Question

A body was found at 6 a.m. outdoors on a day when the temperature was 50oF. The medical examiner found the temperature of the body to be 66oF. What was the approximate time of death? Use Newton's law of cooling, with k = 0.1947.
Midnight (12 a.m.)

5 a.m.

2 a.m.

3 a.m.

Answer 1

Answer:

Midnight (12 a.m.)

Step-by-step explanation:

Newton's law of cooling-

$T(t)=t_a+(t_0-t_a)e^{-kt}$

where,

$t_0$ = the initial temp. = 98.6 F (human body temp.)

k = 0.1947,

$T(t) = 66$ F,

$t_a = 55$ F,

$\Rightarrow 66=50+(98.6-50)e^{-0.1947\cdot t}$

$\Rightarrow 66=50+(48.6)e^{-0.1947\cdot t}$

$\Rightarrow (48.6)e^{-0.1947\cdot t}=16$

$\Rightarrow e^{-0.1947\cdot t}=\dfrac{16}{48.6}$

$\Rightarrow \ln e^{-0.1947\cdot t}=\ln \dfrac{16}{48.6}$

$\Rightarrow -0.1947\cdot t=\ln \dfrac{16}{48.6}$

$\Rightarrow t=\dfrac{\ln \frac{16}{48.6}}{-0.1947}=5.7\approx 6$ hr

Therefore, the time of death was 6 hours before 6 am i.e at midnight (12 a.m.)

Answer 2

The answer is : Midnight~Apex