Newton's Law of Cooling Application A thermometer that has been stored indoor where the temperature is 22 degrees Celsius, is taken outdoor. After 5 minutes it reads 18 degrees. After 15 minutes it reads 15 degrees. What is the outdoor temperature?
I have solved the differential equation dT/dt = k(T-TA) where T is the temperature at the time t and TA is the ambient temperature. I got T-TA = ce^kt.
My thought is that we're looking for the ambient temperature TA in this question, but without the TA I have 3 unknowns (k, TA and the constant c) and I know I'm supposed to solve for the 3 unknowns using the conditions provided in the question. However, I am having difficulties solving them. Am I on the right track? 
 A: Newton's Cooling law:
$\frac{dT}{dt}=k(T_e-T)$; where $T_e$ is the outdoor temperature
If we know $T(0)=T_0$; then the general solution is
$T(t)=T_e+(T_0-T_e)e^{-kT}$
We have to find $T_e$ knowing that
$T_0=22^{\circ}C$ and  
$T(5)=T_e+(22-T_e)e^{-5k}=18$
$T(15)=T_e+(22-T_e)e^{-15k}=15$  
With these two last equations we can find $k$ and $T_e$  
$(22-T_e)e^{-5k}=18-T_e$
$(22-T_e)e^{-15k}=15-T_e$ 
$e^{-5k}=(18-T_e)/(22-T_e)$
$e^{-15k}=(15-T_e)/(22-T_e)$ 
$-5k=ln \frac{18-T_e}{22-T_e}$
$-15k=ln\frac{15-T_e}{22-T_e}$ 
$3ln \frac{18-T_e}{22-T_e}=ln\frac{15-T_e}{22-T_e}$  
$ln (\frac{18-T_e}{22-T_e})^3=ln\frac{15-T_e}{22-T_e}$  
$ (\frac{18-T_e}{22-T_e})^3=\frac{15-T_e}{22-T_e}$
$ (18-T_e)^3=(15-T_e)(22-T_e)^2$
Working on this last equation we end up with the following quadratic equation  
$5T_e^2-172T_e+1428=0$
We find two solutions to this equation:
$T_e=20.4^{\circ}C$ and $T_e=14^{\circ}C$
And we take $T_e=14^{\circ}C$ as the solution because
$T=15^{\circ}C$ must be over $T_e$
With this value we can also find $k=\frac{ln(2)}{5}$
.....................................................
Now I will use a somewhat different approach to solve the problem  
$T=T_e+Ce^{-kt}$ is the general solution where the initial temperature is not stated  
Then we can write
$T(0)=T_e+C=22$
$T(5)=T_e+Ce^{-5t}=18$
$T(15)=T_e+Ce^{-15t}=15$
Now I will make this substitution: $\gamma=e^{-5k}$;$\hspace{10 mm}$Then $\gamma^{3}=e^{-15k}$, so:
$T_e+C=22$
$T_e+C\gamma=18$
$T_e+C\gamma^{3}=15$
Subtracting between the third and second equation and subtracting between the second and first equation we can write:
$C\gamma(\gamma^{2}-1)=-3$
$C(\gamma-1)=-4$
Now dividing between these last two equations
$\gamma(\gamma+1)=\frac{3}{4}\hspace{10 mm}$and$\hspace{10 mm}\gamma=\frac{1}{2};C=8$
So: $T_e+8=22\hspace{10 mm}$ and $\hspace{10 mm}T_e=14^{\circ}C$
