1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You now have:

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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:
    or more obscurely:
    where [itex]A=T_{i}-T_{0}[/itex]
  8. Oct 28, 2006 #7

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook