2. A question on exponential decay (Eton College 01C_MT1 Q14)

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This question comes from a test paper for A-level maths for Michaelmas Term at Eton College.

Q14. At time $t$ secons after the start of an experiment, the temperature of a cooling liquid is proportional to $\textrm{e}^{-kt}$.

(a) Given that the liquid's initial temperature was $80^\circ$C, find a formula for $T$ in terms of $t$ and $k$. [1 mark]
(b) Given also that $T=20^\circ$C when $t=6$, find the exact value of $k$. [2 marks]
(c) Calculate the time at which the temperature will reach $10^\circ$C, giving your answer to three significant fitures. [3 marks]

 

Solution. (a) Since the temperature $T$ is proportional to $\textrm{e}^{-kt}$, we write $$ \begin{align} T&\propto \textrm{e}^{-kt} \\ \Rightarrow\quad T&=A\textrm{e}^{-kt} \end{align} $$

The initial temperature was $80^\circ$C, i.e. $T=80$ when $t=0$ and thus $$ \begin{align} A&=80 \\ \Rightarrow\quad T&=80\textrm{e}^{-kt}.\end{align} $$

 

(b) We have $T=60$ when $t=6$ which gives $$ \begin{align} 60&=80\textrm{e}^{-6k} \\ \Rightarrow\quad \textrm{e}^{-6k}&=\frac34 \\ \Rightarrow\quad -6k&=\ln\left(\frac34\right) \\ \Rightarrow\quad k&=-\frac16\ln\left(\frac34\right)=\frac16\ln\left(\frac43\right) \end{align} $$

 

(c) We want the value of $t$ when $T=10$, i.e. $$ \begin{align} 10&=80\textrm{e}^{-kt} \\ \Rightarrow\quad \textrm{e}^{-kt}&=\frac18 \\ \Rightarrow\quad -kt&=\ln\left(\frac18\right) \\ \Rightarrow\quad \frac16\ln\left(\frac34\right)t&=\ln\left(\frac18\right) \\ \Rightarrow\quad t&=\frac{6\ln\left(\frac18\right)}{\ln\left(\frac34\right)}=43.4\quad{\rm (3\;s.f.)} \end{align} $$