Science:Math Exam Resources/Courses/MATH100/December 2013/Question 05/Solution 1

The solution of the differential equation $\frac{dT}{dt}=k(T-T_s)$ is

$T-T_s=(T_0-T_s)e^{kt}$, where $T$ is the temperature of an object, $T_s$ is the temperature of its surroundings, and $k$ is some constant.

We are given that $T_s=40,T_0=70,T(0.5)=60$, where $t$ is in hours. Thus we begin by solving for $k$:

$\begin{array}{rl}T(t)-T_s& =(T_0-T_s)e^{kt}\\ 60-40& =(70-40)e^{k\cdot 0.5}\\ e^{0.5k}& =\frac{2}{3}\\ \frac{1}{2}k& =\ln \frac{2}{3}\\ k& =2\ln \frac{2}{3}=\ln {\left(\frac{2}{3}\right)}^2=\ln \frac{4}{9}\\ \\ \therefore T(t)& =30e^{t\ln \left(\frac{4}{9}\right)}+40=30e^{\ln \left[{\left(\frac{4}{9}\right)}^t\right]}+40\\ & =30{\left(\frac{4}{9}\right)}^t+40\end{array}$

Finally, solving for $t$ when $T(t)=50$ yields:

$\begin{array}{rl}50& =30{\left(\frac{4}{9}\right)}^t+40\\ \frac{1}{3}& ={\left(\frac{4}{9}\right)}^t\\ \ln \frac{1}{3}& =t\ln \frac{4}{9}\\ t& =\frac{\ln \frac{1}{3}}{\ln \frac{4}{9}}=\frac{\ln 3}{\ln \frac{9}{4}}\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{s}\end{array}$