# Newton's Law of cooling problem when the thermometer brought indoors

At $$2$$PM a thermometer reading $$80^{\circ}$$F is taken outside. Where the air temperature is $$20^{\circ}$$F. At $$2:03$$PM the temperature reading yielded by the thermometer is $$42^{\circ}$$F. Later, the thermometer brought inside. Where the air temperature is at $$80^{\circ}$$F. At $$2:10$$PM the reading is $$71^{\circ}$$F. When was the thermometer brought indoors$$?$$

$$1$$st part, Using cooling law, $$T=(T_0-T_m)e^{-kt}+T_m \quad T_0=\text{initial temp.}\quad T_m=\text{medium/Air temp.}$$ \begin{align} 42&=(80-20)e^{-k\times3}+20\\ k&=-\frac{1}{3}\ln\left(\frac{11}{30}\right)\\ T(t)&=60e^{\frac{1}{3}\ln\left(\frac{11}{30}\right)t}+20 \end{align} $$2$$nd part, Now let after $$x$$ min the thermometer brought indoors then $$T(x)=60e^{\frac{1}{3}\ln\left(\frac{11}{30}\right)x}+20$$ which will be $$T_0$$ for second part where, $$T(t)=T_m-(T_m-T_0)e^{-kt}\quad\text{Using warming Law}$$ \begin{align} T(?)=80-\left(100-60e^{\frac{1}{3}\ln\left(\frac{11}{30}\right)x}\right)e^{-k\times?} \end{align} Now I am lost how to processed. Any help will be appreciated.

I will use $$T(t)=T_m+(T_0-T_m)e^{kt}$$ for both case.
You done correctly,$$T(t)=60e^{kt}+20\quad\text{ Where }k=\frac{1}{3}\ln\left(\frac{11}{33}\right)$$ Indoor:
Here you are also correct. I will just change your notation taking $$\color{green}{T_0}=\underbrace{60e^{kt}+20}_\color{red}{{T(t)}}$$. Then $$T(t)=80+(\color{green}{T_0}-80)e^{kt}$$. Suppose after $$t_1$$ min. the thermometer brought indoor then $$T(10-t_1)=71^{\circ}F$$ where $$k=\frac{1}{3}\ln\left(\frac{11}{33}\right)$$ \begin{align} T(10-t_1)&=71\\ 80+(60e^{kt_1}+20-80)e^{k(10-t_1)}&=71\\ \vdots\\ t\approx 4.95 \end{align} Hence the thermometer brought indoors at $$2:05$$PM
• did you use same $k$ values for last calculations$?$ – NajmunNahar Dec 10 '19 at 19:05