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At 9:00 PM a coroner arrived at a hotel room of a murder victim
At 9:00 PM a coroner arrived at a hotel room of a murder victim. The temperature of the room was 70 Degrees F. It was assumed that the victim had a body temperature of 98.6 degrees F (AT THE TIME OF DEATH)(not at 9:00 PM). The coroner took the victim's temperature at 9:15 PM at which it was 83.6 degrees F and again at 10:00 PM at which it was 80.3 degrees F.
At what time did the victim die?
This problem is supposed to be solved by usinf Newton's Law of Cooling :
T(t) = T_s + (T_0 - T_s)e^(Kt).
Expert Solution
Please see the attached file.
Theory
Newton's Law of Cooling describes the cooling of a warmer object to the cooler temperature of the environment. Specifically we write this law as,
T (t) = Te + (T0 − Te ) e - kt,
where T (t) is the temperature of the object at time t, Te is the constant temperature of the environment, T0 is the initial temperature of the object, and k is a constant that depends on the material properties of the object.
Solution
In this case: Te = 70 oF
First find k
From 9:15 to 10:00 (45 minutes) the body temperature changed from 83.6 to 80.3. Thus:
80.3 = 70 + (83.6 - 70)e-45k
Hence we can find the time of death using the temperature at 10PM
80.3 = 70 + (98.6 - 70)e-0.006176t
80.3 = 70 + 28.6e-0.006176t
Thus the time of death is 165 minutes before 10PM or at ~7:15PM
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