A problem occured to me over my morning coffee.  The waitress brought
the hot coffee in a cup, and the cold cream in a small metal pitcher. 
After I added some cream to the coffee, the mixture was still too hot
to drink for a few minutes.  Would the waiting time for an acceptable
temperature have been different had I not mixed the two liquids right
away, and let them both settle separately from opposite directions
toward the room temperature?
Upon reflection, it seems there are actually two questions.  The first
is practical -
there are two different types of liquids in two different types of
containers (the coffee cup is tapered, so as the total amount of
liquid in the cup increases, the surface area of the liquid grows
larger, perhaps increasing the rate of   evaporation). The second
scenario is for two amounts of the same liquid, such as water, in two
identical containers, at two different temperatures.  Would mixing
make a difference in this case?

Hello pdq2-ga:

Question one -
I will assume that:
- the coffee is hot (say, 90degC)
- the cream is cold (say, 5degC)
- the final temperature shall be 60degC.

Consider case 1, where you don't add the cream to the coffee.
You wait, and let the coffee cool to 65degC, and then you add the
cream. In the same time interval, the cream warms to 10degC. You then
mix the two, and suppose the coffee + cream cools to 60degC.

Consider case 2, where you add the cream to the coffee immediately.
Then, you wait for the coffee + cream to cool to 60degC.

In case 1, the cream absorbs LESS heat from the coffee than in case 2,
because the cream has already warmed a bit. Thus, the effectiveness of
the cream as a heat-absorber is reduced. In addition, in Case 2, the
surface area of the coffee + cream is greater than the surface area of
coffee + cream in Case 1 - meaning Case 2 dissipates heat slightly
faster than Case 1. Therefore, by this analysis, Case 2 would let the
coffee cool faster.

If we do not consider the warming of the cream, then Case 2 is still
faster, because of slightly higher surface area and faster cooling.

Here, I assume that the heat capacity (the ability to store energy by
heat*) for coffee alone is the same as coffee + milk. This is a
reasonable assumption, because both coffee and milk are mostly water,
with solutes such as calcium or proteins or carbohydrates or fat
making up a very small part.


Question two -
Suppose that:
Case 1) You have two containers with 10 L of water at 90degC and you
want it to cool.
Case 2) You have one container with 20 L water at 90degC and you want
it to cool. (i.e. "mixed")

Under Case 2, you have a limited surface area, so the water takes
longer to cool than Case 1, where there is a greater interface between
the water (and its container) with the air - with more surface area in
two containers, Case 1 cools faster.

Consider a larger case, if the 20 L of water was divided into 20
containers. In such a situation, the surface area would be very high,
resulting in even faster cooling.

In the limiting case, if the 20 L of water was divided into individual
molecules and spread over an infinite area, the cooling would take
place extremely quickly.
For a practical example, think of ice cubes. It would take longer to
cool and freeze a glass of water placed outside in the snow than if
the water was poured on the pavement and let to freeze.

All of my solutions assume that the environment is large (i.e. the
containers are separated and their individual cooling has no effect on
the temperature of the ambient air) and has a reasonable temperature
(i.e. 20degC).

Let me know if you have questions about any part of my answer - please
use the "Clarify Answer" function.

Thanks!!

- supermacman-ga


Fine print:
* Heat capacity, strictly speaking, is actually the ability to store
energy per degree-gram. I'm using "heat" in a very liberal way here,
by its popular definition.

---

Clarification of Answer by supermacman-ga on 08 Feb 2004 13:53 PST

> The second scenario is for two amounts of the same liquid, such as
water, in two identical containers, at two different temperatures. 
Would mixing make a difference in this case?


Whoops! I just realized I answered a slightly different question. Here
is the actual answer for your question two.

Yes, mixing would make a difference. It would result in faster cooling.
Consider two identical containers of two identical solutions, one at
90degC, and the other at 10degC. You want the solutions to reach
20degC.

If you mix the two, then they will reach 50degC immediately. The
solution will then cool to 20degC. If the containers are cylinders
(i.e. glasses), then the surface area will nearly double, and the rate
of cooling will be near equal to the rate of cooling as if they were
separate.

If you don't mix, then it longer for the 90degC container to cool to
20degC. In the meantime, the 10degC solution will warm to 20degC.
Mixing would have no effect, but effectively, you would have waited
for the 90degC solution to cool to 20degC. This takes longer, because
there is a large temperature change.

In fact, the fastest way would be to mix the solutions, let their
temperatures equilibrate, and then separate them - relating my
erroneous answer to this one. This causes a nice increase in surface
area that hastens cooling.

I apologize for the error in my original answer.

- supermacman-ga

---

Clarification of Answer by supermacman-ga on 08 Feb 2004 14:03 PST

I went to work and then thought about this question for a little bit
longer. Here's another explanation for the second part of the
question.

Suppose you have two containers, each with one unit of surface area.
One is at 90degC, the other at 10degC. You want them to cool be
20degC.

If you don't mix, then the first container has to dissipate 90degC -->
20degC worth of heat using one unit of surface area. The difference is
70degC.

If you mix, then the combined container (with, say, 1.5 units of
surface area), has to dissipate 50degC --> 20degC worth of heat using
1.5 units of surface area. The difference is 30degC, but since the
mixed solution is twice as massive, it takes twice as long.
Effectively, the difference is 60degC. However, this solution has 1.5
units of surface area to cool 60degC, while the unmixed container has
1 unit of surface area to cool 70degC. Clearly, mixing will result in
faster cooling.

Again, let me know if you need elaboration. Good question!

 - supermacman-ga

---

Request for Answer Clarification by pdq2-ga on 09 Feb 2004 17:31 PST

Thanks to both of you for your reasoning on this problem.  To my
non-technical mind, it seems that the bottom line is that it is better
to mix immediately, because if you don't, the heat-dissipating power
of the liquid that is cooler than room temperature will be reduced, as
it makes its way gradually, alone, towards room temperature before it
is mixed with the hot liquid.  But what about two liquids that are
both hotter than room temperature?  For example:
Room Temperature:  20degC
1st liquid:        50degC
2nd liquid         30degC

Thanks,  pdq-ga

---

Clarification of Answer by supermacman-ga on 09 Feb 2004 19:49 PST

Your situation:
50degC and 30degC cooling to 20degC

The first solution has a temperature difference of 30degC. If it cools
one-tenth of the difference (3degC) in one time unit, it cools to
47degC.
The second solution has a temperature difference of 10degC. It cools
to 29degC in one time unit (one-tenth of its difference, or 1degC).
Mix now, and you get 38degC.

If you mix immediately, you get 40degC. One-tenth is 38degC (the "catch-up" part).

However, here, the surface area of the COOLING solution is REDUCED.
Before mixing, you had TWO containers, BOTH cooling to room
temperature. After mixing, you have one larger container, but without
double the surface area.

Thus, you do not mix for a cooler solution... since you want the
maximum amount of surface area cooling a solution.

This is like my erroneous answer - read that, and you'll see why you
want to maximize surface area.

- supermacman-ga

Excellent, clear answer!  Now I can enjoy my coffee without worrying.

Your interesting question reminds of a very old joke in which Mike
says to Pat, "I just got one of those things -- I think they call it a
Thermos bottle -- that keeps hot things hot and cold things cold." 
Says Pat, "But how do it know?"

markj-ga

For mathematical and physics background to do with heat:

http://www.taftan.com/thermodynamics/CP.HTM
http://scienceworld.wolfram.com/physics/HeatCapacity.html
http://en.wikipedia.org/wiki/Specific_heat_capacity

I think that the researcher's answer to this question is quite wrong. 
Under ordinary circumstances, drinkable temperature will be reached
faster if the coffee and cream are not mixed immediately.  There are 2
reasons for this.

1) The rate at which heat is transfered between a cooling (or warming)
body and its environment is proportional to the temperature
difference.  If the constant of proportionality is the same for the
coffee alone, the cream alone, and the coffee-cream mixture (that is,
if the temperature difference between each liquid and its environment
drops to half its former value in, say, 10 minutes) then the length of
time to reach a given temperature is exactly the same whether you mix
immediately or after waiting.  However, the constants of
proportionality will not all be the same.  It would take the
coffee-cream mixture longer to halve its temperature difference from
the environment than it would take the coffee alone, because the
mixture has more thermal inertia.  It's obvious from experience that
half an inch of hot coffee in the bottom of a cup cools much faster
than a full cup.  So since the rate of cooling of the mixture is
slower than than that of the coffee alone, and since the cream begins
much closer to room temperature than the coffee, it's better to wait
than to mix immediately.

As an example, let's say the coffee is 90C, the cream is 6C and room
temperature is 20C.  Then the coffee is 70 degrees warmer than
equilibrium and the cream is 14 degrees cooler than equilibrium.  That
is, the temperature difference is 5 times greater for the coffee
(that's why I chose the 'weird' temperature of 6C for the cream).  Now
let's say you mix equal amounts of coffee and cream (obviously, no one
drinks this much cream in their coffee, but it makes the calculations
simple, and it doesn't change the outcome).  Assuming similar thermal
properties of the containers, the temperature of the coffee will
decrease 5 times as fast as the temperature of the cream will
increase.  If we wait for the coffee to cool by 10C (one seventh of
its temperature difference from the environement) to 80C, the cream
will have warmed to 8C (also changing its temperature difference by
one seventh), and when we mix, we get 44C.  Now if we mix immediately,
we get 48C, which is 28C warmer than room temperature.  To 'catch up'
to the case where we waited to mix, the mixture has to cool by 4C, or
one seventh of its temperature difference.  Since the coffee-cream
mixture will have a slower cooling rate than either liquid alone, it
will take it longer to achieve this one-seventh cooling factor than it
took the two liquids separately.

2) More steam will rise from the hot coffee than from the cooler
coffee-cream mixture.  Steam has a very large amount of latent heat,
so it carries away lots of heat from the coffee.

Hello racecar-ga,

I see where you're coming from, but:
1) You mentioned the following:
> Since the coffee-cream mixture will have a slower cooling rate than
either liquid alone, it will take it longer to achieve this
one-seventh cooling factor than it took the two liquids separately.

a) Since the temperature difference is one-seventh, and since the
coffee, cream after waiting also cooled one-seventh, and since the
rate of cooling is proportional to the difference in temperature, by
your argument, the time to cool is the same.

Suppose it takes one time unit to cool or warm one-seventh of a
liquid's temperature difference. Thus, it takes one time unit for the
coffee to cool from 90degC to 80degC, and one time unit for the cream
to warm from 6degC to 8degC. Mixing results in a 44degC solution.

Then, consider mixing immediately. We get a 48degC solution. Then
cooling one-seventh of its temperature difference, to 44degC, takes
one time unit. The time is the same for both, by this argument.
However:

b) The coffee-cream mixture has a higher surface area, higher
interface between the environment, and faster cooling rate. Thus,
mixing wins, not because of temperature differences but because of a
surface area increase.


2) You are correct that steam from the coffee will cool down the
solution, but consider this -
We have a 90degC coffee that cools to 80degC by evaporation. It takes
a unit time t to do this.
We also have 90degC coffee that could cool immediately to 80degC by
mixing. It takes zero time to do this.

Given, after mixing, the mass of the coffee + cream would be greater
than the mass of the coffee alone, but since cream is a small part of
the overall coffee + cream, it can be disregarded. As well, the higher
surface area may justify this simplification.

- supermacman-ga

Here's an answer giving a mathematical proof of my answer.

http://faculty.salisbury.edu/~kmshannon/math202/coffee.htm

They examine two scenarios, one where the cream is kept cold, the
other where the cream is left out. In the former, the temperature is
the same; in the latter, the temperature is colder without mixing.
However, the former is applicable to this question only.

Note that this solution does not consider an increase in surface area.
If we do consider the increase surface area, then mixing immediately
means the coffee will cool faster.

This is basically a mathematical confirmation of my intuitive
reasoning based on Newton's Law of Cooling, above.

- supermacman-ga

Hello.

That's a cool link.  But it says the reverse of the comment above.  If
the cream is left in the fridge, the coffee will be cooler if you mix
later, whereas if the cream is left out, the temperatures will be the
same.  Just a typo.  The case here is that the cream and coffee are
both on the table (not in the fridge), so under the assumptions in the
link, the temperature would be the same with or without mixing.  My
opinion that in reality waiting to mix will result in a lower
temperature is based on the assumption that a larger volume of liquid
will cool more slowly.  Supermacman is right that if surface area
increases by more than volume does when you mix, mixing immediately
could result in a lower temperature.  But if you think about it, it's
hard to imagine a shape of cup that would allow this to happen.

I think the best way to answer this for certain is to do an experiment =)

Somehow I am reminded of this old story...

Three college physics students were given this problem: If you have a
3 pound roast at room temperature (70 degrees Fahrenheit), and you put
it in a 350 degree oven, how long will it take for the temperature at
the center of the roast to reach 160 degrees?

The first student determined that a roast is mostly water, and used
the formula for the transmission of heat through water to calculate an
answer.

The second student bought a 3 pound roast, left it out at room
temperature for several hours, then placed it in a 350 degree oven
with a meat thermometer in it. He recorded the amount of time it took
for the thermometer to reach 160 degrees.

The third student called his mother and asked, "Mom, how long do I
cook a 3 pound roast?"

Regarding the analysis at the link quoted by supermacman,

http://faculty.salisbury.edu/~kmshannon/math202/coffee.htm

This derivation is incorrect because among other things one of the
critical assumptions is faulty.  The author assumed that the
proportionality constant would be the same for coffee with cream as
without cream.  This could be approximately true only if the mass of
the cream was very much smaller than the mass of the coffee and the
surface area available for heat transfer was relatively constant.

What the author labels the proportionality constant, k, is actually equal to

  k = (K * A) / (M * C)

  K = the thermal conductivity (in this case, equivalent to the combined
  effect of conduction, convection, and radiation)

  A = the area available for heat transfer

  M = the mass of the liquid

  C = the heat capacity of the liquid

According to the author, the formula to calculate the final
temperature for the two cases would be,

Case 1: Add room temperature cream early
  T(t) = C + [(aH + (1-a)C) - C]e^(-kt)
        = C + [aH - aC]e^(-kt)

Case 2: Add room temperature cream late
  T(t) = a[C + (H - C)e^(-kt)] + (1-a)C
        = C + [aH - aC]e^(-kt)

where,
  C = room and cream temperature
  H = starting black coffee temperature
  a = mass of coffee / mass of cream

The author's equation for Case 1 is incorrect.  If we assume no change
in heat loss area (which may be true if container is a cylindrical
styrofoam cup), then the Case 1 equation would be,

  T(t) = C + [aH - aC]e^[-akt/(a+1)]

where k is the proportionality constant for the original black coffee.

Plug in the numbers and it should be clear that the total heat loss in
time, t, for Case 1 is LESS than for Case 2.

newbery-ga