Science:Math Exam Resources/Courses/MATH100/December 2010/Question 02 (b)/Solution 1

Proceeding as in the hints, let

$\displaystyle y(t)=T(t)-19$

Then, Newton's law of cooling reads

$\displaystyle {\frac {dy}{dt}}={\frac {dT}{dt}}=k(T(t)-19)=ky(t)$

Solving this differential equation gives

$\displaystyle y(t)=y(0)e^{kt}$

In terms of T and using the initial condition $T(0)=3$, we have

$\displaystyle T(t)-19=(T(0)-19)e^{kt}=(3-19)e^{kt}=-16e^{kt}$

After 30 minutes, we know $T(30)=11$ and so

$\displaystyle -16e^{30k}=T(30)-19=11-19=-8$

Solving for k by isolating and taking logarithms gives

$\displaystyle k={\frac {\ln(1/2)}{30}}=-{\frac {\ln(2)}{30}}$

We want to find the temperature after 90 minutes and so

${\begin{aligned}T(90)-19&=-16e^{90\cdot (-\ln(2))/30}\\&=-16e^{-3\ln(2)}\\&=-16e^{\ln(2^{-3})}\\&=-16(1/8)\\&=-2\end{aligned}}$

and thus $T(90)=19-2=17$ degrees celsius