Use Law of Cooling:
[tex]T(t) = (T_0 - T_A)e^{-kt} +T_A[/tex]
T0 = initial temperature, TA = ambient or final temperature
First solve for k using given info, T(3) = 42
[tex]42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})[/tex]
Substituting k back into cooling equation gives:
[tex]T(t) = 60(\frac{11}{30})^{t/3} + 20[/tex]
At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
[tex]T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80[/tex]
Solve for x:
[tex]x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80 [/tex]
Sub back into original cooling equation, x = T(t)
[tex] -9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20[/tex]
Solve for t:
[tex]60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959[/tex]
This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
