# Newton's Law of Cooling

• November 12th 2009, 07:18 AM
Belowzero78
Newton's Law of Cooling
Hi everyone, thanks for taking the time to help me out!

Question: An object of temperature 120 degree F is placed in a medium maintained at a constant temperature m degree F. At the end of 2 minutes the temperature of the object is 90 degree F and after another 2 minutes its temperature is 72 degree F.

(i) Find m
(ii) When will the temperature of the object be 61.20 degree F?
(iii) What is the temperature of the object at the end of 8 minutes?

(i) T(t)=Ta+(To-Ta)e^-kt
I know that Ta=90 F, To can be 120 degree F when T(2)=90 and t=2. I still have two unknowns left and i cant solve for Ta, so i need help here.

T(2)=90=Ta+(120-90)e^-2t

As for the rest if i can get the equation i will be able to solve!
• November 12th 2009, 09:35 AM
galactus
At T(0)=120.

By $\frac{dt}{T-m}=kdt$

Integrate both sides gives:

$ln(T-m)=kt+C$

$T=e^{kt+C}+m$

$T=Ke^{kt}+m$

At T(0) we have T=120

$120=Ke^{kt}+m$......[1]

At T(2) we have 90:

$90=Ke^{2k}+m$....[2]

At T(4) we have 72:

$72=Ke^{kt}+m$....[3]

We have three equations to solve, [1], [2], and [3].

From [1], K=120-m

Sub into [2] and get

$90=(120-m)e^{2k}+m$

Can you finish?.
• November 15th 2009, 01:07 PM
Belowzero78
I'm not exactly sure where to go from here after ive substituted into the equation. I still have m and k left.