is this correct?

Thanks in advance.

Joe
 
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Agent M27 said:

Homework Equations



[tex]\frac{Vf}{Vi}[/tex] = [tex]\frac{\rho}{\rho}[/tex]

[tex]\rho[/tex]f = density of sea water
[tex]\rho[/tex]i = density of ice
You left out the subscipts on the ρ terms in your equation.

The Attempt at a Solution


I know this is a rather simple problem but where I am getting hung up is the ratio of the visible portion of the iceberg. I hate to admit but the simple maths trip me up. I began by isolating the eq so that:

[tex]\rho[/tex]f = [tex]\rho[/tex]i [tex]\frac{Vi}{Vf}[/tex]

from which I get [tex]\frac{9}{10}[/tex] [tex]\rho[/tex]i
Think about your answer. It says that the sea water is (more, less?) dense than the ice, and therefore the iceberg should (float, sink?).
 
As you said, It's a very simple problem.

[tex]\rho_{0}V_{0}g=\rho g (V_{0}-V)[/tex]

[tex]\rho_{0}[/tex] and [tex]V_{0}[/tex] are density and volume of the iceberg and [tex]\rho[/tex] is density of water and [tex]V[/tex] is visible volume of the iceberg.

You can easily get the sought ratio.

[tex]\frac{\rho_{0}}{\rho}=\frac{V_{0}-V}{V_{0}}=1-\frac{V}{V_{0}}[/tex]
 
From my answer it says that the sea water is less dense than that of ice, which is actually incorrect when I think about a glass of ice water. So then my answer ought to be:

[tex]\rho[/tex]f = [tex]\frac{10}{9}[/tex] x [tex]\rho[/tex]i

which correctly shows that the sea water is more dense than ice, is this the correct answer?

Thanks redbelly. Also I didn't put the subscripts in the original EQ because I have issues when I apply subs scripts in a fraction for some silly reason.

Joe
 
Looks good. One way to remember this is that the total mass of the iceberg (or other floating object) equals the mass of the displaced water (or other fluid). Since mass is equal to ρV, you can set up the equation by equating the two masses.
 

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