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Newtons cooling law problem given in precalc

  1. Oct 21, 2009 #1
    I'm going to write this up the same way my teacher gave me the sheet

    1. The problem statement, all variables and given/known data
    At 9am on october 19, 2009 a body was found in room 327 at University Center. The room is kept at a constant temperature of 72 degrees. The medical examiner was called and he arrived in eight minutes. The first thing he did was to take the temperature of the body. It was 83 degrees. Thirty minutes later the temperature of the body was taken again and it was now 78 degrees. Help the police by telling them when the person was murdered.

    Variables

    t= time
    y= temp. in degrees Fahrenheit of an object in a room
    T= temperature in degrees Fahrenheit of the room
    T= 72
    yinitial(0)= 98.6 (assumed)
    yfirst(t+8?)= 83
    ysecond(t+38?)= 78
    I put the question marks cause I'm not sure whether that is the right time.

    2. Relevant equations
    This is the equation as it was given to me

    ln|y-T|= kt + C

    but everywhere else I look has a different formula than this one the closest i could find/come up with (i cant remember at this point) is

    y-T=e^(kt + C)

    3. The attempt at a solution

    I have tried the equation (and just about every other one I could find or think of) so many times that I cannot figure them out on the paper anymore and I am basically lost as to where to start because every way I've tried by plugging the numbers back in to try and get the temp at the time comes out wrong. Can someone please help?? Thanks to whomever can
     
  2. jcsd
  3. Oct 21, 2009 #2

    Mark44

    Staff: Mentor

    The two equations above are equivalent. IOW, they say exactly the same thing.
    Show us what you've tried. Assume that the body was found at time t = X, where X represents the number of minutes after the person was murdered. You know the temperature of the body at three different times: t = 0 (98.6 deg. F), t = M + 8 (83 deg. F), and t = M + 30 (78 deg. F). Use these times and temperatures in your equation to find k, C, and M.
     
  4. Oct 22, 2009 #3
    This still makes no sense this is what I tried and got stuck at:

    ln|98.6-72|=k*0+C

    so k and 0 cancel and i get

    3.280911216=C

    so then since I dont know M (which i think you originally used x) but from going from 908 to 938 and there being a 5 degree drop I assumed thats constant and figure 30 minutes after he died the temp would be at 93.6 so i did:

    ln|93.6-72|=k*30+3.280911216

    and get

    3.072693315=k*30+3.280911216

    so I tried both dividing first and subtracting first and when i checked both dividing first gave me wild numbers and subtracting actually gave me the right answer so:

    3.072693315=k*30+3.280911216
    -3.280911216 -3.280911216
    gives me
    -.208217901/(30)=k*30/(30)
    gives me
    k=-.0069405967
    so then to check it i did
    y-72=e^(-.0069405967*30+3.280911216)
    and got something around 93.59999333 or so rounding up to 93.6 so I thought it had worked so then i tried this to check it

    ln|83-72|=-.0069405967*t+3.280911216
    so then
    2.397895273=-.0069405967*t+3.280911216
    -3.280911216 -3.280911216
    getting this
    -.883015943/(-.0069405967)=-.0069405967/(-.0069405967)*t
    equaling
    127.2247879=t
    so to check it i divide 127 ish by 30 to see how many halfs there are (4.240826263) then multiply that by 5 (21.20413132) then I add 83 to see if it gets back to 98.6 since that was the original, and end up with 104.2041313.... sooo im completely lost again
     
  5. Oct 22, 2009 #4

    Mark44

    Staff: Mentor

    OK, this is what I get.
    No, that's not right. You're assuming that the temperature is going to drop at a constant rate. It's not. The temperature drop is rapidest at the time when the body's temperature is largest in comparison with the ambient room temperature. In the half-hour between 9:08 and 9:38 the body's temperature dropped 5 degrees. In the half-hour right after death, the temperature drop would have been greater than this.

    Because of this assumption, all your work from here on is incorrect.
    You got C. Let's see if you can work through the rest of the problem. I'll set it up slightly different than I set it up before.

    This time, let M = number of minutes between death and time 9:08
    Then M + 30 = number of minutes between death and time 9:38

    At 9:08, the body temp was 83 degrees, so
    ln(83 - 72) = kM + C

    At 9:38, the body temp was 78 degrees, so
    ln(78 - 72) = k(M + 30) + C

    You have C, and you now have two equations in two unknowns. Can you solve for k and M?

    I get that he was killed sometime between 6am and 7am.
     
  6. Oct 22, 2009 #5
    ok well see through me trying to do all of this this was the problem, i dont know how to solve for both variables, i feel like itll end up being this after subtracting c from ln(83-72)
    -.883015943= k
    -----------
    m
    or

    -.883015943= m
    -----------
    k

    (m and k are being divided out)

    and if i tried the second temp itd be something similar and i have no idea how to go any further than that
     
    Last edited: Oct 22, 2009
  7. Oct 22, 2009 #6

    Mark44

    Staff: Mentor

    Once you have solved for k in one equation (with M on the other side), substitute for k in the other equation.
     
  8. Oct 22, 2009 #7
    do I just use k=(-.883015943)/m ?

    i tried doing that and this is what happened
    ln(78-72)=((-.883015943)/m)*(m+30)+C

    so it becomes

    1.791759469=-.883015943+30+c ?

    or would it be

    1.791759469=(-.883015943)*(m+30)+C

    or would the 30 multiply to the -.883015943?

    like i can never remember when variables cancel does it just become 1 or (in this case)
    1m?
    doing this it turns out as either

    35.66090963=m

    by subtracting c+30 first then dividing by -.8830.... or

    -35.31004725=m
    by dividing -.8830... first then subtracting C+30.

    but neither of these seem right at all

    after 7 hours and 30 or so minutes of working on this problem (total) this is just getting rediculous
     
    Last edited: Oct 22, 2009
  9. Oct 22, 2009 #8

    Mark44

    Staff: Mentor

    Yes.
    Yes, this is fine.
    No. What happened to m?
    No, it's not that one, either. This time you lost one of the m's.
    If you can't remember when variables cancel, then you should NEVER cancel them, at least not until you understand what's really happening with this operation.

    Cancelling means removing factors that are equal to 1, as in this example.
    [tex]\frac{Ma + Mb}{M}~=~\frac{M(a + b)}{M}~=~\frac{M}{M}\frac{a + b}{1}~=~ a + b[/tex]
    But cancellation cannot be done in the following example.
    [tex]\frac{Ma + b}{M}~\neq~a + b[/tex]
    M is not a factor of the numerator.
     
  10. Oct 22, 2009 #9
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