Differential equations: Newton's Law of Cooling

Newton's law of Cooling:
$\dfrac{dT}{dt} = k(T - T_s)$

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

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

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

$\ln (T - T_s) = \ln Ce^{kt}$
 

Hence,
$T - T_s = Ce^{kt}$
 

At 9:00 am (t = 0 ), the thermometer was taken outdoors where T_s =15°F and the thermometer reading was T = 70°F.
$70 - 15 = Ce^{k(0)}$

$C = 55$
 

Therefore,
$T - 15 = 55e^{kt}$
 

At 9:05 am (t = 5), the thermometer reads 45°F:
$45 - 15 = 55e^{k(5)}$

$k = -0.121\,227\,160\,7$
 

At 9:10 am (t = 10), the thermometer is taken back indoors.
$T - 15 = 55e^{-0.121\,227\,160\,7(10)}$

$T = 31.364^\circ F$   ←   thermometer reading just before it was taken back indoors
 

Inside the room, the surrounding temperature T_s is 70°F, and we will reset the t to zero, hence,
$T - T_s = Ce^{kt}$

$31.364 - 70 = Ce^{-0.121\,227\,160\,7(0)}$

$C = -38.636^\circ F$
 

Therefore, the equation when the thermometer is inside the room is
$T - 70 = -38.636e^{-0.121\,227\,160\,7t}$
 

At 9:20 am, 10 minutes after the thermometer was taken indoors
$T - 70 = -38.636e^{-0.121\,227\,160\,7(10)}$

$T = 58.505^\circ F$
 

For T = 70°F
$70 - 70 = -38.636e^{-0.121\,227\,160\,7t}$
 

Since logarithm of zero do not exist, we can approximate the solution. Use T → 70^-, say T = 69.99999 then solve for t. It will actually do the job.
 

I leave it to you then.