A couple of years ago, my Chemistry teacher wondered out loud about whether putting milk in her tea later or earlier would cool it faster so she would be able to drink it during break time. I had a strong hunch that it would be putting it in later due to the higher rate of cooling due to the larger heat difference between the room and the tea. But I had no proof – until now (well a couple of years ago when I did the experiment).

$$T_T = \text{initial tea temperature}$$

$$T_M = \text{milk temperature}$$

$$T_R = \text{room temperature}$$

$$C_T = \text{heat capacity of tea} = \text{mass} \cdot \text{specific heat capacity}$$

$$C_M = \text{heat capacity of milk} = \text{mass} \cdot \text{shc}$$

$$t = \text{time}$$

$$T_M < T_R$$

$$C_M, C_T > 0$$

$$k > 0$$

Temperature of mixture of 2 liquids:

$$M(T_1, T_2) = \frac{T_1 C_1 + T_2 C_2}{C_1 + C_2}$$

Temperature of object against time:

$$C(T_T, t) = T_R + (T_T – T_R)e^{-kt}$$

When is \(M(C(T_T, t), T_M) \leq C(M(T_T, T_M), t)\)?

$$\frac{( T_R + (T_T – T_R)e^{-kt} ) C_T + T_M C_M}{C_T + C_M} \leq T_R + \left ( \frac{T_T C_T + T_M C_M}{C_T + C_M} – T_R \right ) e^{-kt}$$

$$(T_R + (T_T – T_R)e^{-kt}) C_T + T_M C_M \leq T_R(C_T + C_M) + (T_T C_T + T_M C_M – T_R(C_T + C_M) ) e^{-kt}$$

$$T_R C_T + (T_T – T_R)e^{-kt} C_T + T_M C_M \leq T_R(C_T + C_M) + (T_T C_T + T_M C_M – T_R(C_T + C_M) ) e^{-kt}$$

$$e^{-kt}((T_T – T_R) C_T – (T_T C_T + T_M C_M – T_R(C_T + C_M)) \leq T_R(C_T + C_M) – T_M C_M – T_R C_T $$

$$e^{-kt}(T_T C_T – T_R C_T – T_T C_T – T_M C_M + T_R C_T + T_R C_M) \leq T_R C_T + T_R C_M – T_M C_M – T_R C_T $$

$$e^{-kt}(- T_M C_M + T_R C_M) \leq C_M (T_R – T_M)$$

$$e^{-kt} \cdot C_M (T_R – T_M) \leq C_M (T_R – T_M)$$

Division is allowed because \(C_M > 0, T_R > T_M \therefore T_R – T_M > 0\)

$$e^{-kt} \leq 1$$

$$-kt \leq \ln(1)$$

$$kt \geq 0$$

$$k \gt 0 \therefore$$

$$t \geq 0$$

Thus, as long as the milk is colder than room temperature, your tea will always cool quicker if you add milk after letting it cool rather than adding milk and then waiting for it to cool. This result also means that if you want to keep a cup of tea warmer for longer (and you are going to add milk), you should add milk as soon as possible (whilst also factoring in optimum tea extraction temperature).

Now all this result shows is that it will be cooler, but without any knowledge of the variables, it’s impossible to say whether the difference is perceptible or even measurable. So I did the experiment.

I made two cups of tea, removing the teabag then measuring the temperature over 5 minutes using a cheap digital thermometer from Amazon connected to an Arduino. In one I added milk immediately while with the other I added it at the end. The raw data can be found here. I found that there was about a 3°C difference – definitely measurable but not particularly noticeable.

So, if you need your tea to cool down faster, put the milk in later. However, I’ve found the inverse much more useful – keep tea warmer for longer by adding milk sooner.

N.B. There are a number of things that I didn’t account for in the theoretical section: I modelled the cooling as a very simple exponential decrease depending only on the temperature of the liquid and the room temperature. This is quite a simplification and may break down in some situations. It doesn’t capture any of the following:

- Specific heat capacity of mixture
- Difference in surface area of liquid (i.e. when the milk is added first, that cup has a larger surface area)
- Difference in mass of the liquid (as above)
- The effects of evaporative cooling
- Probably many other factors