Newton's Law of Cooling Help

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  • #1
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I have a question about Newton's Law of cooling. Basically I understand that the equation,
http://album6.snapandshare.com/3936/45466/853596.jpg [Broken]
Comes from the DE, dT/dt = K(T-To)

Using this, I am to solve this problem:

A thermometer is taken from an inside room to the outside, where the air temperature is 5 °F. After 1 minute, the thermometer reads 55 °F, and after 5 minutes the reading is 30 °F. What is the initial temperature?

So to start, I solved for e^k...

http://album6.snapandshare.com/3936/45466/853597.jpg [Broken]

So now that I have e^k, what do I do? My guess is that A is the initial Temperature? But I am not sure and my text does not really explain it too well. So, basically, I guess I am asking, what is the constant A in the general formula mean? And if it is not the initial temperature, or initial condition, then what can I do next with this problem?
 
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  • #2
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Rewrite it as [tex] y(t) = y_{0}e^{kt} [/tex] where [tex] y = T - 5 [/tex].

So [tex] y(t) = y_{0}(\frac{1}{2})^{\frac{1}{4}t} [/tex]

[tex] y(1) = y_{0}(\frac{1}{2})^{\frac{1}{4}} [/tex]

Solve for [tex] y_{0} [/tex] and then get [tex] T_{0} [/tex]
 
  • #3
arildno
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You now have:
[tex]A=50*(\frac{1}{2})^{-\frac{1}{4}}[/tex]

The initial temperature is now found by computing T(0)

As for what A is, it is the DIFFERENCE between the initial temperature and the ambient temperature.
 
  • #4
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courtrigrad said:
Rewrite it as [tex] y(t) = y_{0}e^{kt} [/tex] where [tex] y = T - 5 [/tex].

So [tex] y(t) = y_{0}(\frac{1}{2})^{\frac{1}{4}t} [/tex]

[tex] y(1) = y_{0}(\frac{1}{2})^{\frac{1}{4}} [/tex]

Solve for [tex] y_{0} [/tex] and then get [tex] T_{0} [/tex]

So what you are saying here is that [tex] y_{0} [/tex] in your equation is [tex] T_{0} [/tex], which is the initial temperature?
 
  • #5
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arildno said:
You now have:
[tex]A=50*(\frac{1}{2})^{-\frac{1}{4}}[/tex]

The initial temperature is now found by computing T(0)

As for what A is, it is the DIFFERENCE between the initial temperature and the ambient temperature.

I don't think I am understanding this at all here... Sorry to put you through this, but, if A is the difference between the initial temperature and the ambient temperature, what is the variable for the initial temperature if [tex]T_{0}[/tex] is not the initial temperature, but the ambient temperature that arises as time gets very large or goes to infinity?
 
  • #6
arildno
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I don't get your question!
Let's start with the diff.eq, with an assigned initial temperature [itex]T_{i}=T(0)[/tex], and an ambient temperature [itex]T_{0}[/tex]
We have the diff.eq:
[tex]\frac{dT}{dt}=k(T-T_{0}), T(0)=T_{i}[/itex]
Introduce the new variable:
[tex]y(t)=T(t)-T_{0}\to\frac{dy}{dt}=\frac{dT}{dt}, y(0)=T_{i}-T_{0}[/tex]
Thus, we have the diff.eq problem:
[tex]y(t)=ky, y(0)=T_{i}-T_{0}\to{y}(t)=(T_{i}-T_{0})e^{kt}[/tex]

Thus, solving for T(t), we get:
[tex]T(t)=T_{0}+(T_{i}-T_{0})e^{kt}[/tex]
or more obscurely:
[tex]T(t)=T_{0}+Ae^{kt}[/tex]
where [itex]A=T_{i}-T_{0}[/itex]
 
  • #7
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[tex]T(t)=T_{0}+(T_{i}-T_{0})e^{kt}[/tex]

Wow... This really made it clear here. Sorry for the obscure questions, but you really nailed it for me here. I am going to try a few problems in my text and see how they work out. Thanks again!!
 
  • #8
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Ok, so I worked it out and I got ~ 64.5°. If anyone has the time, would you mind checking this for me as I don't have the answer to this in my text. Thanks!!
 
  • #9
arildno
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I haven't worked it out, but:
Start having confidence in yourself!
I'm sure you managed it all right.
 
  • #10
I don't understand how we find k in problems like this where no initial temperature is given.
Do you have to compare the temps at t=1 and t=5?
 

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