A dead body was found within a closed room of a house where

A dead body was found within a closed room of a house where the temperature was a constant 65 degree F. At the time of discovery the core temperature of the body was determined to be 800 F. One hour later a second measurement showed that the core temperature of the body was 75 degree F. Assume that the time of death corresponds to t = O and that the core temperature at that time was g8.6 degree F. Determine how many hours elapsed before the body was found. [Hint: Let t1 > 0 denote the time that the body was discovered.] (Round your answer to one decimal place.) ____________ hr

Solution

dT/dt = -k * (T - T[a])
dT / (T - T[a]) = -k * dt
ln(T - T[a]) = -kt + C
T - T[a] = e^(C - kt)
T - T[a] = C * e^(-kt)
T = T[a] + C * e^(-kt)

T[a] = 65

T = 65 + C * e^(-kt)

T[1] = 75
T[0] = 98.6
t[0] = 0

98.6 =65 + C * e^(0)
33.6 = C

T = 65 + 33.6 * e^(-kt)
75 = 65 + 33.6 * e^(-k*t)

10 = 33.6 * e^(-kt)
10/33.6 = e^(-kt)
ln(10/33.6) = -kt

11 = 33.6 * e^(-k * (t + 1))
11/33.6 = e^(-k * (t + 1))
ln(11/33.6) = -k * (t + 1)

ln(14/33.6) / t = -k
ln(11/33.6) / (t + 1) = -k

ln(10/33.6) / t = ln(11/33.6) / (t + 1)
(t + 1) * ln(10/33.6) = t * ln(11/33.6)
t * ln(10/33.6) + ln(14/33.6) = t * ln(11/33.6)
ln(10/33.6) = t * (ln(11/33.6) - ln(14/33.6))
ln(10/33.6) = t * (ln(11) - ln(33.6) - ln(14) + ln(33.6))
ln(10/33.6) = t * (ln(11) - ln(10))
ln(10/33.6) = t * ln(11/10)
t = ln(10/33.6) / ln(11/10)
t = 2.9621135166920002367875574138984

The person was found 2.96 hours after they had died, so 3.0 hours when rounded.