Let’s say that I have 2 small cubes of ice (frozen distilled water in case it matters, but I doubt it will). I place each cube into a tray that has an open top and four sides that rise higher than the cube of ice.
The only difference between the trays is that one has a grated bottom, allowing the melted water to drain away from the cube, and the other has no grate, meaning that the cube sits in the water that has melted from it.
Assume both cubes were frozen in a standard freezer and both are subjected to standard room temperature for the test (75 or 76 degrees Farenheit? no matter I’m sure). Which melts faster, the cube with more surface area exposed to air, or the one more exposed to recently melted water?
Is this just a matter of what conducts heat better, air, or water? I would guess that the drained cube would melt faster because of the constant exposure to room temperature air, and that the “floating” cube would melt slower because it is being increasingly surrounded by water that is just below frezing. So, Which is it?
I don’t know for sure, but my General Chemistry II teacher told us that water from melted ice stays at 0 degrees celcius. That would make me assume that the cube in the water would melt slower, while the one loosing the water would melt faster…
The facile explanation is that air is a much better insulator than water, so the ice exposed to air is heated less effectively (and thus melts more slowly) than the ice sitting in its own melt water. I initially though of dry ice as an example: it can sit for quite a long time in air but sublimates rapidly when submerged in water. Or consider how preferable it is to be dry when hiking in winter. But I’m pretty sure that’s not the whole story.
Certainly the different heat conductivities of ice and water can explain what happened in the experiment araminty linked to: the spaces between the ice in the colander got cold and stayed cold, while the spaces between the ice in the bowl got filled with water, which conducted heat from the outside.
However, dnooman poses a much nicer experiment, one in which there are no air pockets in the melting medium, which is thus not fully addressed by the colander/bowl experiment.
Both the ice cube on the grate and the ice cube in the closed container will require the same amount of energy to melt, and the rate of melting will depend on the rate of energy transfer from the air to the ice. Since there is no complex pocket-forming stacking to worry about, we just have to think about the rate of heat transfer to the ice or ice+water system.
We know that that rate of heat transfer increases with the temperature difference, but as Hirka T’Bawa pointed out, both the exposed ice and the ice water are at zero degrees. (This is where the dry ice example is misleading: dry ice submerged in liquid nitrogen would behave very differently than dry ice in water!) Let’s also assume that any heat transferred to the ice water in the closed case will be quickly conducted to any remaining ice. So far, the two cases are equal.
But because the rate of energy transfer should also be proportional to the surface area exposed to the warmer medium (the air), the case with the larger surface are should melt faster. Thus, the ice in the closed container should melt faster, but it does so not because water is a better conductor than air, but rather because it maintains a larger surface area in contact with the air.
The collander would also let air circulate downward through the ice as the ice cools it and makes it contract and get heavier. This mechanism should favor melting ice in the collander.
>Thus, the ice in the closed container should melt faster, but it does so not because water is a better conductor than air, but rather because it maintains a larger surface area in contact with the air.
Aren’t there two issues here that are intertwined? The ice in the closed bowl builds a water pool around its lower portion, and the combined surface area of the water pool and the uncovered ice can be bigger than the surface of the ice in the collander. The water isn’t a perfect conductor, but if you suppose the air around the lower part of the ice in the bowl is also a pool, it’s obvious having a higher conductivity pool of water would help the melting more.
>teacher told us that water from melted ice stays at 0 degrees celcius
What? This doesn’t make sense. Perhaps the point was that if you mix ice and water and then insulate the whole mess so that there’s no heat transfer, it will equilibrate at 0. But as stated it’s obviously wrong - go melt some ice and then try experiments with the water. You can boil it or do anything else with it, as it’s just water. Or go swimming on a warm beach. That ocean is in contact with ice somewhere.
“…room temperature…75 degrees…” Hey can I come live with you? Ambient temp here in my Chicago suburb is 1F, and the greenhouse policewoman won’t let me bump up the thermostat.
Trying to answer this question without actually doing the experiment makes me think of a medieval scientist debating Aristotle instead of looking directly at nature…
Is what you are asking this: “Will an ice cube melt faster in an icewater bath or exposed to (totally still) air with an ambient temp of 75?”
The colander experiment is a great try, but wouldn’t you also need to control these variables:
A spherical chunk of ice to minimize surface area (is that an oxymoron?) and some other shape with a large surface area relative to its volume.
An icewater bath where the water is kept at 32F and the ice is kept submerged to separate air v water effects.
A 75 degree air “bath” which allows the ice sphere to melt and the liquid water to drain away, and also minimizes convective currents.
And finally, would the volume of the sphere matter?
Or is this one of those deals where the answer is obvious and the witless among us are reminded of why we avoided advanced physics?
Where’s that high-school kid and how 'bout he does this experiment? I have to go to work…
Why would “surface area” be an oxymoron??? :dubious:
As for the rest, unless there was only a slight difference between the results in the experiment proposed by the OP, all the other things you suggest aren’t necessary, though they would certainly be fun to try and see what happens.
Well, it does but only if you stayed awake for the entire class.
If ice at 0[sup]o[/sup]C has a little heat added, the energy from the heat will be used to melt the ice but will not raise the temperature. So you will have water at 0[sup]o[/sup]C. However, this is somewhat theoretical as you would have to have carefully controlled conditions for this to actually happen.
The way to carefully control the conditions is to put a little ice in the water. The physics will then take care of itself.
And of course, after the ice has all melted, or the water is taken away from the ice, you can boil it or do whatever else you want with it, but as long as there’s still ice sitting in the water, the water will remain at 0 Celsius.
What are these containers made of? The R value of the container may come into play. Ice in thick styrofoam container? Grated pocket melts first. Ice in copper container with massive radiating fins on the outside? Maybe not so much.
I have NO idea what I meant by “is that an oxymoron?” Sorry. I musta put sumpin in there and only backed out part of it… :smack:
As for the rest:
A spherical chunk of ice to minimize surface area <snip of my own blunder>and some other shape with a large surface area relative to its volume.
I was thinking the total melting time is affected by volume of ice and surface area and that these are independent variables. If they vary independently, is it not possible that a sphere would win one contest and sheet of ice, the other?
An icewater bath where the water is kept at 32F and the ice is kept submerged to separate air v water effects.
Wouldn’t you have to use an icewater bath? Otherwise only the liquid right next to the ice is at 32F. Heat will be lost by this layer of water to adjacent layers and the rate of melting will be partially dependent on the ambient temperature sucking heat out of the portion of water closest to the periphery of the water bath.
A 75 degree air “bath” which allows the ice sphere to melt and the liquid water to drain away, and also minimizes convective currents.
If you don’t minimize convective loss in the air the way you minimized convection loss in the water (by maintaining a constant water temp) wouldn’t you get two different results… e.g. blowing a constant stream of 75 degree air would melt the ice much more rapidly than still air, would it not?
And finally, would the volume of the sphere matter?
Obviously, the temperature of the air matters: 1000 degree air would always beat the water bath right? The amount of surface are relative to volume also matters, right? Very hot air will melt a sheet of ice much faster than it melts a sphere of the same volume, I’d guess. Since volume varies with the cube of the radius and surface area varies with the square (do I even have THAT right?), and heat loss is affected by the surface area relative to volume, wouldn’t the volume come into play?
Hey…there’s a reason I took Physics at a third tier University over the summer, and it wasn’t 'cuz I was good at it. “Chief Pedant” does not imply any actual ability to think.
So if I empty the contents of an ice tray into a swimming pool, the pool water will be 0 °C?
Lets assume I have a cube of ice in an 4 oz. glass of water. It is 0 °C at the solid-liquid interface. Inside the ice cube, the temperature will be less than 0 °C. There will also be a temperature gradient in the cube; the center of the cube will be at the lowest temperature, while the outside surface of the cube will be at 0 °C. Same with the liquid water… the liquid water molecules farthest away from the cube will have the highest temperature, while the liquid water molecules touching the cube will be at 0 °C.
