Welcome to the third installment of Problem of the Week! We are very pleased with the number of responses we have gotten so far and we super duper promise to put up some kind of leader board soon. In fact, I super duper promise to relegate that assignment to Alemi. We intend to keep this up as long as we can and give out prizes for high scores maybe...? They will be lame internet prizes...?
The solution to the last problem of the week will be up shortly. Adam is the only one who knows the answer and he was busy all weekend taking a magic ring to Mordor.
One more housekeeping note. Since we want everyone to have a clean shot at answering the question, we would prefer the solutions to be sent by email instead of posted in the comments. However, we certainly don't want to stop discussion on the problem, so if you don't want any hints you might want to avoid the comments!
Now for the problem...
Dr. Cherrycoke has constructed a fitting trigger for his death machine / koala feeder. A single ice cube is placed in a glass of water so that the water is completely filling the glass up to the brim. The glass of water is then placed on a very sensitive detector. If even a single drop of water spills out of the glass as the ice cube melts, James Bond will find himself on the wrong end of a murderous marsupial mauling.
Vontavious exits the room with his minions, confident that his death trap will be triggered and Bond will be vanquished once and for all.
Does Mr. Bond survive? If not, roughly how long does it take until he checks in to the big MI6 in the sky? Be sure to back up your answer!
Here's some discussion: not enough information is provided to solve the problem.
ReplyDeleteHere's the issue: water doesn't have a constant density as a function of temperature, so the usual "displace a volume of water equivalent to its mass" isn't enough because the melting of the ice cube will change the temperature of the water, yada yada.
Interestingly enough (because of the crazy temperature dependance of the density of water) it turns out that, depending on the starting conditions, it could go either way!
I do like the koalas, though.
The Koalas get him every time. No exceptions.
ReplyDeleteThe ice will float.
There will be a point where the ice meets water and air. At this point, the water is at its triple point. There will be water vapour. It will escape into the room.
As the ice melts, it will cool the water. The water vapour will recondense (at least partially)on the outside of the glass. Koalas. (Actually, you're thinking of Drop Bears, but never mind...)
--Charles
I like where this is going.
ReplyDeleteAdditionally, I offer no clarifications or justifications. There was certainly a solution I had in mind that relied on several assumptions, but I like it better vague.
Any answer that relies on physically motivated assumptions will be considered.
Arbitrarily granted bonus points to anyone who gives the best "quantum of hits."
ReplyDeleteBond dies due to condensation forming on the side of the glass.
ReplyDeleteSure, you've considered that the density of water may change with temperature, but you also need to consider how the surface tension changes with temperature. So does the evaporation rate exceed the rate of thermal expansion? We need to know the ambient conditions (both temperature and humidity) and the proportions of the container.
ReplyDeleteAnd let's suppose it's a double-walled container. There may not be any condensation.
I've had no discussion with Corky on this problem, but I'll offer a suggestion to the folks wanting to know more information. Think like a physicist, and make it simple (to first order). I personally would make the following assumptions in solving the problem:
ReplyDelete1) Assume that the relative humidity of the room is such that there is no evaporation to consider.
2) Assume the glass of water is large enough (or the ice cube small enough) that the temperature change of the water is negligible (no thermal expansion, density fluctuations, etc). This also assumes perfect, fast, mixing of water in the glass.
3) Assume that the glass/water is in thermal equilibrium with the room.
4) Assume that there is no condensation on the glass.
Now, I know that in a life or death situation, some of the assumptions might become important (we are talking about a single drop of water). However, my assumptions would allow me to get a good first order estimate of what was going to happen before I actually watched the outcome on Mr. Bond! As a professor of mine said to me once, if you can't solve the problem as stated, make assumptions until it becomes a problem you can solve.
If you want the second order solution to this question, then pick what you think are some logical assumptions that aren't quite that easy (for example, pick 8oz water at 20C and a 1oz ice cube, and let the density of water fluctuate), and give us that solution!
Or, you could actually carry out the experiment, and send us a video as your solution. This is one where the experiment would be quite easy to do.
The long answer:
ReplyDelete- If more ice forms, the volume increases.
- If the water is really about to overflow, it will when more ice forms.
To cause an overflow, we need more ice to form. Factors affecting this will be:
- Initial temperature of the ice
- Initial temperature of the water
- Heat flow from the environment into the water
- Heat flow from the water into the ice
- Effectiveness of convective mixing of the water
If the net effect of this is a faster flow of heat from the water around the ice into the ice, than from the rest of tha water into the water around the ice, then the mass of ice will increase, and the overall volume will increase, and the koalas get fed.
So far I have assumed that we are within 'reasonable' conditions where koalas would thrive. At extremes of pressure (or temperature) we may end up with ice in a less-familiar phase. The phase diagram for water goes to some very odd places, and several phases of ice are denser than water.
"Up to the brim" is very vague about a meniscus. If the water surface is level, there is some wiggle room as a meniscus forms.
If the meniscus is already at the limit of surface tension. The presence of ice will ensure that the meniscus remains at zero degrees, so the surface tension will not decrease until all the ice is gone and the temperature at the surface start to rise.
Unless, of course, the amount of ice were very small. In this case, the ice melting will not reduce the meniscus significantly. If the room is now above water temperature, as the water heats it will expand, and the surcase tension will decrease, and the water will spill.
So, ignoring the condensation I mentioned above, the answer is "It depends".
As often happens, there's a surprising amount of physics going on in this seemingly simple situation. I'm sure I've missed some.
-- Charles
Re: Jesse at 11/23/2010 2:34PM
ReplyDelete1) The problem is asking us - in essence - if the water level goes up, and you suggest we ignore all the physical effects that might make the water level go up or down? Boo.
2) "Think like a physicist". Wow. That's kind of insulting, no? And in this case, I think, inaccurate. The only reason that I know about the temperature dependence of the density of water is that Caltech's Jeff Kimble pointed it out in the context of this very problem. I can assure you: that dude thinks like a physicist. I apologize for the name-dropping, but you started it with the name-calling.
3) I realize personal tastes may differ, but to me, the whole mass-displacement thing is kind of boring. But the crazy-ass non-monotonic temperature dependence of water's density? I find that fascinating. What the heck is going on there?
Re: AC
ReplyDeleteI apologize if my post was misconstrued. My point wasn't to ignore all the effects. My point was to start out with the simplest situation you can, and then build on that to something more complicated. I certainly didn't mean to be insulting, but I think that's more or less exactly what a physicist usually does (certainly what most of us do). I would wager that most physicists, if asked this question, would initially make the assumptions that I've mentioned (or something similar), and get a quick, first order answer. Then, if they needed something more accurate, they would go back, and add in what they considered to be the most important effect (perhaps the density fluctuation with temperature), and see what that does. If that perturbed the system thoroughly, then add in another effect, such as evaporation, and see if what that did.
I'm hopefully that this is just one of those cases of things typed on the internet 'sounding' different than they did in my head. The point I was trying to make was that, to get a quick, first order, answer before Mr. Bond dies (or doesn't) there's a set of simplifying assumptions that makes this an easy problem. Once we've established a simple model, we can add back in the complexity as we see fit. That's what I meant, and perhaps should have explicitly said, in my post. That is thinking like a physicist. If you wanted to stereotype a little, you could say that we always start with a harmonic oscillator, and add complexity from there (no matter what we're modeling)!
I certainly wouldn't tell anyone to ignore the interesting physics going on here, but to provide some direction for people attempting to work the problem, I thought I would give an idea of how I would start it. To first order, the most important effect here is the mass displacement. If, instead of an ice cube, he put a rock in the glass, and then very carefully removed it, we wouldn't be having any of this discussion of the other effects. It's only once you realize that effect gives no change that you become free to consider these second order effects. I say second order, because even from 0 C to 100 C, the fluctuation in density is only ~4%, and the fluctuation around 4 C (that weird non-monatonic stuff you mention) is on the order of .01%. My goal was to guide our less experienced readers to a basic conception of the problem, at least.
For what it's worth, the "quick, first-order answer" is unsatisfying because it results in no displacement from the initial condition. Therefore, any small perturbation coming from the less-than-ideal assumptions is what it will take to tip the scenario in one direction or the other.
ReplyDeleteHowever, even given the initial 'ideal' assumptions, the ice is supported somewhat by its buoyancy in the surrounding air, so you could argue that there's actually a very small increase in water level as the ice melts, which is usually ignored, and is certainly more negligible than the other non-ideal assumptions discussed.
Now I haven't looked that the numbers yet, but I know a couple of things. Firstly, water is often described as 'nearly incompressible', which leads me to believe that the change in density is negligible compared to the rate of evaporation. And I also know that liquid water is densest at 4 degrees C. So the initial temperature of the water in the glass certainly comes into play.
My guess is that it will not spill over because evaporation overwhelms thermal expansion, but it depends on the proportions of the container. A tall, narrow glass will allow thermal expansion to change the water level more readily, and a short, wide glass presents a larger surface area for evaporation to occur. Although if the only heat transfer occurs at this surface, then it's a wash. Introducing the concept of surface tension adds the dimensionality of height into the mix, because the meniscus supports an absolute height.
The wording of the article says if a "single drop of water spills out of the glass...". This removes from consideration the concept of condensation on the container; at least that's what Corky had in mind. Let's assume the container insulates well enough to prevent condensation on its outer surface.
Re: Jesse at 11/24 8:57am
ReplyDeleteFair enough. I apologize for my misreading/overreaction.
This comment has been removed by the author.
ReplyDeleteHi! I am making simplified assumptions but Mr. Bond will survive.
ReplyDeleteThe ice will melt but the melted ice will occupy the volume taken by the submerged part of the initial ice so not a drop will spill out of the glass. Assume no condensation, no random fluctuations.
(sorry about the double post ^^)
Re: Drew at Nov 24th,
ReplyDelete>For what it's worth, the "quick, first-order answer" is unsatisfying because it results in no displacement from the initial condition. Therefore, any small perturbation coming from the less-than-ideal assumptions is what it will take to tip the scenario in one direction or the other.
I completely agree.
>My guess is that it will not spill over because evaporation overwhelms thermal expansion
Not to be a broken record, but again: not enough information is provided to solve the problem.
If you want to talk about evaporation/condensation, you have to know the temperature of the water/ice and the humidity (i.e. dew point) in the room to determine whether the water level will go up or down.
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ReplyDelete