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Sagot :
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.
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.
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