Question

Police arrive at a murder scene at 2:30 AM and immediately record the body's temperature which was 91°F. At 3:00 AM, after thoroughly inspecting and fingerprinting the area, they again took the temperature of the body which had dropped to 86°F. The temperature of the crime scene has remained at a constant 71°F. Determine when the person was murdered. (Assume that the victim was healthy at the time of death. That is, assume that the temperature of the body at the time of death was 98.6°F.)

Answer 1

Let T(t) = temperature of the body t hours after the temperature was first recorded

By Newton's Law of cooling, T(t) = T_s + (T₀ - T_s)e^-kt, where T_s is the temperature of the surroundings, T₀ is the temperature the first time that it is recorded, and k is a constant to be determined.

So, we have T(t) = 71 + (91 - 71) e^-kt

T(t) = 71 + 20e^-kt

Next, we need to find k:

We are given that the temperature at 3 AM was 86°. Since 3 AM is 0.5 hours after 2:30 AM, we get:

T(0.5) = 71 + 20e^{-k (0.5)} = 86

= 20e^{-k (0.5)} = 15

= e^{-k (0.5)} = 0.75

= -0.5k = ln (0.75)

k = 0.5754

So, T(t) = 71 + 20e^-(0.5754)t

We are trying to find the time of death. In other words, since it is assumed that the body temperature was normal at the time of death, we need to find t so that T(t) = 98.6°. i.e.

71 + 20e^-(0.5754)t = 98.6

  20e^-(0.5754)t = 98.6 - 71

  $e^{-(0.5754)t} = \frac{27.6}{20}=1.38$

- (0.5754) t = ln(1.38)

$t=-0.55976 \approx -0.56$

(the negative sign indicates that the time of death was 0.56 hours prior to the time that the temperature of the body was first taken)

Thus, the time of death was 0.56 hours before 2:30 AM.

0.56 hours = 0.56 * 60 = 33.6 min $\approx 34$ min

Time of death = 2:30 AM - 34 minutes = 1:56 AM

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