01 - Thermometer reading after 6 minutes of being outside

According to Newton’s Law of cooling, the time rate of change of temperature is proportional to the temperature difference.
$\dfrac{dT}{dt} = -k(T - T_s)$

$\dfrac{dT}{dt} = -k(T - 10)$

$\dfrac{dT}{T - 10} = -k \, dt$

$\ln (T - 10) = -kt + \ln C$

$\ln (T - 10) = \ln e^{-kt} + \ln C$

$\ln (T - 10) = \ln Ce^{-kt}$

$T - 10 = Ce^{-kt}$

$T = 10 + Ce^{-kt}$
 

When t = 0, T = 70°F
$70 = 10 + C$

$C = 60$
 

Hence,
$T = 10 + 60e^{-kt}$
 

When t = 3 min, T = 25°F
$25 = 10 + 60e^{-3k}$

$15 = 60e^{-3k}$

$\frac{15}{60} = e^{-3k}$

$e^{-k} = (\frac{1}{4})^{1/3}$
 

Thus,
$T = 10 + 60(\frac{1}{4})^{t/3}$
 

After 6 minutes, t = 6
$T = 10 + 60(\frac{1}{4})^2$

$T = 13.75^\circ F$           answer