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Dead bodies - exponential decay

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data
    A body is found at 2:00pm at a temperature of 26°C, with a surrounding temperature of 18°C.
    Two hours later the temperature of the body is 21°C, when did the body die?
    T=Ae^(kt)+Ts
    where T is the temperature of the body
    A is the initial temperature
    k is a constant
    t is time in minutes
    Ts is the surrounding temperature
    In order to do this assumed at t=0 the body would be 37°C because that is roughly a persons body temperature before death. When I substituted that in though I couldn't get k. How do I get k? Do I need any assumptions? Am I allowed to find k by doing the following
    21=26e^(120k)+18 and re-arrange from there?
     
  2. jcsd
  3. Mar 26, 2013 #2
    You did identify A to be the initial temperature. So is 260C really the initial temperature?:wink:
     
  4. Mar 26, 2013 #3
    I guess not, so how do I go about it then? Substitute in 37°C for the initial,
    T=37e^(kt)+18, and if we let T=26 then
    26=37e^(kt)+18? Then I can't find k or t, so It'd have to be
    21=37e^(120k)+18
    re-arrange to find k, but then that defeats the purpose of the other temperature given.
     
  5. Mar 26, 2013 #4

    Mentallic

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    You have two equations:

    [tex]26 = 37e^{kt_1}+18[/tex]

    and

    [tex]21 = 37e^{kt_2}+18[/tex]

    What is the relationship between t1 and t2? In other words, [itex]t_2=t_1+?[/itex]
    Once you find this, you'll have two equations in two unknowns.
     
  6. Mar 26, 2013 #5
    Does it have anything to do with the 2 hours that had passed? Like t2=t1+120? or do we make the subject of both equations t and find it from there?
     
  7. Mar 26, 2013 #6

    epenguin

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    Yes you can combine those two last equations - do something obvious. I hope it's clear to you that representing as you have, if time is going forward k is a negative number. (Don't know where you got equation from but it is more usual to write as e-kt with k a positive number.) You will have to recall or revise how to combine terms with indices, and natural logarithms.
     
    Last edited: Mar 26, 2013
  8. Mar 26, 2013 #7
    You're on the right track:smile:
     
  9. Mar 26, 2013 #8

    Mentallic

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    You're right, t2=t1+120 where t is measured in minutes, but I'd stick to hours so you have t2=t1+2.
    Now, if you replace t2 by that expression then you'll have two equations and two unknowns, mainly k and t1. You're looking to find t1 since that represents the time (in hours or minutes, depending on the units you used) before 2:00pm.

    And yes, begin by making the subject of both equations t1 so that you can find k by equating each equation.

    In other words,

    [tex]t_1 = f_1(k)[/tex]

    [tex]t_1 = f_2(k)[/tex]

    Then

    [tex]f_1(k)=f_2(k)[/tex]

    And you can solve k here. Once that's done, just plug k into either f1 or f2 (depending on which you think might be easier to simplify) to find t1
     
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