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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.

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I will use $T(t)=T_m+(T_0-T_m)e^{kt}$ for both case.
Outdoor:
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

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  • $\begingroup$ did you use same $k$ values for last calculations$?$ $\endgroup$ – NajmunNahar Dec 10 '19 at 19:05
  • $\begingroup$ Yes. I mention it now @NajmunNahar $\endgroup$ – emonHR Dec 10 '19 at 19:07

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