Answer: Coffee and Cream

In order to solve this problem about cooling coffee by adding cream to it, we’ll need to make some simplifying assumptions.

  • The cream is essentially the same stuff as the coffee, only cooler. It has the same density and same heat capacity. We don’t have to worry about the entropy of mixing.
  • When sitting unattended, the coffee cools according to Newton’s cooling law.
  • When we add cream to the coffee, the system reaches equilibrium very quickly. We don’t have to worry about how long it takes the coffee and cream to come to the same temperature, since this is short compared to the time it takes the coffee to cool off.

Newton’s Cooling Law

Newton’s cooling law says that the greater the difference in temperature between the coffee and the room around it, the faster heat will flow from one to the other. Specifically, if the heat in the cup of coffee is Q, then

\frac{dQ}{dt} = k (T_c - T_r)

where k is some constant, T_c is the temperature of the coffee, and T_r is the temperature of the room. k will be negative, meaning that heat flows out of the coffee when it’s hotter than the room. If the cup is a very good insulator, k is a small number, and if not, k is larger.

Newton’s law implies that if the coffee starts out at temperature T_0, its temperature at a later time will be

T_c = T_0 \, e^{-t/\tau} + T_r

where \tau is a time, which roughly measures how long it takes the coffee to cool.

Two Scenarios

Even with these simplifying assumptions, it’s difficult to see straight to the answer to this problem. It might be better to add the cream right away, because the temperature the coffee drops is more dramatic if there’s a bigger difference between the coffee and the cream. On the other hand, the coffee cools from 100C to 99C much more quickly than from 50C to 49C, so if we add the cream right off the bat, we’re skipping over the “cheap” temperatures.

Further, adding the cream to the coffee may change the way it cools. The coffee now has more mass and more heat, so it may cool differently.

We’ll analyze two extreme scenarios. In the first, adding the cream to the coffee doesn’t change the rate it cools. That is, adding the cream does not change \tau. This is the most optimistic scenario, since in real life I would expect that adding cream doesn’t change the surface area as much as the mass, and hence the coffee cools a little more slowly than before.

In the second scenario, adding the cream to the coffee does not affect the rate of heat transfer, that is, it does not affect k. This is a pessimistic scenario, because in real life I would expect adding the cream to add a little more surface area to the coffee, and hence to let it cool a little more quickly.

First Scenario: \tau unchanged

In this scenario, the coffee cools for a while, then has an abrupt temperature drop when the cream is added, then cools a while longer. The abrupt drop saves us time – however much time it would have taken for the coffee to cool down that much on its own. Let’s calculate that time saved, and see when it’s largest.

If I want the coffee to cool from temperature T_a to T_b it happens when

T_b = T_ae^{-t/\tau}+T_r

Solving for t, the time saved,

t = -\tau\ln\left(\frac{T_b - T_r}{T_a}\right).

When I add the cream, the coffee falls to a temperature

T_b = \frac{MT_a + mT_r}{M+m},

which is the weighted average of the temperatures of the coffee and cream (the cream is at temperature T_r, the same temperature as the room.) M is the mass of the coffee, and m is the mass of the cream.

Plugging this expression for T_b into the previous one for the time the coffee would have taken to cool from T_a to T_b, I get

t = -\tau \ln\left(\frac{1-T_r/T_a}{1+m/M}\right).

Because the natural logarithm is strictly increasing, this is maximized whenever the argument of the log is minimized (\tau is positive and -\tau is negative). And the argument of the logarithm is clearly increasing in T_a, so in this scenario, where the coffee cools just as well after you add cream, you should add the cream later rather than sooner.

Second Scenario: k unchanged

This scenario works the same way as the previous one, except that after we add the cream, the mass increases without changing the coffee’s ability to give off heat. Hence, the coffee will cool a little more slowly now, and it may become advantageous to wait before adding the cream.

First lets’ calculate \tau', the new characteristic cooling time after the cream has been added. For fixed k, \tau is linear in the mass, since the heat is also linear in the mass. Hence, \tau' = \frac{M+m}{M}\tau.

This time is longer than \tau, meaning the coffee cools more slowly. That works against the plan of adding the cream early on, since if you add the cream early on you’ll have it in there slowing down the cooling process for more time. We already said it’s better to add the cream later without this effect, so it’s still better even with it.

In the comments to the problem a reader posted a link to Jonathan Afililo’s experiment about this. I only read the abstract/pictures, but he seems to have the same result, only backed up by data. There the author was considering it “good” to keep the coffee hot, whereas I was considering it “good” to cool it down.

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