To determine the approximate time of death, we'll use Newton's Law of Cooling. Given the constants and final body temperature, we can rearrange the formula to solve for the time [tex]\( t \)[/tex] since death. Here are the steps we follow:
1. Understand Newton's Law of Cooling:
[tex]\[
T(t) = T_A + (T_0 - T_A) e^{-kt}
\][/tex]
Where:
- [tex]\( T(t) \)[/tex] is the temperature of the body at time [tex]\( t \)[/tex]
- [tex]\( T_A \)[/tex] is the ambient temperature (55°F)
- [tex]\( T_0 \)[/tex] is the initial body temperature (98.6°F)
- [tex]\( k \)[/tex] is the cooling constant (0.1947)
- [tex]\( t \)[/tex] is the time since death in hours.
2. Given values:
[tex]\[
T(t) = 66^\circ F \quad (body \ temperature \ found)
\][/tex]
[tex]\[
T_A = 55^\circ F \quad (ambient \ temperature)
\][/tex]
[tex]\[
T_0 = 98.6^\circ F \quad (initial \ body \ temperature)
\][/tex]
[tex]\[
k = 0.1947
\][/tex]
3. Formulate the equation:
Plug the given values into the Newton's Law of Cooling formula:
[tex]\[
66 = 55 + (98.6 - 55) e^{-0.1947t}
\][/tex]
4. Isolate the exponential term:
[tex]\[
66 - 55 = (98.6 - 55) e^{-0.1947t}
\][/tex]
[tex]\[
11 = 43.6 e^{-0.1947t}
\][/tex]
5. Solve for the exponential term:
[tex]\[
\frac{11}{43.6} = e^{-0.1947t}
\][/tex]
[tex]\[
0.2522935779816514 = e^{-0.1947t}
\][/tex]
6. Use the natural logarithm to solve for [tex]\( t \)[/tex]:
Take the natural logarithm of both sides:
[tex]\[
\ln(0.2522935779816514) = -0.1947t
\][/tex]
[tex]\[
-1.375911 \approx -0.1947t
\][/tex]
7. Solve for [tex]\( t \)[/tex]:
[tex]\[
t = \frac{-1.375911}{-0.1947} \approx 7.073 \text{ hours}
\][/tex]
8. Determine the time of death:
Since the body was found at 12 a.m. midnight, we conclude:
[tex]\[
\text{Time since death} = 7.073 \text{ hours}
\][/tex]
To find the approximate time of death, subtract this from 12 a.m.:
[tex]\[
\text{Time of death} = 12 \text{ a.m.} - 7.073 \text{ hours} \approx 5 \text{ p.m.}
\][/tex]
Therefore, considering all the calculations:
The approximate time of death was around 5 p.m. Hence, the correct answer is:
C. 5 p.m.
