Question

question 2 of 10
a body was found at midnight (12 a.m.) in a warehouse where the temperature was 55°f. the medical examiner found the temperature of the body to be 66°f. what was the approximate time of death? use newtons law of cooling, with k = 0.1947.
t(t)=t_a+(t_o - t_a)e^{-kt}
a. 3 p.m.
b. 5 p.m.
c. 4 p.m.
d. 6 p.m.

Explanation:

Step1: Assume normal body - temp

The normal body temperature $T_o = 98.6^{\circ}F$, the ambient temperature $T_A=55^{\circ}F$, and at time $t$ (when the body was found), $T(t) = 66^{\circ}F$, $k = 0.1947$. Substitute these values into Newton's law of cooling formula $T(t)=T_A+(T_o - T_A)e^{-kt}$.
So, $66 = 55+(98.6 - 55)e^{-0.1947t}$.

Step2: Simplify the equation

First, simplify the right - hand side:
$66=55 + 43.6e^{-0.1947t}$.
Subtract 55 from both sides:
$66 - 55=43.6e^{-0.1947t}$, which gives $11 = 43.6e^{-0.1947t}$.
Then, divide both sides by 43.6:
$\frac{11}{43.6}=e^{-0.1947t}$, so $e^{-0.1947t}=\frac{11}{43.6}\approx0.2523$.

Step3: Take the natural logarithm

Take the natural logarithm of both sides:
$\ln(e^{-0.1947t})=\ln(0.2523)$.
Using the property $\ln(e^x)=x$, we get $- 0.1947t=\ln(0.2523)$.
Since $\ln(0.2523)\approx - 1.377$, then $t=\frac{-1.377}{-0.1947}\approx7$ hours.
Since the body was found at midnight (12 a.m.), the time of death was approximately 7 hours before midnight, which is 5 p.m.

Answer:

B. 5 p.m.