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:
dTdt=k(T−Ts)
dTT−Ts=kdt
ln(T−Ts)=kt+lnC
ln(T−Ts)=lnekt+lnC
ln(T−Ts)=lnCekt
Hence,
T−Ts=Cekt
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=Cek(0)
C=55
Therefore,
T−15=55ekt
At 9:05 am (t = 5), the thermometer reads 45°F:
45−15=55ek(5)
k=−0.1212271607
At 9:10 am (t = 10), the thermometer is taken back indoors.
T−15=55e−0.1212271607(10)
T=31.364∘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−Ts=Cekt
31.364−70=Ce−0.1212271607(0)
C=−38.636∘F
Therefore, the equation when the thermometer is inside the room is
T−70=−38.636e−0.1212271607t
At 9:20 am, 10 minutes after the thermometer was taken indoors
T−70=−38.636e−0.1212271607(10)
T=58.505∘F
For T = 70°F
70−70=−38.636e−0.1212271607t
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∘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.