How Long Until My Coffee Cools?

Newton's law of cooling, one cup at a time

Oct 2, 20262 min read#Calculus #ODEs

A freshly brewed pour-over starts at about $90^\circ\mathrm{C}$ , and I like to drink mine around $60^\circ\mathrm{C}$ . The question is a humble one: how long do I have to wait?1In practice I never wait. My tongue has paid the price many times.

A model that is too simple

Newton gave us a handy rule of thumb: an object cools at a rate proportional to the difference between its temperature and that of its surroundings. As an equation,

\frac{dT}{dt} = -k\,\bigl(T - T_{\text{room}}\bigr).

Here $T_{\text{room}}$ is the room temperature and $k>0$ is a constant.2It depends a lot on the cup. Ceramic, paper and a thermos can differ by an order of magnitude; a good thermos has a $k$ so small it feels like cheating. The minus sign says that coffee hotter than the room gets cooler, and the bigger the gap, the faster it falls.

Definition 1 (cooling constant)

We call $k$ the cooling constant, measured in $\text{min}^{-1}$ . It bundles the cup, the table, the airflow and everything else we are too lazy to model.

Solving it

Lemma 2 (cooling curve)

If $T(t)$ satisfies the equation above with $T(0)=T_0$ , then for all $t\ge 0$ ,

T(t) = T_{\text{room}} + \bigl(T_0 - T_{\text{room}}\bigr)\,e^{-kt}.

Proof.Let $u = T - T_{\text{room}}$ , so $u' = -ku$ . Then $\bigl(u\,e^{kt}\bigr)' = (u' + ku)\,e^{kt} = 0$ , hence $u\,e^{kt}$ is constant and equals its value at $t=0$ , namely $T_0 - T_{\text{room}}$ .

In other words, the temperature gap decays exponentially. Drag the measurement at the 5-minute mark to feel the difference a cup makes:

Figure 1. Room at 22°C, coffee starts at 90°C. The dashed line is the ideal sipping temperature.

Doing the numbers for my cup

I measured once: the room was $22^\circ\mathrm{C}$ , the coffee started at $90^\circ\mathrm{C}$ , and after 5 minutes it read $75^\circ\mathrm{C}$ . By Lemma 2,

e^{-5k} = \frac{75-22}{90-22} = \frac{53}{68} \quad\Longrightarrow\quad k = \frac{1}{5}\ln\frac{68}{53} \approx 0.0498.

Solving $T(t) = 60$ then gives

t = \frac{1}{k}\ln\frac{68}{38} \approx 11.7\ \text{min}.

So the answer is about 11 minutes 40 seconds: just enough time to answer two emails, or to stare at the cup.

from math import log

k = log(68 / 53) / 5   # cooling constant, 1/min
t = log(68 / 38) / k
print(f"{t:.1f} min")  # 11.7 min

Bonus: when should the milk go in?

Suppose you will wait 10 minutes before drinking, and you want to add a splash of cold milk. Add it now, or after the 10 minutes?

Corollary 3

If the goal is the coolest possible coffee, add the milk last.

The intuition: hot coffee has a large temperature gap and sheds heat quickly. Adding milk first makes it “not hot enough” from the start and wastes the fastest-cooling stretch. The rigorous proof is left as an exercise.3”Left as an exercise” is the oldest form of laziness in mathematics, and I intend to keep the tradition alive.