Newsgroups: rec.puzzles
From: hoey@AIC.NRL.Navy.Mil (Dan Hoey)
Date: 2 Sep 94 20:46:59 GMT
Subject: Re: physics/milk.and.coffee

dwr2...@tam2000.tamu.edu (Dave Ring) writes:

> Assume Newton's law of cooling (the heat exchange rate is proportional
> to the temperature difference). Assume the milk temp stays constant until
> added. Let room temp be 0, initial coffee temp be 1 and initial milk
> temp be Tm. Assume the milk volume is small.

I'm going to do this with two parameters.  Let V be the ratio of the
milk+coffee volume to the coffee volume, and let H be the amount of
heat in the milk (relative to the room temperature).  So Tm=H/(V-1).
This means that adding milk to coffee at temperature T will change the
temperature to (T+H)/V.  Note that the temperature of the milk is not
dependent on when it's added--this is why it the problem is not
symmetrical between coffee and milk.

> Assume same density, heat capacity, thermal conductivity, etc...
> Assume temp always uniform throughout the liquid. Assume the cup is
> an open-ended insulating cylinder (so no change in surface area).

I think we need to assume that the air above the coffee is
continuously agitated, so that a warm layer doesn't form.  I suspect
that the presence of such a layer, and its disturbance when the cream
is added, are the most serious departures of this model from reality.

> Assume the time period is such that without the milk, the coffee would
> cool to T=1/e .

I'm going to call that temperature L, so I can solve for other final
temperatures.

> Puzzle: Find the range of Tm at which you should add the coffee
> (before/during/after) to get the warmest coffee.

Note that the original puzzle did not provide "during" as an option.
In fact, I don't know how to solve that problem, so I'm simply going
to compare before and after.

Solution follows:

Without milk, the coffee at time t will have temperature
T(t)=exp(t log L)=L^t.  At time 1 it will reach temperature L, and if
the cream is then added then the temperature will be (L+H)/V.

Adding milk first, the temperature will start at (1+H)/V and (due to
the increased volume) will drop at a rate 1/V as fast for any given
temperature.  So the temperature at time t will be
((1+H)/V) exp((t log L)/V).  So at time 1 the temperature will be
((1+H)/V) L^(1/V).  Let R=L^(1/V), so this temperature is (1+H)R/V.

The milk should be added first when (1+H)R/V > (L+H)/V , or
H < (R-R^V)/(1-R).  This is positive, so the FAQ is correct that
chilled milk should also be added first.  But room temperature milk
should also be added first.  The critical milk temperature above which
the milk should be added last is Tm = (R-R^V)/((V-1)(1-R)), or
(L^(1/V)-L) / ( (1-L^(1/V)) (V-1) ).  As V approaches 1, the critical
temperature approaches L log L/(L-1).  In the case of L = 1/e ~ 0.368,
the answer is that you add the milk last when Tm > 1/(e-1) ~ 0.582.
That is pretty warm milk.  In fact, the limiting critical temperature
is always warmer than the temperature of the coffee that has been
cooled without milk, and I think that's the case even when V>1.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil