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Re: physics/pedagogy of coffee-mixing



This is a new problem for me also, so I looked at it algebraically.
No assumptions are needed. Call the cups 1 (originally coffee) and 2,
and the amount of coffee, which is the same as the amount of tea, N.
Then after both transfers

1: c_1 + t_1 = N
2: c_2 + t_2 - N

However, the total amount of coffee and the total amount of tea is also
N, so

3: c_1 + c_2 = N
4: t_1 + t_2 = N

(Equation 4 is not independent; it is eq. 1 + eq. 2 - eq. 3 .)

Subtract eq. 3 from eq. 1 and you have

t_1 - c_2 = 0

so the amount of tea in the first cup (t_1) is equal to the amount of
coffee in the second cup (c_2). No information about the transfer is
needed.

There must be a good way to show that in English, but I don't see it at
the moment.

Stefan Jeglinski wrote:

> Suppose you have a cup of coffee and a cup of tea. In step 1,
> you transfer one spoonful of liquid from the coffee-cup to the
> tea-cup. In step 2, you transfer one spoonful of liquid from
> the tea-cup back to the coffee-cup. Question: Is there more
> tea in the coffee, or more coffee in the tea?

I never heard this question. So let me show how I would answer.
Model each cup as an urn with 1000 balls, black for the coffee and
red for the tea. Suppose one spoonful means 10 balls and that good
mixing occurs after the first transfer. Also that volumes of coffee
and tea balls are identical. Pure water balls are white and can be
ignored in this problem.

After the first transfer we have 990 black balls in one cup and a
mixture of 1010 balls in another. What happens after the second
transfer? It is unlikely that the second spoon will contain more
than one or two black balls. On that basis my answer is that
there is more coffee in the tea than tea in the coffee.

If you use this ball model, I think your answer is incorrect. Say you
take 1000 coffee balls and 1000 tea balls. Also say a a spoonful is
100 balls. After the first spoonful you have

coffee: 1000c-100c
tea: 1000t+100c

Now we do the usual assumptions of complete mixing, uniform
distribution, etc., and transfer a spoonful back. Since a spoon is
still 100 balls, we surmise that the spoon dipped in the tea will
have 91t+9c balls. Transferring it to the coffee yields

coffee: 1000c-100c+91t+9c
tea: 1000t+100c-91t-9c

In the end, you have:

coffee: 909c+91t
tea: 909t+91c

The same of each in the other.

This result applies to the ball model, and is subject to the
assumptions of mixing etc, which I have not rigorously justified.
Although the assumptions may seem intuitive, I'm not sure they are
obvious.

Stefan Jeglinski (delurking after a year)

--
Maurice Barnhill (mvb@udel.edu)
Department of Physics and Astronomy
University of Delaware
Newark, DE 19716