That first sip of coffee in the morning

It is a good idea to enjoy a cup of coffee before starting a busy day.

Suppose the coffee fresh out of the pot with temperature \alpha^{\circ} C is too hot, we can immediately add cream to reduce the temperature by \delta^{\circ} C instantly, then wait for the coffee to cool down naturally to \omega^{\circ} C before sipping it comfortably. We can also wait until the temperature of the coffee drops to (\omega+\delta)^{\circ} C first, then add the cream to further reduce it instantly to \omega^{\circ} C.

Typically, \alpha = 90, \omega = 75, and \delta = 5.

If we are in a hurry and want to wait the shortest possible time, should the cream be added right after the coffee is made, or should we wait for a while before adding the cream?


The heat flow from the hot water to the surrounding air obeys Newton’s cooling and heating law, described by the following ordinary differential equation:

\frac{d}{dt}\theta(t) = k (E-\theta(t))

where \theta(t), a function of time t, is the temperature of the water, E is the temperature of its surroundings, and k>0 is a constant depends on the heat transfer mechanism, the contact are with the surroundings, and the thermal properties of the water.

Fig. 1 a place where Newton’s law breaks down

Under normal circumstances, we have

E \ll \omega < \omega+\delta < \alpha-\delta < \alpha\quad\quad\quad(1)

Based on Newton’s law, the mathematical model of coffee cooling is:

\begin{cases} \frac{d}{dt}\theta(t) = k (E-\theta(t)) \\ \theta(0)= \theta_0\end{cases}\quad\quad\quad(2)

Fig. 2

Solving (2), an initial-value problem (see Fig. 2) gives

\theta(t) = E + e^{-kt}(\theta_0 - E).

Therefore,

t = \frac{\log\left(\frac{E-\theta_0}{E-\theta(t)}\right)}{k}=\frac{\log\left(\frac{\theta_0-E}{\theta(t)-E}\right)}{k}.\quad\quad\quad(3)

If cream is added immediately (see Fig. 3),

Fig. 3 : cream first

by (3),

t_1=\frac{\log\left(\frac{(\alpha-\delta)-E}{\omega-E}\right)}{k}.

Otherwise (see Fig. 4),

Fig. 4: cream last

t_2=\frac{\log\left(\frac{\alpha-E}{(\omega+\delta)-E}\right)}{k}.

And so,

t_1-t_2 = \frac{\log\left(\frac{(\alpha-\delta)-E}{\omega-E}\right)}{k}- \frac{\log\left(\frac{\alpha-E}{(\omega+\delta)-E}\right)}{k}=\frac{1}{k}\left(\log\left(\frac{(\alpha-\delta)-E}{\omega-E}\right)-\log\left(\frac{\alpha-E}{(\omega+\delta)-E}\right)\right)\quad(4)

Fig. 5

Since

\frac{(\alpha-\delta)-E}{\omega-E}- \frac{\alpha-E}{(\omega+\delta)-E}=\frac{(\alpha-\delta-E)(\omega+\delta-E)-(\omega-E)(\alpha-E)}{(\omega-E)(\omega+\delta-E)}=\frac{-\delta(\omega+\delta-\alpha)}{(\omega-E)(\omega+\delta-E)}\overset{(1)}{>0}

implies

\log\left(\frac{(\alpha-\delta)-E}{\omega-E}\right)-\log\left(\frac{\alpha-E}{(\omega+\delta)-E}\right)>0,\quad\quad\quad(5)

from (4) , we see that

t_1-t_2 > 0;

i.e.,

t_1 > t_2

Hence,

If we are in a hurry and want to wait the shortest possible time, we should wait for a while before adding the cream!


Exercise-1 Solve (2) without using a CAS.

Exercise-2 Show that \frac{\alpha-\delta-E}{\omega-E}\cdot\frac{\omega+\delta-E}{\alpha-E} >1.

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