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Newton's law of cooling and your body

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data
    a body has temperature 27C at twelve oclock. The room temperature is constant at 16C. Two hours later the body was found to have a temperature of 24C. The temperature of a normal human body is 37C.Using Newton's law of cooling estimate the time of death.


    2. Relevant equations
    Newton's law of cooling says that the rate of heat loss from a body is proportional to the temperature difference between it and the surroundings.
    dQ/dt=hA(Te-Tb)
    where Te is the temp of the environment 16C=289K and Tb is the temp of the body 24C=297K. I am not sure what the heat transfer coefficient h is or how to determine this. Also not sure how to use the fact that normal body temperature is 37C

    Could someone please give a bit of a start in the right direction
    Thank you
     
  2. jcsd
  3. Oct 24, 2009 #2

    Doc Al

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    Staff: Mentor

    Think of Newton's law of cooling in this form:
    dT/dt = -k(T - Te)

    What's the general form of the solution to that equation?
     
  4. Oct 24, 2009 #3
    After some googling, the heat transfer coefficient for an idle body is [tex] 2.1 W/m^2\cdot K[/tex] and the approximated surface area is [tex]1.7 m^2 [/tex].

    As for solving this, you'd need to integrate your equation (which btw should be [tex]dT/dt[/tex], not [tex]dQ/dt[/tex] and find the temperature at a later time [tex]t[/tex] is

    [tex]
    T(t)=T_a+\left(T_0-T_a\right)\exp[-hAt]
    [/tex]

    where [tex]T_a[/tex] is the ambient temperature and [tex]T_0[/tex] is the initial temperature of the body.
     
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