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{\bf Question}

When a thermometer is placed in a liquid, the rate at which the
indicated temperature $\theta$ rises is proportional to the
difference between the indicated temperature and the true
temperature $T$ of the liquid. Initially the thermometer indicates
$15^{\circ}C$; 30 seconds later the indicated temperature is
$20^{\circ}C$, and a further 30 seconds later it is $21^{\circ}C$.
Show that the differential equation governing the rise of the
thermometer reading is

$$\ds\frac{d\theta}{dt}=k(\theta-T),\ \ k=\rm{constant}$$

Solve this with the given data to determine the true temperature
of the liquid, which is assumed to be constant.

\medskip

{\bf Answer}

Let true temperature be $T$. Let thermometer reading be $\theta$,
time be $t$.

At $\begin{array} {ll} t=0 &amp; \theta=15^{\circ}C\\t=30 secs &amp;
\theta=20^{\circ}C\\t=60 secs &amp; \theta=21^{\circ}C \end{array}$

What is the differential equation?

$\begin{array} {rrcl} &amp;\rm{rate of temp rise} &amp; \propto &amp;
\theta-T\\ \Rightarrow &amp; \ds\frac{d\theta}{dt} &amp; \propto &amp;
\theta-T\\ \Rightarrow &amp; \ds\frac{d\theta}{dt} &amp; = &amp; k(\theta-T)
\end{array}$

$k$=constant of proportionality

This is variables separable

Thus

$\ds\int\ds\frac{d\theta}{\theta-T}=k\ds\int\,dt \Rightarrow
\ln(\theta-T)=kt+c$

Must find $k,\ c$ and then $T$. Apply boundary conditions.

At

$$\begin{array} {lcl} t=0 \Rightarrow \ln(15-T)=c\ \ \ (1)\\ t=30
secs \Rightarrow \ln(20-T)=30k+c\ \ \ (2)\\ t=60 secs \Rightarrow
\ln(21-T)=60k+c\ \ \ (3) \end{array}$$

As before, consider $2 \times (2)-(3)$

$$2\ln(20-T)=60k+2c$$

$$\un{\ln(21-T)=60k+c}$$

$$2\ln(20-T)-\ln(21-T)=c$$

Thus with (1) we have

$$2\ln(20-T)-\ln(21-T)=\ln(15-T)$$

$$\rm{or}\ \ln[(20-T)^2]-\ln(21-T)-\ln(15-T)=0$$

or $\ln\left[\ds\frac{(20-T)^2}{(21-T)(15-T)}\right]=0$

$\Rightarrow \ds\frac{(20-T)^2}{(21-T)(15-T)}=1$

$\Rightarrow (20-T)^2=(21-T)(15-T)$

$\Rightarrow 400+T^2-40T=315-36T$

$\Rightarrow 85=4T$

$$\un{\Rightarrow T=21.25^{\circ}C}$$


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