Cooling coffee with milk

Relevant wiki: Newton's Law of Cooling

Let $\Delta T = T_\text{coffee} - T_\text{room}$ be the temperature difference between the coffee and the environment. Newton's cooling law yields $\begin{aligned} \frac{d Q}{dt} &= C_\text{coffee} \frac{d \Delta T}{d t} = - h A \Delta T \\ \Rightarrow \quad \Delta T &= \Delta T_0 \exp(-t/\tau_\text{coffee}), \quad \tau_\text{coffee} = \frac{C_\text{coffee}}{ h A} \end{aligned}$ with the heat capacity $C_\text{coffee}$ of the coffee, the heat transfer coefficient $h$ , the surface area $A$ and the time constant $\tau_\text{coffee}$ for cooling. Note, that the milk coffee has a larger heat capacity $C_\text{mix} = \frac{6}{5} C_\text{coffee}$ , so that the time constant $\tau_\text{mix} = \frac{6}{5} \tau_\text{coffee}$ is correspondingly larger. Since the pure coffee needs 15 minutes to cool down, we can determine the time constants $\begin{aligned} \Delta T &= 60^\circ \text{C} \cdot \exp(-15\,\text{min}/\tau_\text{coffee}) = 40^\circ \text{C} \\ \Rightarrow \quad \tau_\text{coffee} &= \frac{15}{\log(3/2)} \approx 37\,\text{min} \\ \Rightarrow \quad \tau_\text{mix} &= \frac{18}{\log(3/2)} \approx 44.4\,\text{min} \end{aligned}$ The mixing temperature $T_\text{mix}$ of the milk coffee results from the energy conservation $\begin{aligned} Q &= C_\text{mix} T_\text{mix} = C_\text{coffee} T_\text{coffee} + C_\text{milk} T_\text{room} = C_\text{coffee} \left(T_\text{coffee} + \frac{1}{5} T_\text{room} \right) \\ \Rightarrow \quad \Delta T_\text{mix} &= \frac{5}{6} \Delta T, \quad \Delta T_\text{mix} = T_\text{mix} - T_\text{room} \end{aligned}$ Therefore, the temperature-time curve results $\Delta T = \begin{cases} \Delta T_0 e^{-t/\tau_\text{coffee}} & t < t_\text{mix} \\ \frac{5}{6} \Delta T_0 e^{-t_\text{mix}/\tau_\text{coffee}} e^{-(t -t_\text{mix})/\tau_\text{mix}} & t \geq t_\text{mix} \end{cases}$ with the time $t_\text{mix}$ when the milk is added. The required cooling time $t_\text{cool}$ , at which a temperature difference $\Delta T = 40^\circ\,\text{C}$ is reached, yields for $t_\text{cool} > t_\text{mix}$ $\begin{aligned} \Delta T &= 40^\circ\,\text{C} = 50^\circ\,\text{C} \cdot e^{-t_\text{mix}/\tau_\text{coffee}} e^{-(t_\text{cool} -t_\text{mix})/\tau_\text{mix}} \\ \Rightarrow \quad t_\text{cool} &= \tau_\text{mix} \log\frac{5}{4} - \left(\frac{\tau_\text{mix}}{\tau_\text{coffee}} - 1\right) t_\text{mix} \end{aligned}$ Since $\tau_\text{mix} > \tau_\text{coffee}$ , the cooling decreases with increasing $t_\text{mix}$ . For the cases $t_\text{mix} = 0 \,\text{min}$ and $8 \,\text{min}$ we get the results, $t_\text{cool} \approx 9.9 \,\text{min}$ and $8.3 \,\text{min}$ , respectively. In conclusion, if the coffee is left to cool for 8 minutes before adding the milk, the coffee reaches the drinking temperature 96 seconds earlier.