Bob made a cup of coffee, and Bob likes to drink coffee only once reaches 60 degrees Celsius and will not burn him. Initially at time \(t=0\) minutes, Bob measured the temperature and the coffee was 89 degrees Celsius. One minute later, Bob measured the coffee again and it had 85 degrees. The temperature of the room (the ambient temperature) is 22 degrees. When should Bob start drinking?

Let \(T\) be the temperature of the coffee in degrees Celsius, and let \(A\) be the ambient (room) temperature, also in degrees Celsius. Newton’s law of cooling states that the rate at which the temperature of the coffee is changing is proportional to the difference between the ambient temperature and the temperature of the coffee. That is,

\begin{equation*}
\frac{dT}{dt} = k(A-T) ,
\end{equation*}

for some positive constant \(k\text{.}\) For our setup \(A=22\text{,}\) \(T(0) = 89\text{,}\) \(T(1) = 85\text{.}\) We separate variables and integrate (let \(C\) and \(D\) denote arbitrary constants):

\begin{equation*}
\begin{aligned}
\frac{1}{T-A} \, \frac{dT}{dt} & = -k , \\
\ln (T-A) &= -kt + C , \qquad \text{(note that } T-A > 0 \text{)} \\
T-A &= D\, e^{-kt} , \\
T &= A + D\, e^{-kt} .
\end{aligned}
\end{equation*}

That is, \(T = 22 + D\, e^{-kt}\text{.}\) We plug in the first condition: \(89 = T(0) = 22 +
D\text{,}\) and hence \(D = 67\text{.}\) So \(T = 22 + 67\, e^{-kt}\text{.}\) The second condition says \(85 = T(1) =
22 + 67\, e^{-k}\text{.}\) Solving for \(k\) we get \(k = - \ln \frac{85-22}{67} \approx 0.0616\text{.}\) Now we solve for the time \(t\) that gives us a temperature of 60 degrees. Namely, we solve

\begin{equation*}
60 = 22 + 67 e^{-0.0616t}
\end{equation*}

to get

\(t = - \frac{\ln \frac{60-22}{67}}{0.0616} \approx 9.21\) minutes. So Bob can begin to drink the coffee at just over 9 minutes from the time Bob made it. That is probably about the amount of time it took us to calculate how long it would take. See

Figure 1.9.