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Newton's Law of Cooling differential equation

  1. Sep 5, 2008 #1
    From Newton's Law of Cooling, we can use the differential equation

    dT/dt= -k(T-Ts)

    where Ts is the surrounding temperature, k is a positive constant, and T is the temperature.

    Let [tex]\tau[/tex] be the time at which the initial temperature difference T0-Ts has been reduced by half. Find the relation between k and [tex]\tau[/tex].


    I have found the temperature equation for the object to be

    T(t) = Ts + (T0-Ts)e-kt

    If the initial temperature difference is reduced by half, i tried

    T(t) = Ts + 0.5(T0-Ts)e-kt

    but couldn't solve it any further. Could someone please shed some light on what i should do next?
  2. jcsd
  3. Sep 6, 2008 #2
    This seems to be right, but here:

    you made some mistake with the parameter t. If we set notation

    \Delta T(t) = T(t) - T_s,

    then the equation you want to solve is

    \Delta T(\tau) = \frac{1}{2}\Delta T(0).
  4. Sep 7, 2008 #3
    Oh thanks for replying, i was able to get the correct answer.

    k [tex]\tau[/tex] = ln2

    However, I'm still a bit unclear of why T([tex]\tau[/tex])-Ts and T0-Ts canceled out during calculation. Or do T([tex]\tau[/tex]) and T0 equal? Also why isn't the 0.5 placed like

    T([tex]\tau[/tex]) = Ts + 0.5(T0-Ts)e-k[tex]\tau[/tex], but instead

    0.5[T([tex]\tau[/tex])-Ts] = (T0-Ts)e-k[tex]\tau[/tex]

    Thanks again!
    Last edited: Sep 7, 2008
  5. Sep 7, 2008 #4
    The use of latex like this doesn't seem to be working very well. At least my browser makes that look like [tex]k^{\tau}=\textrm{ln}(2)[/tex], although you are most certainly meaning [tex]k\tau = \textrm{ln}(2)[/tex].

    It looks like you are getting the correct result, because you know what it is, and you are arriving at it "by force". :wink: Just try to chop the problem into smaller pieces, so that you can control it. We want this:

    \Delta T(\tau) = \frac{1}{2} \Delta T(0)\quad\quad\quad\quad (1)

    where the temperature difference at each instant is given by

    \Delta T(t) = T(t) - T_s = (T_0-T_s) e^{-k t}.

    So onto the left side of the equation (1) we get

    (T_0 - T_s)e^{-k\tau}

    and on the right side

    \frac{1}{2} (T_0 - T_s)e^{-k\cdot 0}.

    See how things go then?
  6. Sep 7, 2008 #5
    Oh yeah i see it now, thanks a lot jostpuur. As for the latex inputs, I think I will go look for some tutorials on this forum. :rofl:
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