Differential equations (1 Viewer)

[Antwan23q]

Hi, we just started this topic today, can anyone help with this question

A body, initially at room temperature 20ºC, is heated so that its temperature would rise by 5ºC/min if no cooling took place. Cooling does occur in accordance with Newton's law of Cooling and the maximum temperature the body could attain is 120ºC. How long would it take to reach a temperature of 100ºC?

im just guessing, but so far, ive got this

heating: dT/dt=5, .: T=5t+20
Cooling: T=20+Ae^-kt
so total temp would be
T=5t+20-(20+Ae^-kt)
T=5t-Ae^-kt

t=0, T=20 .: A=-20
so
T=5t+20e^-kt

and i dont know from there

 

[Pace_T]

T = A + Be^-kt

t--> infinite T--> 120 ; A = 120
t= 0 T=0 ; B = -100

T = 120 - 100e^-kt

dT/dt = 100e^-kt

t= 0, dT/dt = 5

5= 100k
k = 0.05

T = 120-100e^-0.05t
T= 100, t= ?
100 = 120-100e^-0.05t

e^-0.05t = 0.2
t = ln(0.2)/-0.05
t= 32.2 minutes

 

[Antwan23q]

thanks alot man. I didnt know bout that infinity part, because the teach didnt finish explaining it

 

[acmilan]

The inifinity part means that t can get as large as possible, and since e has a negative power, anything to a negative large number goes towards zero, so after a very long time the Be^-kt part of T = A + Be^-kt will go to zero, making T = A, or atleast as close as possible to it without actually reaching it.

If you graphed that T against t, there would be a horizontal assymptote at T = A, where the graph will get very close to it as time goes on but never reach it.

 

[Riddles]

exactly what he said