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Help with a Newtons Cooling Law Problem

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A thermometer reading 70o F is placed in an oven preheated to a constant temperature. Through a glass window in the over door, an observer records that the thermometer reads 110o F after 1/2 minute and 145o F after 1 minute. How hot is the oven.


    2. Relevant equations
    T(t)=(To-TR)ekt+TR


    3. The attempt at a solution
    Ok so I let said
    TR= Oven Temperature
    To=70
    To(.5)=110
    To(1)=145

    I then set up the general equation as
    T(t) = (70-TR)ekt+TR

    Then using the two times and temperatures I was given I created these two equations.
    145=(70-TR)ek+TR
    110=(70-TR)e.5k+TR

    I then moved over the TR on the end over to the left side giving me these two equations
    145-TR=(70-TR)ek
    110-TR=(70-TR)e.5k

    Dividing the top equation by the bottom i get
    (145-TR)/(110-TR)=e.5k

    Taking the natural log of both sides and then dividing by .5 gets me
    k= 2*ln((145-TR)/(110-TR))

    To me this seems like I may have overcomplicated something or made a mistake as plugging that value of k back into the equation seems like it would make solving for TR incredibly difficult, not even sure how I would be able to solve for TR using that value for k.

    So what I'm asking is if the steps I took were the right ones to do and if I should go ahead and try to solve for TR using that value for k or if there is a simpler way. Thanks in advance.
     
    Last edited: Sep 18, 2009
  2. jcsd
  3. Sep 18, 2009 #2
    Welcome to PF, ZeromusX99.

    You did all the hard work already!


    From this last equation, you can substitute your expression for e.5k into 110-TR=(70-TR)e.5k and solve for TR.
     
  4. Sep 18, 2009 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Whoa! This would say the final temperature is either plus or minus infinity.

    You have a sign error.


    Using your result, k= 2*ln((145-TR)/(110-TR)), what is e.5k? (It isn't all that complex.)

    Note well: This is a valid approach but it will of course yield the wrong answer because of your sign error.


    Aside: There is a much easier approach. What happened in the second half minute?

    EDIT
    Your approach is fine. It will just happen to yield a negative value for k.
    The comment about an easier approach is still valid.
     
  5. Sep 19, 2009 #4
    Thanks for the suggestion Billy Bob substituting for e.5k worked, guess I was just over thinking things.

    Also, thanks to your suggestion D H of looking at the second half minute, where temperature increased by 35, I found the simpler way of doing the calculation TR, I'll go ahead and post it here

    Instead of dividing I just subtract the equations I initially listed to get
    35 = (70-TR)(e-k-e-.5k

    Then do another subtraction of equations using my T(.5) and T(0)
    110 = (70-TR)(e-.5k+TR
    70 = (70-TR)+TR

    I get
    40 = (70-TR)(e-.5k-1)

    dividing my
    35 = (70-TR)(e-k-e-.5k)
    by
    40 = (70-TR)(e-.5k-1)
    then gets me
    7/8 = (e-k-e-.5k)/(e-.5k-1)

    pulling an e-.5k from the numerator then gives me
    7/8 = e-.5k(e-.5k-1)/(e-.5k-1)
    (e-.5k-1) cancels leaving me with
    7/8 = e-.5k

    natural of both sides gives me ln(7/8) = -.5k so
    k= -2ln(7/8)

    Plugging this into my equation
    40 = (70-TR)(e-.5k-1)
    gives me
    40 = (70-TR)(eln(7/8)-1)
    which is just
    40 = (70-TR)((7/8)-1)
    leaving me with
    40 = (70-TR)(-1/8)
    multipling both sides by -1/8 gives me
    -320 = 70 - TR[/SUB
    Add TR and 320 to both sides then gives me
    TR[/SUB = 390

    Hehe next time I'll remember to make it -kt as well
     
  6. Sep 19, 2009 #5

    D H

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    Staff Emeritus
    Science Advisor

    Correct.

    You didn't quite get the gist of my hint.
    Write Newton's law of cooling as

    [tex]T(t)-T_R = (T_0-T_R)e^{-kt}[/tex]

    During the first half minute the temperature rose from 70o F to 110o F:

    [tex]110-T_R = (70-T_R)e^{-kt/2}[/tex]

    from which

    [tex]e^{-kt/2} = \frac{110-T_R}{70-T_R}[/tex]

    During the next half minute the temperature rose from 110o F to 145oo F. Thus

    [tex]145-T_R = (110-T_R)e^{-kt/2}[/tex]

    Substituting the expression for [itex]\exp(-kt/2)[/itex] in the above,

    [tex](145-T_R)(70-T_R) = (110-T_R)^2[/tex]

    Simplifying,

    [tex]\aligned
    10150-215T_R+T_R^2 &= 12100-220T_R+T_R^2 \\
    (220-215)T_R &= (12100-10150) \\
    T_R &=390
    \endaligned[/tex]
     
  7. Sep 19, 2009 #6
    Ah I understand now thanks for all the help D H
     
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