Question

a body was found at 6 a.m. in a warehouse where the temperature was 50°f. the medical examiner found the temperature of the body to be 66°f. what was the approximate time of death? use newton’s law of cooling, with k = 0.1947. t(t) = t_a + (t_o - t_a)e^(-kt) a. 2 a.m. b. midnight (12 a.m.) c. 5 a.m. d. 3 a.m.

Explanation:

Step1: Identify known values

We know that \( T_A = 50^\circ\text{F} \) (ambient temperature), \( T(t) = 66^\circ\text{F} \) (body temperature at discovery), \( k = 0.1947 \), and the normal body temperature \( T_o = 98.6^\circ\text{F} \). The formula is \( T(t)=T_A+(T_o - T_A)e^{-kt} \).

Step2: Substitute values into the formula

Substitute the known values into the formula:
\[ 66 = 50 + (98.6 - 50)e^{-0.1947t} \]
Simplify the equation:
\[ 66 - 50 = 48.6e^{-0.1947t} \]
\[ 16 = 48.6e^{-0.1947t} \]

Step3: Solve for \( e^{-0.1947t} \)

Divide both sides by 48.6:
\[ e^{-0.1947t}=\frac{16}{48.6}\approx0.3292 \]

Step4: Take the natural logarithm of both sides

Take the natural log of both sides:
\[ \ln(e^{-0.1947t})=\ln(0.3292) \]
Using the property \( \ln(e^x)=x \), we get:
\[ -0.1947t=\ln(0.3292) \]

Step5: Solve for \( t \)

Calculate \( \ln(0.3292)\approx - 1.115 \). Then:
\[ t=\frac{-1.115}{-0.1947}\approx5.73 \]
Wait, but this seems off. Wait, maybe I mixed up \( T(t) \) and \( T_o \). Wait, the body's initial temperature (at death) is \( T_o = 98.6 \), and at time \( t \) (time since death) the temperature is \( T(t) = 66 \), and the ambient is \( T_A = 50 \). Wait, but the time of discovery is 6 a.m. So we need to find \( t \) (time since death), then subtract from 6 a.m.

Wait, maybe I made a mistake in substitution. Let's re - do:

The formula is \( T(t)=T_A+(T_o - T_A)e^{-kt} \), where \( t \) is the time elapsed since death.

So \( 66 = 50+(98.6 - 50)e^{-0.1947t} \)

\( 66 - 50=(98.6 - 50)e^{-0.1947t} \)

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

\( e^{-0.1947t}=\frac{16}{48.6}\approx0.3292 \)

\( \ln(e^{-0.1947t})=\ln(0.3292) \)

\( - 0.1947t=\ln(0.3292)\approx - 1.115 \)

\( t=\frac{1.115}{0.1947}\approx5.73 \) hours? No, that can't be. Wait, maybe the normal body temperature is 98.6, but maybe I messed up the formula. Wait, Newton's law of cooling is \( T(t)=T_A+(T_0 - T_A)e^{-kt} \), where \( T_0 \) is the initial temperature of the object (body at death, 98.6), \( T_A \) is ambient (50), \( T(t) \) is temperature at time \( t \) (time since death). We need to find \( t \) when \( T(t) = 66 \).

Wait, but if \( t\approx5.73 \) hours, that would mean death was about 5.73 hours before 6 a.m., which is around 12:16 a.m., but that's not one of the options. Wait, maybe I used the wrong \( k \) or messed up the formula. Wait, maybe the formula is \( T(t)=T_0e^{-kt}+T_A(1 - e^{-kt}) \), which is the same as \( T(t)=T_A+(T_0 - T_A)e^{-kt} \). Wait, maybe the problem has a different normal body temperature? No, 98.6 is standard. Wait, maybe the \( k \) is different? Wait, the problem says \( k = 0.1947 \). Wait, maybe I made a calculation error.

Wait, let's recalculate \( \ln(16/48.6) \). \( 16\div48.6\approx0.3292 \). \( \ln(0.3292)\approx - 1.115 \). Then \( t=\frac{1.115}{0.1947}\approx5.73 \). But the options are 2 a.m., 12 a.m., 5 a.m., 3 a.m. So maybe the normal body temperature is taken as 98.6, but maybe the problem assumes \( T_0 = 98.6 \), \( T_A = 50 \), \( T(t)=66 \), \( k = 0.1947 \). Wait, maybe I inverted the formula. Let's try again.

Wait, maybe the time \( t \) is the time from death to discovery (6 a.m.). So we need to find \( t \), then 6 a.m. minus \( t \) is the time of death.

Wait, let's check the options. Let's try \( t = 6 \) hours (midnight to 6 a.m.). Then \( T(t)=50+(98.6 - 50)e^{-0.1947\times6} \). \( 0.1947\times6 = 1.1682 \). \( e^{-1.1682}\approx0.310 \). Then \( 50 + 48.6\times0.310\approx50 + 15.07 = 65.07\approx66 \). Oh! Wait, my previous calculation of \( t \) was wrong. Wait, \( 0.1947\times6 = 1.1682 \), \( e^{-1.1682}\approx0.31 \), \( 48.6\times0.31 = 15.066 \), \( 50+15.066 = 65.066\approx66 \). So \( t\approx6 \) hours. So 6 a.m. minus 6 hours is midnight (12 a.m.). So the time of death is midnight.

Ah, I see, I made a mistake in the calculation of \( t \). Let's do it properly:

\( T(t)=T_A+(T_o - T_A)e^{-kt} \)

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

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

\( e^{-0.1947t}=\frac{16}{48.6}\approx0.3292 \)

Take natural log:

\( - 0.1947t=\ln(0.3292)\approx - 1.115 \)

\( t=\frac{1.115}{0.1947}\approx5.73\approx6 \) hours (since we approximated). So 6 a.m. minus 6 hours is 12 a.m. (midnight).

Answer:

B. Midnight (12 a.m.)