Coroners estimate time of death using the rule of thumb that a body cools about

Question

Coroners estimate time of death using the rule of thumb that a body cools about 2 degrees F during the first hour after death and about 1 degree F for each additional hour.

Assuming an air temperature of 78 degrees F and a living body temperature of 98.6 degrees F, the temperature T(t) in degrees F of a body at a time t hours since death is given by T(t) = 78 + 20.6 e^{-kt}.

1. For what value of k will the body cool by 2 degrees F in the first hour?

k=

2. Using the value of k found above, after how many hours will the temperature of the body be decreasing at a rate of 1 degree F per hour?

after ______ hours

3. Using the value of k found above, show by calculating both values that, 24 hours after death, the coroner's rule of thumb gives approximately the same temperature as the formula.

T(24)=

degrees F rule of thumb gives

T = _____ degrees F

Explanation / Answer

Coroners estimate time of death:

Given
T(t) = 64 + 34.6 e^(-kt)

1) given:
(t, T(t)) = (1, 96.6)
so
(96.6) = 64 + 34.6 e^(-k(1))
       32.6/34.6 = e^(-k)
  ln(32.6/34.6) = ln(e^(-k))
  ln(32.6/34.6) = -k ln(e)
-ln(32.6/34.6) = k
         0.05954 k

2) T(t) 64 + 34.6 e^(-0.05954t)
   T'(t) -2.0601 e^(-0.05954t)
so
                                     -... -2.0601 e^(-0.05954t)
                      1 / 2.0601 e^(-0.05954t)
                 ln(1 / 2.0601) -0.05954t
ln(1 / 2.0601) / -0.05954 t
                                      ... 12.1390 hrs

3) Coroner's estimate:
after 24 hours, body has cooled by 2 + 23 degrees = 25°
so 98.6 - 25 = 73.6°
and per T(t):
T(24) 64 + 34.6 e^(-0.05954(24))
T(24) 64 + 34.6 e^(-1.42899)
T(24) 72.3°
So
coroner would estimate ToD 1h 18m earlier than T(t)

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