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Newton's Law of Cooling (Calculating Time of Death)

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem states that you discovered the body at 1pm Thursday in a freezer where the temperature was 10F. Temperature of the corpse at discovery was 40F. I have to find how many hours ago the victim died.

    2. Relevant equations

    You are given the formula T = Ta + (98.6 - Ta)(0.97)^t where Ta is air temperature.

    3. The attempt at a solution

    So I plugged the numbers in: 40 = 10 + (98.6-10)(0.97)^t but I have no idea how to solve for t. Any help?
     
  2. jcsd
  3. May 1, 2012 #2
    Move the constants to one side and take logs on both sides you should get:

    [tex] t log(0.97) = log \frac{30}{88.6} [/tex]

    Evaluate that for t.
     
  4. May 1, 2012 #3
    Ok so I get the log of 30/88.6 and divide that by the log of 0.97 to get t, which equals 35.55 hours. Did I do that right?
     
  5. May 1, 2012 #4
    Im just looking for confirmation i solved the problem right.
     
  6. May 1, 2012 #5

    sharks

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    Gold Member

    It appears correct to me. But the answer is actually... 1:45 am on Wednesday?
     
    Last edited: May 1, 2012
  7. May 1, 2012 #6
    Ok thanks. Yes, true ;)
     
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