Question

A dead body was found within a closed room of a house where the temperature was a constant 65° F. At the time of discovery the core temperature of the body was determined to be 85° F. One hour later a second measurement showed that the core temperature of the body was 80° F. Assume that the time of death corresponds to t = 0 and that the core temperature at that time was 98.6° F. Determine how many hours elapsed before the body was found.

Answer 1

Answer:

1 hr 52 minutes

Step-by-step explanation:

As per Newton law of cooling we have

$T(t) = T_s +(T_0-T_s)e^{-kt}$

where T0 is the initial temperature of the body

Ts = temperature of surrounding

t = time lapsed

k = constant

Using this we find that T0 = 98.6 : Ts= 65

Let x hours be lapsed before the body was found.

Then we have

$T(x) = 65 +(98.6-65)e^{-kx} = 85\\e^{-kx}=\frac{20}{33.8} =0.5917$

Next after 1 hour temperature was 80

$T(x+1) = 65+33.6(e^{-k(x+1)}=80\\e^{-k(x+1) =0.4464$

Dividing we get

$e^k = 1.325408\\k = 0.2817$

Substitute this in

$e^{-kx} =0.5917\\x=\frac{ln 0.5917}{-k} \\=1.863$

approximately 1 hour 52 minutes have lapsed.