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Newton's law of cooling to find time of death

by Lee.Robbo
Tags: cooling, death, exponential, help!, newton's law
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Oct26-11, 04:06 PM
P: 1
1. The problem statement, all variables and given/known data

The victim, Peter Sloane (a senior physics lecturer), was discovered at 9.43pm with a liver temperature of 22.26C in the Oliver Lodge coffee room, with the window open. The temperature of the room matched that of the outside. After checking weather reports, it was found that at precisely 9.43pm, the temperature outside was 17C, however, it had steadily been dropping by approximately 1C every hour before this. The temperature for the rest of the day had been constant at 20C until it started to drop. Cause of death was determined to be heavy metal poisoning. (No puns please, that's how it's stated...) Using the equation for Newton’s law (given*previously), determine how*long Peter had been dead when he was found, and consequently, his time of death.

2. Relevant equations

Newton's law of Cooling:
T(t) = Te+ (T0− Te)e^−kt
Where T(t)= temperature with respect to time
Te= Temperature of surrounding environment
T0= Initial temperature (when t=0)
k= constant, worked out previously as 0.0070636
t= time

3. The attempt at a solution

I have no idea where to start. The paper gives nothing about the initial body temperature, but even if I take that to be 37 degrees C, I'm pretty sure I can't just take natural logs, because Te isn't constant over the previous 3 hours, and it's due in tomorrow morning... Help!
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Oct26-11, 08:33 PM
P: 11,682
Interesting problem. Essentially you want to run Newton's Law of Cooling "backwards", and with a time-varying environmental temperature, until you reach normal body temperature.

I think you'll have to write and solve a differential equation for the period where the environmental temperature is changing. Your time zero will be the moment of discovery of the body, and time increases going backwards (so the time variable represents "hours ago").

Your constant k = 0.0070636, what are the units attached to it? Seconds or hours?
Oct26-11, 08:43 PM
P: 308
What you've listed for Newton's law of cooling is actually not Newton's law of cooling. Rather, it is simply a solution of the actual law under the assumptions that ambient temperature is constant.

Newton's law of cooling is actually:
[tex]\frac{dT}{dt} = -k(T-T_e)[/tex]

In order to solve it, you'll have to derive a function for [itex]T_e = T_e(t)[/itex] using the information given in the problem, and plug this function into the equation above. You'll have to solve the ODE to obtain [itex]T(t)[/itex].

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