Since the temperature decreases, we can write the differential equation:
$$\begin{align}\frac{dT}{dt} =k(T-5) \end{align}$$
where k is a positive constant.
Separating variables and integrating, we get:
$$\begin{align} \frac{1}{T-5}dT =kdt\end{align}$$
$$\ln |T-5|=kt+C_1$$
$$T-5=Ce^{kt} $$
$$T=Ce^{kt}+5 $$
Using the initial condition \(T(0)=20\), we find that \(C=15\)
Therefore, the solution to the differential equation (1) is
$$T(t)=15e^{kt}+5$$
Using the other given condition \(T(1)=12\), we find that
$$12=15e^k+5$$
$$e^k=\frac{7}{10} \therefore $$
$$k=\ln\frac{7}{10} $$
Thus the solution to the differential equation (1) becomes:
$$\boxed{T(t)=15 e^{\left ( \ln \frac{7}{10} \right ) t} +5 }$$
Setting \(T=6\), we finally get
$$6=15e^{(\ln \frac{7}{10})t}+5$$
$$1=15e^{(\ln \frac{7}{10})t}$$
$$\frac{1}{15}=e^{(\ln \frac{7}{10})t}$$
$$(\frac{1}{15})^{\frac{1}{\ln \frac{7}{10}}} = t $$
$$t=\frac{\ln{\frac{1}{15}}}{\ln \frac{7}{10}}$$
$$t=\frac{\ln 1-\ln15}{ \ln7-\ln 10} \approx 1.23\text{ minutes}$$
Therefore, it will take approximately 1.23 minutes for the thermometer to read C.
