Hi AD1: 1-I was not talking about the buoyancy force, I was talking about the relative proportion of the cubes submberged. I may be wrong about this, but my logic is as follows consider a cube of sides x the force compressing the water underneath a point on the bottom of a cube as x units. The distance the cube will sink into the water is directly related to that force in a given density of water. Now consider a cube of sides 2x, the force compressing the water underneath each point on the bottom of the cube is 2x units, and the cube must sink twice as deep to have the same proportion of itself submerged as its cousin. Now, perhaps, it can be established that the distance a cube of x is submerged is exactly 1/2 that of a cube 2x, but the logic escapes me. 2.If you have doubts about different shaped pieces of ice having different portions submerged, please check out links on icebergs. Cheers!
I was not talking about the buoyancy force My apologies. You did say, however: "I am pretty sure that a cube with sides of 2x in length will have different proportion of itself above the water than a cube with side of length x, based on things like buoyance etc." [Emphasis my own.] I may be wrong about this, but my logic is as follows consider a cube of sides x the force compressing the water underneath a point on the bottom of a cube as x units. The distance the cube will sink into the water is directly related to that force in a given density of water. Now consider a cube of sides 2x, the force compressing the water underneath each point on the bottom of the cube is 2x units, and the cube must sink twice as deep to have the same proportion of itself submerged as its cousin. Are the masses of the two cubes equivalent? Are their densities equivalent? As I understand it, the proportion of an object submerged in the water is proportional to the density of the object and irrespective of its volume and shape. This is Archimedes' Principle. The force exerted on a body immersed in a fluid is equivalent to the weight of the displaced fluid. It follows from this that the proportion of an object kept above the water level is dependent only upon the relative densities of the fluid and the body. That's how I understand it, anyway.
My apologies to AD1: After careful reading of Archimedes principle, I find that I am wrong, the density is the principle which guides the proportion of a cube which is submerged.(In my defense, I hadn't taken this since junior high). I was neglecting to consider that icebergs whose submerged portions do vary could also vary substantially in density. As such, it follows that the height of the water after melting of the ice cube is unchanged in a closed system, anyway. Cheers! added in edit: In practise, the ice cube melting will serve to cool the water it is floating in which will slight alter the water level.
In practise, the ice cube melting will serve to cool the water it is floating in which will slight alter the water level. Yes. But imagine performing the experiment with a glass of room temperature water and measuring the initial water level before an appreciable volume of ice has melted. Then allow the ice to melt and the water to once again rise to room temperature. Pretending none of the water evaporated, has the water level changed from the initial water level?
I found the following information. One important property of ice is that it expands upon freezing. At zero degrees Celsius, it has specific gravity 0.9168 as compared to specific gravity 0.9998 of water at the same temperature. As a result, ice floats in water. Using the above, I calculated that the water level would be the same after the ice melted. Obviously my intuition (see my previous post) was incorrect. Perhaps one of you could check my logic and my arithmetic. I assumed the following. 10,000cc of water in a container with a 4X5 base and a height of 500cc One cc of ice put into container. Temperature zero Celsius so that the above densities could be used for calculations. IceWeight = .9168 grams VolumeDisplaced = 0.916983396679336 — ( This is .9168 / .9998 ) NewHeight = 500.045849169834 — ( 500 + 0.916983396679336 / 20 ) MeltedIceVolume = 0.916983396679336 — ( .9168 / .9998 ) The volume of the melted ice is the same as the volume of water it displaced. The amount above the water is a misleading issue. FinalHeight = 500.045849169834 ---- ( 500 + .9168 / .9998 ) I hope there are no typo's above. I am guessing that the same conclusion would result if different amounts of water & ice were used and/or the sahpe of the contaner was changed. With some logic, it looks as though the above could be proven as a general theorem, making suitable assumptions about temperature.
Someone mentioned that if the ice caps melted the sea level would not rise.. thats kinda right. If the north pole melted the sea level would not rise; the amount above sea level is balanced out by the difference in greater space occupied by frozen water. This is possible because the north pole is just a floating chunk of ice. However, if the south pole were to melt the sea level would rise. That's because the ice of the south pole lies on an actual land mass (or a lot of it at least). Since it would be unlikely that the north pole would melt without the south pole also melting polar ice melting = bad.
Absolutely coz the more seen above the water, the more unseen under the water! They both cancel out and it is the SAME proportion above the water always (contrary to some peoples opinion on this thread!).
With some logic, it looks as though the above could be proven as a general theorem, making suitable assumptions about temperature. Call the Volume of the ice cube V<sub>1</sub> and its volume once it has melted V<sub>2</sub> and the densities of the respective states ρ<sub>1</sub> and ρ<sub>2</sub>. Its mass is constant, so ρ<sub>1</sub>V<sub>1</sub> = ρ<sub>2</sub>V<sub>2</sub> (1) The proportional change in the water's volume (V<sub>2</sub>/V<sub>1</sub>) as it changes from solid to liquid is (ρ<sub>1</sub>/ρ<sub>2</sub>). And V<sub>2</sub> = V<sub>1</sub>(ρ<sub>1</sub>/ρ<sub>2</sub>) (2) Calling the volume of the submerged portion of the ice cube V<sub>s</sub>, the buoyancy force acting on the ice cube as it floats in the water is F<sub>b</sub> = ρ<sub>2</sub>V<sub>s</sub>g (3) This is equal to the weight of the ice cube, so ρ<sub>2</sub>V<sub>s</sub>g = ρ<sub>1</sub>V<sub>1</sub>g (4) Both sides multiply by g, so ρ<sub>1</sub>V<sub>s</sub> = ρ<sub>1</sub>V<sub>1</sub> (5) What is the volume of the submerged portion of the ice? It is: V<sub>s</sub> = V<sub>1</sub>(ρ<sub>1</sub>/ρ<sub>2</sub>) (6) The righthand side is identical to equation (2). Therefore V<sub>2</sub> = V<sub>s</sub> (7) And that means that when the ice melts, it shrinks to a volume equal to that of its submerged portion. How's that for an attempt?
I wonder about icebergs in the ocean. I think they are almost pure water, while the ocean is salty and more dense than ordinary water. The arithmetic might not work out the same for fresh water ice floating in salt water. The ice would float higher in a more dense liquid.
Will a saturated solution of salt water freeze out and produce salt crystals at a temperature well below zero?
What, u mean ice will form in one part of the container and salt will remain in solution in liquid water in another part? In that case I am right, the salt will have to precipitate out coz the water is already saturated.
Correct me: Solid water reaches it's minimum density at a few degs below freezing point, and liquid water reaches its maximum density at 4 deg C, provided that for both cases the pressure is constant.
OK I would have thought that like most solids, the minimum density of ice would be just below its melting point. In other words as u cool ice, it becomes more and more dense. Im not sure about this. Whats ur point anyway?
Riddle You have glass of water, with an ice cube floating in it. you mark the water level. After the ice cube melts, will the level of the water go up, down or will it stay the same? Water is unique in the fact that it expands when frozen. Considering this I would assume that the water level drops after the ice has melted given the ice cube displaces more water than its liquid state. Of course a certain percentage of the ice cube would be above the water level before melting. The difference in the volume of the frozen water below the water level and the volume of the entire "iceberg" as liquid should determine your answer. Right??
John Connellan: I have no ideas about freezing a saturated salt water solution water, other than the common knowledge that the salt water freezing point is below the fresh water freezing point. I thought that the salt froze along with the water rather than precipitating out of the solution. Id est: I expect that freezing salt water would result in salty ice cubes, but I would not bet money on my opinion.
Yes that was my first thought too, that the salt would get caught up in the crystal lattice but an experienced colleague (and physics teacher) said that the water would freeze without salt in it!
