You are just served a hot cup of coffee and want it to be as hot as possible later.

Additional material

[ Editor: a 1994 discussion of this entry, started by Neil Harding
<etlnei@etlxd13a.ericsson.se>, who questioned the solution above.  His
question, quoted in the replies, was: ]

>I've been looking through the puzzle archive and I came across
>physics/milk.and.coffee.p which says:
>
>  You are just served a hot cup of coffee and want it to be as hot as
>  possible later.  If you like milk in your coffee, should you add it
>  when you get the cup or just before you drink it?
>
>In summary, the answer given is:
>
>  Warm milk.  Adding milk later is better.
>  Room temp.  And it doesn't matter.
>  Cold milk.  And adding milk now is better.
>
>This doesn't seem right to me. Consider the case when the coffee is piping
>hot and the milk is at room temp. The coffee will be losing heat at a rate
>that depends on its difference to room temperature. If the milk is added
>straight away the temperature of the coffee will drop so it will lose heat
>less rapidly. This means that there will be more heat left in the coffee
>when the time comes to drink it.
>
>I could be wrong of course, what do you think?

[ Editor: Dave Ring, in mail to the archive maintainer, quoting the
same question and the first paragraph of the solution: ]

I have to agree that the archive is in error here. If the amount of
milk is small then the difference between the final results will be
small, and the effect of the change in the constant of proportionality
cannot be neglected.

It seems more sensible to me to assume the rate of _heat transfer_ is
proportional to the temperature difference and this constant does not
change when the milk is added. Such would be the case if the cup were
an open ended insulating cylinder. (constant surface area)

With this assumption, one wants the coffee to have minimal average
temperature over the time period, so that cold milk should be added
early, hot milk should be added later, and one needs a calculation to
determine the cutoff milk temperature, where the two methods agree.

I get m = (y - 1/e) / (f - fy)   where m is the ratio of milk to
(initial) coffee temperatures, f is the ratio between initial milk and
coffee volumes, and y = exp[-1/(1+f)]

For small f this is 1/(e-1) (not zero as in the archive entry)

-- Dave Ring <dwr2560@tam2000.tamu.edu>, August 1994

[ Editor: Dean Inada's reply to Neil Harding: ]

Remember that adding the milk also causes a drop proportional to its
temperature difference.  If the milk drops the temperature difference
to, say, 1/2 and waiting drops the difference to 1/3, 1/3 * 1/2 = 1/2 *
1/3.
However, I would question the assumption that heat loss is proportional
to the temperature difference.
If the coffee is steaming, I'd expect its loss rate to be more than
linear with the temperature difference.
Also, coffee+milk in the cup should have more heat capacity than just
coffee, without proportionally increasing the surface area, so I would
expect the constant of proportionality between temperature difference
and cooling rate to be lower for the combined system.
In the other two cases, I would question whether the milk container was
better insulated than the coffee container.  With coffee in a thermos,
and cold milk in a little creamer, perhaps it's better to let the milk
warm to room temperature before adding it.

-- Dean Inada <dmi@dmi.cerfnet.com>

In <1994Aug26.045402.10867@sol.UVic.CA> nagarwal@uglw.UVic.CA (Nainesh
Agarwal) writes:
>I agree with you completely. Atleast that is what I have always
>known the answer to be. The reason is exactly as you stated.
>Perhaps there is something in this question that we are missing.
>Then again maybe not. :-)
>
>I again could be wrong, and would love to have someone
>clear up the problem.

I have heard this problem several times over the years.  The "correct"
answer would seem to be based on a quantitative analysis vs. a
qualitative one.  There are many factors at play here, each pulling the
solution in a particular direction.  For instance the fact that the
solution will cool more quickly the greater the differential between
the solution temperature and the ambient temperature is pulling the
solution towards the answer which would put the milk in sooner.
However, if the cup has slanted sides, then the addition of milk early
will increase the radiating surface and the conducting surface (through
the side of the cup) and increase the cooling effect.

And then of course, a darker colored liquid will radiate more quickly,
so that would have you put the milk in sooner.  And so it goes...  What
other factors can you think of?  All the factors must be thought of,
then a quantitative analysis performed.

-- Stan Rosenfeld <vegstan@ix.netcom.com>

As I understand it Neil Harding (the original questioner) was
accepting the archive solution's assumption that loss of heat to the
environment, governed by Newton's Law of Cooling, is the only
significant effect. He was then asking whether/why it's true that if
the coffee's hot and the milk's at room temperature it makes no
difference when you add the milk. Leaving aside the question of
whether this _is_ the only significant effect this is still an
interesting question. My stab at why Neil's counter-argument is wrong:

Assume the coffee is hotter than both the milk and the room to start
with. It's true that if you add the milk at the start the coffee-milk
mixture will cool less rapidly than the coffee on its own would.
However, as you add the milk the fall in temperature from the coffee
to the coffee-milk mixture will be larger if you add the milk at the
start than if you add it at the end, because the difference in
temperature between the coffee and milk is larger if you add the milk
at the start.

So either the temperature drops a lot and then falls slowly (milk
added at start), or it falls fast and then drops a small amount (milk
added at end). It's not obvious which of these strategies will give a
hotter final mixture, and you need to solve the equations to find
out. The archive solution does this, and finds that if the milk's at
room temperature the effects exactly balance, so it doesn't matter
when you add the milk.

-- Thomas Bending <tdb@maths.qmw.ac.uk>

There is no net heat loss to the outside when you mix the coffee and
the milk, so the mixing itself is not really relevant. Whether the
conditions are such that Newton's law of cooling applies exactly or
not, it is always going to be the case that the higher the temperature
of the coffee, the greater the heat loss, so it is always better to add
the milk immediately. (I was sort of assuming that the milk is cooler
than the coffee, but it doesn't really have to be - if not, the same
argument applies with milk and coffee interchanged.)

-- Derek Holt <dfh@maths.warwick.ac.uk>

In article <1994Aug31.094211.24137@dcs.warwick.ac.uk>
dfh@maths.warwick.ac.uk (Derek Holt) writes:

> There is no net heat loss  to the outside when you mix the coffee and the
> milk, so the mixing itself is not really relevant. Whether the conditions
> are such that Newton's law of cooling applies exactly or not, it is
> always going to be the case that the higher the temperature of the coffee,
> the greater the heat loss, so it is always better to add the milk
> immediately. [...]

I think you're confusing "heat loss" and "temperature loss". It is
true that the higher the temperature of the liquid, the greater the
rate of _temperature_ loss. However, if we add the milk immediately
this increases the mass of the liquid involved in the cooling. Hence
although the rate of _temperature_ loss during the cooling is now
smaller, the rate of _heat_ loss is not necessarily smaller, and in
fact if the milk was initially at room temperature the rate of _heat_
loss will be the same whether we add the milk immediately or later.

-- Thomas Bending <tdb@maths.qmw.ac.uk>

In article <TDB.94Aug31121642@galois.maths.qmw.ac.uk> tdb@maths.qmw.ac.uk
(Thomas D Bending) writes:
>I think you're confusing "heat loss" and "temperature loss".

No, I think that is exactly what you are doing.
> It is
>true that the higher the temperature of the liquid, the greater the
>rate of _temperature_ loss.

No, it is the rate of loss of energy (i.e. heat) which increases with
temperature differential.

Let me quote from a very old article of Martin Gardner on the subject:

"On the basis of Newton's law of cooling (which states that the rate of
heat loss is proportional to the difference between the temperature of
the hot material and the temperature of the ambient), and taking into
consideration the significant and easily overlooked fact that the
volume of the coffee increases after the cream is added, it turns out
that an immediate mixing of the liquids always conserves heat. This is
true regardless of whether the cream is at ambient temperature or
below. Other factors such as changes in the rate of radiation due to
the lightened colour of the liquid, an increased surface area in cups
with sloping sides, and so on, have a negligible influence."

-- Derek Holt <dfh@maths.warwick.ac.uk>

The answer to this problem certainly depends on the assumptions. It
turns out the archive answer was wrong for the set of assumptions it
makes, so the 'doesn't matter' answer should be disregarded. I have
written to Chris, explaining why the archive answer is wrong. But it
makes a good puzzle, so I will just give a hint here:

Assumptions:

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.
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).
Assume the time period is such that without the milk, the coffee would
cool to T=1/e . I hope I didn't forget anything.

Puzzle: Find the range of Tm at which you should add the milk
[Editor: "coffee" in the original] (before/during/after) to get the
warmest coffee.

-- Dave Ring <dwr2560@tam2000.tamu.edu>

dwr2560@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>

***SPOILER ALERT***

Basic answer:
The only loss of energy is heat loss, and a hot liquid will lose heat
more rapidly than a cool one. If the milk is added first, the coffee
will be cooler at any given time, and lose heat slower. So there will
be more energy at the end, and thus the coffee will be warmer.

This analysis depends on several assumptions, for example, the milk is
assumed to be much cooler than the coffee, and it does not lose heat
when sitting.

If the milk starts out _hotter_ than the coffee, the same argument
demonstrates that the milk should be added later. In order to determine
at what temperature the best strategy changes, we need a mathematical
analysis.

Typical assumptions:

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 (M) is small.
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).
Assume cooling rate is such that without the milk, the coffee would
cool to T=exp[-t] where t is the time period.

Adding the milk later results in a final temp
Tf = ( 1*exp[-t] + M*Tm ) / (1+M)

Adding first: Tf = exp[ -t/(1+M) ] ( 1*1 + M*Tm ) / (1+M)

Note the slower cooling rate due to the larger mass of liquid.

Setting these equal and taking M small, we get Tm = t / ( exp[t] - 1 )

Above this temperature, we add the milk later. Below it, before.

Adding the milk in the middle is never the best strategy. If the milk
is added when it is hotter than the coffee, then it would be better to
wait a little longer (the energy argument again.) If the milk is added
when it is cooler than the coffee, then it would have been better to
add it a little sooner. At the same temp you can improve either way. So
before and after are the only possibilities.

-- Dave Ring <dwr2560@tam2000.tamu.edu>

[ Editor: Dave Ring also mailed this text to the archive maintainer as
a proposed solution. ]

I am the person who started this thread and I must say that I'm very
pleased with the response. I had given thought to this problem long
before I discovered rec.puzzles, so when I saw it in the archive I
thought I already knew the answer. I was surprised to find that the
answer given was different to mine.

To summarise:

I thought that it would be better to add the milk straight away but
the archive answer said that if the milk's at room temp then it doesn't
matter.

My reason for wanting to add the milk straight away is that the coffee
will lose heat at a rate that depends on its difference to room
temperature. If the milk is added straight away the temperature of the
coffee will drop so it will lose heat less rapidly. This means that
there will be more heat in the coffee when the time comes to drink it.

I was looking at *heat* loss, whereas the answer in the archive looks
at *temperature* loss. A number of people have put the *temperature*
loss method into words. e.g.:

(From a post by Thomas Bending)
... either the temperature drops a lot and then falls slowly (milk
added at start), or it falls fast and then drops a small amount (milk
added at end).
(end of quote)

I believe that either method is valid and should result in the same
answer.  The reason that the archive comes to a different conclusion
is in an assumption it makes, i.e.

(from the archive solution)
Assume [...] that the amount of milk is sufficiently small that the
constant of proportionality is not changed when you add the milk.
(end of clip)

The archive solution is based on rate of loss of temperature, so the
assumption could be re-written as: "Assume that the time taken for the
temperature to drop by a given fraction is the same before and after
the milk is added".

This means that the time taken for the coffee's temperature to drop by
a half is the same before and after the milk is added.

If the temperature drops by a half, that means that half the heat has
been lost. In the case when the milk is at room temperature, the only
source of heat is from the coffee, so "half the heat" is the same
amount regardless of whether the milk is added or not. If it takes the
same amount of time to lose half the heat, the rate of heat loss must
be the same.

Therefore the assumption from the archive can be re-written as "Assume
that the rate of heat loss is not changed when you add the milk".
This is obviously not a valid assumption.

-- Neil Harding <etlnei@etlxd13a.ericsson.se>