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Homework Help: Newton's Law of Cooling Help

  1. Oct 28, 2006 #1
    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?
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 28, 2006 #2
    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]
     
  4. Oct 28, 2006 #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.
     
  5. Oct 28, 2006 #4
    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?
     
  6. Oct 28, 2006 #5
    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?
     
  7. Oct 28, 2006 #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]
     
  8. Oct 28, 2006 #7
    [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!!
     
  9. Oct 29, 2006 #8
    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!!
     
  10. Oct 29, 2006 #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.
     
  11. Feb 19, 2010 #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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