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.
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 Ts =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 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.
If I am not mistaken sir, "to
In reply to Newton's law of Cooling: by Jhun Vert
If I am not mistaken sir, "to the nearest degree" means that the temperature reading will be $69.5^{\circ} F$ ?
I thought it was rounding off
In reply to If I am not mistaken sir, "to by Infinitesimal
I thought it was rounding off, hehehe.