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 \[ u (20) =70e ^{[0.1 \ln (4/7)] (20)} +20=70e^{2\ln (4/7)}+20\approx 42.86{}^{\circ}{\rm C} \]