Here’s the “water and wine” problem:
Take two glasses, and
. Initially,
is full of water and
is full of wine. Remove a teaspoon of water from
and put it in
, then a teaspoon of the mixture in
and put it in
. In what proportions are the water and wine now mixed in
and
? What are the proportions of water and wine in each glass after
iterations?
This problem is quite recreational. I really like it. Give it a try!
I found and solved this problem in August 2006. I believe the problem statement is somewhat loose: a mathematician might believe he has all the information he needs, but a chemist will likely find the problem to be ill-posed. Some assumptions would help to clarify things.
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Assumptions
For the sake of clarity and simplicity, let us make some assumptions:
- conservation of volume: when we add two liquids together, the volume of the mixture can be less than the sum of the individual volumes. In fact, this happens when we add water and ethanol together. And we all know that wine contains ethanol. Nevertheless, in order to make the problem tractable, let us assume that volume is preserved, so that when we add water and wine together, the volume of the mix is the sum of the individual volumes.
- perfect homogeneity: water and ethanol are miscible. We will assume that wine will perfectly dissolve in water (and vice-versa), so that we obtain a perfectly homogeneous solution. Thus, any teaspoon of liquid we take from either glass will always contain water and wine in the same proportions.
- no leakage: we assume that no liquid drips from the teaspoon while we are transferring liquid from one glass to the other. We also assume that the water does not evaporate. Thus, the total volumes of water and wine in the system remain constant no matter how many transfers we carry out. In other words, the “system” (the two glasses) is closed: no liquid comes in, no liquid is leaked.
- big teaspoon: let us assume that the teaspoon’s capacity is at least greater than each glass’ capacity. Thus, we can transfer all the liquid in one glass to the other glass, and no liquid will be spilled. Lastly, let us assume that the capacity (i.e., maximum volume that the glass can hold) of each glass is greater than the total volume of liquid in the “system”, so that we can transfer all the liquid in one glass to the other without any spilling.
Given these assumptions, when we take a teaspoon of liquid from glass and add it to glass
, the volume of liquid in glass
after the mix is the sum of the volume of liquid in glass
before the mix and the volume of liquid in the teaspoon.
It has been quite a few years since I last studied Chemistry, so please forgive me if my explanation is not very accurate, or if I misused technical terms.
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Tags: Puzzles, Water and Wine Problem

August 9, 2008 at 17:07 |
If water and wine were not miscible, the problem would not be interesting. For example, if we used olive oil instead of wine, things would be strange because olive oil is not miscible in water. We would not have an homogenous solution, we would have two separate phases: water on the bottom of the glass, and a layer of olive oil on top of it (water is denser than oil at room temperature).
August 9, 2008 at 23:01 |
I had posed the same problem at my blog a few months ago, and in solving it, I posed the analogous discrete problem and also offered the corresponding solution – there are two, actually. That post gives away the solution, and so your readers may read it at their own peril!