> For goodness sake - I definitely need it (chemistry help, that is). I am in Chem 102 at my college and was wondering if anyone could help me out with a couple of things.
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> We are studying "Chemical Equilibrium" now. We have to write a paragraph by Friday explaning what would happen to this system: A cube of ice is placed in a beaker and water is poured in until the level of the water is at the very top of the beaker. (The ice will be sticking about 1/3 of the way out of the water.) The beaker is at room temperature. So, the question is: What happens to this system as the ice melts?
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> Here is what I know so far: Ice is less dense than water, so as the ice cube melts the water level will fall. But I am also thinking (since we are studying equilibria) that it has something to do with the water vapor also. Our professor didn't tell us if the beaker was covered or not, so I'm not exactly sure what will happen to the system as vapor is produced. Also I don't know the point at which vapor will start to form - I have a feeling that we need to know this.
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> Ga"any help would be so totally appreciated"halia

OK, here's my take on this. I'm going to use degrees Celsius for my explanation, since that's what I'm used to, but here are a few quick tips for those unfamiliar with ºC:

32ºF = 0ºC (freezing point of water)
70ºF ~= 20ºC (room temperature)

OK, let us assume that the water is initially at room temperature. The ice cube might as well be at 0ºC or -1 or -2-- it doesn't matter too much.

To simplify the system, I will make one assumption: The system is closed. A closed system with respect to mass means no losses to evaporation; w.r.t energy, it means that no outside heating forces are applied.

As Shandar has pointed out, the ice cube displaces an equal mass of water. Hence, the volume of the ice cube which is UNDER water is equal to the volume of water at ROOM temperature which has the same mass as the entire ice cube. Basically, this means that if the ice cube melted into a volume of room temperature water, the beaker's water level would not change. However, since we assumed a closed system w.r.t. energy, the temperature of the water is dropping as the ice cube melts.

Now, the heat of fusion (amount of energy required to move between the liquid and solid states) of water is significantly larger than the specific heat capacity of water. Thus, even though the ice cube is small compared to the whole beaker (in all likelihood) the water will drop in temperature quite a bit in order to provide the energy necessary to melt the ice cube. Water is at its most dense at 4ºC; it gets less dense as the temperature increases (as with most liquids) but ALSO as it decreases towards the freezing point. As a result of this, both the ice cube AND the rest of the water will be decreasing in density as time progresses and they both move towards that 4ºC mark. As a result of this, the total volume of liquid in the beaker will drop slightly as the process goes on. One could argue that the system's equilibrium temperature might be higher than 4ºC, and hence the water which HAD been the ice cube would expand as it approached the equilibrium temperature; however, that point is irrelevant, because we already know that the entire system at room temperature would keep the beaker's water at level at the point we started at, and the equilibrium temperature could not be higher than room temperature.

The other case that could be argued would be that the system's equilibrium would be at a temperature BELOW 4ºC, but I don't feel like doing that much thinking right now. I'm pretty sure we would need to know specific volumes of the ice cube and the beaker to work that situation out fully. I also think that it is less likely to come up. :-)

OK, so we now know that in a closed system, the water level will drop slightly as the ice cube melts. But what about in an open system?

In the open system, there will be two additional factors influencing the level. The first is evaporation. Obviously, this will only serve to reduce the level, although it would be very gradual. (I don't know the vapor pressure of water at this range of temperatures off hand, but it is fairly low.) Of course, depending on the humidity of the room, there could be no evaporation (100% humidity) or some condensation filling the beaker (>100% humidity). Neither of these is likely, though.

The other would be heat influx from the room. If we assume that the room is infinitely large (and compared to the size of the beaker it might as well be) and at a constant temperature of 20ºC, then the beaker will gradually rise in temperature back to 20ºC (the mathematician in me now points out that it will only reach that temperature after an infinite time, but the engineer in me says "Who cares? It'll be 99.9% of the way there in a finite amount of time, and that's good enough") and we already know that at 20ºC, if no evaporation has occured, the beaker will be back at the level it started at.

Summing up: In a real system, the beaker's water level would drop at first, and after the cube had melted completely, the level would begin to rise again, but due to evaporation (resulting from humidity <100%), would not reach the original point (unless the room had precisely 100% humidity). If the room had humidity higher than 100%, the beaker would eventually overflow from condensation filling it further. The overflow would have to beat surface tension before anything spilled, though. Thanks to Issachar for reminding me about surface tension. :-)

Well, it's all there-- pick and choose which portions are important to this class. I'm guessing that some of it won't be-- I drew on some knowledge I didn't learn until my final year of engineering (but which would probably be taught in second or third year chemistry).

Don "Why do I have the feeling I carried this too far?" Monkey