since liquid water is more dense than ice, the water level would go down because more water would be able to fit in less space.
There is no change in the level. The ice floats because it displaces a volume of water which weights as much as the ice. When the ice melts, it turns into water, that will occupy the same volume. (I am assuming that prior to melting the water in contact with the ice will be at 0 degrees C).
Could be a trick question involving the other stages of ice (frozen under high pressure), which I assume are more dense.
Ok, I will take a crack at it. Correct me if I am wrong Folks. Let's take a differential element of water, where the top cube of the water lies at the free surface of the water. (i.e. the cube is just under the water). This cube has dimensions, dx, dy, dz. The volume of this cube is: [tex] dV = dx dy dz [/tex] the associated mass is: [tex] m = \rho_w dx dy dz [/tex] when water expands to ice, it increases by volume at about 9%, but the mass remains constant, therefore: [tex] m = \rho_w dx dy dz = \rho_{ice}( 1.09 dx dy dz )[/tex] So that means: [tex] \rho_w = 1.09 \rho_{ice} [/tex] We can now do a simple force balance to see our result: (weight of ice must balance buoyancy force) [tex] dF_b = \rho_w g dV' = \frac {\rho_w}{1.09} g (1.09 dx dy dz) [/tex] **Edit: Forgot the (1.09) infront of the dxdydz, thanks Gokul! Now we simplify and get: [tex] dV' = dxdydz[/tex] Where dV' is the volume of displaced water by the ice. Because dV' is equal to dV=dxdydz, (which was the original volume of differential water), the water level will stay the same when the ice melts. I just wrote this proof down myself, so if its got a mistake point it out!
Your mistake is assuming that the ice floats yet is totally submerged. Yes, the volume of ice is greater than the volume of an equal mass of water--but only part of the ice is submerged and thus displacing water. (See SGT's post for a simple argument.) On the other hand, if the ice is not floating but resting on the bottom of the beaker (exerting a non-zero force on the bottom of the beaker), then the water level will rise when the ice melts. In such a case the ice obviously weighs more than the displaced water, otherwise it would be floating.
Not quite. I never assumed that. Take a second glance. I started with a differential element of water just under the surface to see the volume it would occupy. Then I compared that volume to the displaced volume of the ice (that had the same mass as the water). I assumed the ice was some fractional distance from the surface, kdz, where k turns out to be 1/(1.09)^2.
Simply put, the ice will float at a height where it displaces an amount of water exactly equal to its mass. If the volume of that block of ice happened to increase (for whatever reason), or even decrease (for whatever reason), without changing it mass, it is the volume above the waterline that will grow or shrink. The volume below the water line will not change, and thus the displaced amount of beaker water will not change. (It couldn't change! The volume displaced in the beaker water is directly created by the displacement from the mass of the block of ice, which hasn't changed in mass!) Since the amount of wtaer displaced in the beaker does not change, it has no effect on the water level.
Error in the last line, on the RHS. [tex]dF_{weight} = \frac {\rho_w}{1.09} g dV_{ice} = \frac {\rho_w}{1.09} g dV_{water} \cdot 1.09 =\rho_w g dx dy dz [/tex]
I think my revision thanks to Gokul has put into equations what you have put into text. Though I think there is need of more equations to back up these arguments at the start of the thread. You guys were right, but it was too informal for my taste. The way in which the differential element grows or shrinks is not needed in this analysis. All you need to worry about is how much volume the ice displaces, and compare that to how much volume that same element of ice would occupy as a liquid. The transient is not of concern.
See attached pic. It's so simple that it should be intuitively obvious that the water level doesn't change. (If anyone thinks my diagram can be improved let me know. But I can't see anything else it needs.)
(Hm. Inb retrospect, my post may have sounded almost patronizing. I didn't mean to suggest it's simple to understand, merely that the diagram is simple.) Does the diagram not work for you?
Meh, don't worry about it. I can't see it until doc approves it. From my perspective, it seemed like everyone was stating the obvious, without actually proving it :tongue2: Something cant be just because you say so!