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I Find This Initial Temperature

  1. Sep 29, 2016 #1
    Suppose that you take a thermometer outside where it is 100°.

    T(5min)=80° T(15min)=90°

    What is the initial temp of the thermometer?

    Given equation

    dT/d t= k(T-Te)

    Derived Equation
    ⌠(T-Te)^-1 (dT/dt)dt =⌠ kdt
    ln(T-Te)=kt + c
    T=ce^kt + Te

    so i basically got the answer by knowing c must be negative and that when i use a square root on the magnitude of c i through out the positive value.
    Wolfram Alpha
    my real question is how can this become -3 ln(-20/x) +ln(-10/x) becomes -2ln(-(1/x)) -ln(800)?

    Attempt to find C

    1) 80=ce^k5 +100 ---- -20=ce^k5 ----- -20/c=e^k5 ------ ln(-20/c)=k5 ----- -3ln(-20/c)=-k15

    2) 90=ce^k15 +100 ---- -10=ce^k15 ----- -10/c=e^k15 ------ ln(-10/c)=k15


    ln(-10/c) - 3ln(-20/c) ----- ln((-10/c) x (c^3 / -8000)) -------- ln(c^2/800) ---- 2ln(c) -ln(800)=0


    c= +/- sqrt(800)
  2. jcsd
  3. Sep 30, 2016 #2

    Simon Bridge

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    Are you saying that the expression:
    ##-3\ln\frac{-20}{x} + \ln \frac{-10}{x} = -2\ln\frac{-1}{x} - \ln 800## ... is not true?
    ... or you don't know how to get there from
    ... which I am reading as ##T(t)=Ce^{kt} + T_e##?

    The rest of your post appears to show you doing the calculation (though you need to say what finding "C" does for you.)
    Last edited: Sep 30, 2016
  4. Sep 30, 2016 #3
  5. Sep 30, 2016 #4

    Simon Bridge

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    Do you know the relationship between the initial temperature and C?
    ... that last bit does not make sense.
    Off post #1, the way to understand how to get "there" must start from knowing what "x" stands for.
    However, if you can find the initial temperature without going "there", then why bother?
  6. Sep 30, 2016 #5
    Your solution to the differential equation should read:$$(T_e-T)=(T_e-T_0)e^{-kt}$$
    So, $$20=(T_e-T_0)e^{-5k}\tag{1}$$
    $$10=(T_e-T_0)e^{-15k}\tag{2}$$If you cube Eqn. 1, you get:$$8000=(T_e-T_0)^3e^{-15k}\tag{3}$$
    What do you get if you divide Eqn. 3 by Eqn. 2?
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