An object is heated to $90{}^{\circ}{\rm C}$ and allowed to cool in a room with a constant ambient temperature of $20{}^{\circ}{\rm C}$. If after $10$ min the temperature of the object is $60{}^{\circ}{\rm C}$, what will its temperature be after $20$ min?

Solution When $t=0$, $u(0) =u_{0}=90{}^{\circ}{\rm C}$, and when $t=10$ min, $u\left( 10\right) =60{}^{\circ}{\rm C}$. Given that the ambient temperature $T$ is $20{}^{\circ}{\rm C}$, we substitute these values into equation (5). $$\begin{eqnarray*} \begin{array}{rcl@{\qquad}l@{\hspace*{-6pc}}} u(t) &=&( u_{0}-T) e^{kt}+T \\ 60 &=&( 90-20) e^{10k}+20 & {\color{#0066A7}{\hbox{\(u=60 \hbox{ when }t=10; \ T=20;\ u_{0} =90\)}}} \\ \dfrac{40}{70} &=&e^{10k} \\ k &=&\dfrac{1}{10}\ln \dfrac{4}{7} = 0.1 0 \ln \dfrac{4}{7} \end{array} \end{eqnarray*}$$

The temperature $u$ is $$u(t) =70 e^{[0.1 \ln (4/7)]t} +20$$

Then when $t=20$, the temperature $u$ of the object is $$u (20) =70e ^{[0.1 \ln (4/7)] (20)} +20=70e^{2\ln (4/7)}+20\approx 42.86{}^{\circ}{\rm C}$$