Differential equations: Newton's Law of Cooling

Need help on this one. DE Newton's Law of Cooling

At 9 A.M., a thermometer reading 70°F is taken outdoors where the temperature is
l 5°F. At 9: 05 A.M., the thermometer reading is 45°F. At 9: 10 A.M., the thermometer is
taken back indoors where the temperature is fixed at 70°F. Find (a) the reading at
9: 20 A.M. and (b) when the reading, to the nearest degree, will show the correct
(70°F) indoor temperature.

Jhun Vert's picture

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

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

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

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

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

$T - T_s = Ce^{kt}$

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

$C = 55$

$T - 15 = 55e^{kt}$

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 Ts 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.

Infinitesimal's picture

If I am not mistaken sir, "to the nearest degree" means that the temperature reading will be $69.5^{\circ} F$ ?

Jhun Vert's picture

I thought it was rounding off, hehehe.

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