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At what time did the murder take place? (Differential Equations)

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Newton’s law of cooling states that the rate of change of temperature of an object is proportional
    to the difference between the temperature, T, of the object and that of its surroundings,
    Ts. Derive the solution

    T(t) = Ts + (T0 − Ts) e−kt,
    where T0 is the temperature at t = 0 and k is a constant whose meaning should be identified.
    The corpse was found in an air-conditioned room. The forensic scientist measured the body
    temperature of the victim at 2am and found it to be 25C; by 3am it had fallen to 21C. The
    temperature of a living body is 37C and the temperature of the room was 19C.
    At what time did the murder take place?

    (Hints: measure time from the time of death, td; write down equations for the body temperature
    at 2am & 3 am.)

    2. Relevant equations

    for y' + ay = b

    y=b/a + Ce-ax


    3. The attempt at a solution

    Almost at a complete loss on this one.

    for the first part of this question i have

    T' =k(T - Ts)
    T'/k = T-Ts
    T'/k - T = -Ts

    so using y=b/a + Ce-ax for y'+ay=b

    T= Ts + Cekt

    I'm not sure this is correct since I'm missing a negative sign before the k in the exponential. also what is k? I know its some kind of proportionality constant but what name does it have? I'm also not sure how to show that C = (T0-Ts).

    for the second part i have

    25=19+Ce-kt1
    21=19+Ce-kt2


    Ce-kt1=6
    Ce-kt2=2

    dividing one by the other gives 3=e-k(t2-t1) -- t2-t1=1hr

    also
    37=19+Ce-kt0

    Ce-kt0=18

    dividing this by Ce-kt1=6 gives

    e-k(t1-t0)=3 where t0 is the time of death.

    since e-k(t1-t0)=e-k(t2-t1)=3
    t1-t0 must = t2-t1 = 1hr

    therefore the murder must have taken place one hour before the first temperature reading was taken at 2am. so the murder must have taken place at 1am.
    this seems correct to me but my lecturer seems to think it took place at 12am. Can anyone see where i might have gone wrong or is this one of those rare occasions when the lecturer is incorrect?
     
  2. jcsd
  3. Jun 2, 2009 #2

    nvn

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    Homework Helper

    IRNB: Your lecturer appears to be incorrect. Your answer is correct.
     
  4. Jun 2, 2009 #3

    LowlyPion

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    Roughing it out I get:

    6 = 18/ekt1

    ekt = 3 or kt = 1.1

    2 = 18/ek(t1 + 1)

    ek(t1 + 1) = 9 or k*(t1 + 1) = 2.2

    ==> k = 1.1 and t = 1

    Same result. Perhaps a polite inquiry of your lecturer ...?
     
  5. Jun 2, 2009 #4
    Thanks guys. Perhaps my lecturer was just in hurry or something and made a mistake.

    can anyone help out with the first part of the question? does the negative sign simply mean that the temperature is decreasing and the rate at which it changes also decreases with time? also does anyone have any ideas on how to show that C=T0-Ts?

    Thanks again guys.
     
  6. Jun 2, 2009 #5

    LowlyPion

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    Homework Helper

  7. Jun 2, 2009 #6
    thanks a lot LowlyPion that did help.

    this case is now closed. :P
     
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