Thanks for the help, guys. I ultimately did convince the guy that he was wrong, though for some reason he seemed to take it personally. :dubious: At least I succeeded in fighting my own ignorance.
I ran the experiment with tap water last night. There was no detectable change in water level when the ice melted.
Nope, mass. That’s why a boat floats. (An object with density lower than water will, of course, sink, and displace *less * than its own mass of water, just because once it becomes submerged it just can’t displace any more.) For a boat (or an ice cube), once the object has sunken low enough to displace its weight in water, it stops sinking and is said to be floating.
Water is one of the very very few substances that actually has the ability to expand (thus becoming less dense) upon solidification. Most substances shrink in volume when changing from liquid to solid. From a very nitpicky standpoint, the volume of the ice/beverage mixture will decrease as the ice melts. However, it would be by a very negligible amount when you take into account the average size of a drinking glass. Not to mention that there are any number of other variables such as the carbonation of the beverage, the amount of alcohol/fruit juice/sugar, etc. in the beverage, the ambient temperature and thus evaporation of liquid from the glass…
I could just go on and on and on… 
Not nitpicky at all. But the volume of the drink is a different question from the water level. if you have a glass of icewater filled to the brim, the total volume of water (liquid and solid) in the glass will be greater than the volume of the glass, because the ice cubes will poke a bit above the surface. When the ice melts, the glass will still be full exactly to the brim (and will not have overflowed), but now there’s no ice sticking up, so the total volume will be less.
Good point. I didn’t consider the ice protruding from the glass.
Slightly more than a nitpick: Weight, not mass.
You would get different results on the moon - the weights of the floating object and the water it displaces are still equal, but one-sixth of what it would be on earth.
If you look at the mass however, the floating object would have six times the mass of the water it displaces.
I think you mean “greater”.
Nonsense. Both the ice and water will weigh one sixth what they do on Earth, but that’s irrelevent to this question. No matter what the local gravity[sup]1[/sup] is, the ice will still displace its mass in water. But also its weight because this is one situation where the difference in mass and weight is irrelevent.
[sup]1[/sup] As long as there is significant gravity, that is. In microgravity/free fall, it would not be correct to say the ice is floating in the water.
Let’s make this question a little harder, what if the drink was on a treadmill such that…
Actually, I meant “specific gravity” instead of “density”.
How can their weights and masses be equal on earth, but on the moon their masses are still equal but their weights are different? :dubious:
Well, the specific gravity of a substance is the ratio of its density to the density of water, so to be correct the statement would still need to say “greater”.
Dang. I need a nap. :o
:smack: the buoyancy principle applies on the earth as well as the moon and if the weights are the same the masses must be too. What good stuff I was smoking!
Ok, I thought I had something: I was picturing a battleship floating on a lunar ocean and I assumed that it would float much higher. But the lunar water is “lighter” too and the boat sinks down just as far as it does on Earth?
This really does answer the imminent secondary question in one.
It sure does answer mine.
Thanks, Polycarp.
Yes, exactly right. It is weight versus weight rather than mass versus mass, because we are talking about forces here - but the equation holds good unless the force of gravity goes to zero (and hence all the weights involved go to zero).
In other words, if you miraculously conjure up a battleship-sized dock full of water with a battleship floating in it and an antigravity device under it, the battleship floats at the exact same height whatever the reading on your antigravity device - except that once you turn the device up to 32 ft.sec[sup]-2[/sup] the battleship will now be equally happy “floating” at any height, from fully submerged to perched on the surface like a water-boatman beetle, because neither it nor the water are exerting any force on each other at all. Of course, at this point the water will show no great inclination to stay in the dock, either.
(Meanwhile, a pendulum clock on board the ship will run slower and slower as the antigravity increases, and eventually stop. It is weight that causes the pendulum to reverse direction, and mass that keeps it swinging in the same one.)
To be fair, you also have to stipulate that the gravitational field is uniform. This is a pretty good approximation on the Earth or the Moon, for objects the size of martinis or battleships, but one could contrive an arrangement of ice cubes, cocktail glasses, and microscopic black holes such that things wouldn’t float at the same level.