A cup of freshly poured coffee is at 90 C and is left in a room at 20 C. After 10 minutes the coffee has cooled to 65 C. Find the temperature after 20 minutes, and find how long it takes to reach 30 C.

Step-by-step solution

1. 1

   Set up the differential equation. Newton's law of cooling says the rate of cooling is proportional to how far the object is above its surroundings, not to its temperature itself:

   $\frac{dT}{dt}=-k\,(T-T_a),\qquad T_a=20^\circ\text{C}$

   Separating variables and integrating gives the standard solution

   $T(t)=T_a+(T_0-T_a)e^{-kt}=20+70e^{-kt}$

   since $T_0-T_a=90-20=70$.

2. 2

   Use the 10-minute reading to find $k$. This is the only unknown in the model, and one data point pins it down:

   $65=20+70e^{-10k}\;\Longrightarrow\;e^{-10k}=\frac{45}{70}=\frac{9}{14}$

   $k=-\frac{1}{10}\ln\frac{9}{14}=\frac{1}{10}\ln\frac{14}{9}\approx0.04418\ \text{min}^{-1}$

3. 3

   Get the 20-minute temperature without ever using the decimal $k$. Because $20=2\times10$, the exponential simply squares:

   $e^{-20k}=\left(e^{-10k}\right)^{2}=\left(\frac{9}{14}\right)^{2}=\frac{81}{196}$

   $T(20)=20+70\cdot\frac{81}{196}=20+\frac{405}{14}=20+28.93=48.93^\circ\text{C}$

   Keeping the exact fraction avoids the rounding drift you get from $70e^{-0.0442\times20}$.

4. 4

   Solve for the time to reach $30^\circ\text{C}$. Set $T(t)=30$:

   $30=20+70e^{-kt}\;\Longrightarrow\;e^{-kt}=\frac{10}{70}=\frac{1}{7}\;\Longrightarrow\;kt=\ln 7$

   $t=\frac{\ln 7}{k}=\frac{10\ln 7}{\ln(14/9)}=\frac{19.459}{0.44183}\approx44.0\ \text{minutes}$

5. 5

   Sanity-check the two answers against each other. The coffee loses $25^\circ$ in the first 10 minutes but only about $16^\circ$ in the second 10 minutes ($65\to48.9$) — decreasing losses are exactly what an exponential approach to room temperature predicts. And $30^\circ\text{C}$ is only $10^\circ$ above the room, so it should take a long tail of time to get there; $44$ minutes is consistent. The coffee never actually reaches $20^\circ\text{C}$ in finite time, because $e^{-kt}>0$ for all $t$.