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Part 1
A body was found in the basement of the Underwater Basket Weaving Building at​ 12:00 noon​ today, where the temperature is a steady 65 degrees Fahrenheit.  When​ found, the core temperature was 92.3 degrees Fahrenheit.  Two hours​ later, at​ 2:00 PM, the core temperature had fallen to 87.1.   Assuming that the body temperature was 98.6 at the time of​ death, use​ Newton's law of cooling to find the time of death.  
ROUND TO 2 DECIMAL PLACES.
The time of death was about 
  
enter your response here hours before the body was found.

To solve this problem using Newton's Law of Cooling, we apply the following formula:

\[
T(t) = T_{\text{env}} + (T_0 - T_{\text{env}}) e^{-kt}
\]

Where:
- \( T(t) \) is the temperature of the body at time \( t \).
- \( T_{\text{env}} \) is the surrounding environment's constant temperature (65°F in this case).
- \( T_0 \) is the body's initial temperature at the time of death (98.6°F).
- \( k \) is the cooling constant.
- \( t \) is the time elapsed since death (in hours).

### Step 1: Create two equations to solve for \( k \).

At noon (time = 0):
\[
92.3 = 65 + (98.6 - 65)e^{-k \cdot 0}
\]
\[
92.3 = 65 + 33.6
\]
\[
92.3 - 65 = 33.6
\]

So this is consistent, and we move to the next equation, after 2 hours.

At 2:00 PM (time = 2 hours):
\[
87.1 = 65 + 33.6e^{-2k}
\]
We can now solve for \( k \).

### Step 2: Solve for \( k \).

Subtract 65 from both sides:
\[
87.1 - 65 = 33.6e^{-2k}
\]
\[
22.1 = 33.6e^{-2k}
\]

Divide both sides by 33.6:
\[
\frac{22.1}{33.6} = e^{-2k}
\]
\[
0.6586 \approx e^{-2k}
\]

Take the natural logarithm of both sides:
\[
\ln(0.6586) = -2k
\]
\[
-0.4172 = -2k
\]

Solve for \( k \):
\[
k = \frac{0.4172}{2} = 0.2086
\]

### Step 3: Solve for the time of death \( t \).

We now use the formula to find \( t \), the time since death.

At noon, the body temperature was 92.3°F. Plug this into the Newton's law of cooling formula:
\[
92.3 = 65 + 33.6 e^{-0.2086t}
\]

Subtract 65 from both sides:
\[
92.3 - 65 = 33.6 e^{-0.2086t}
\]
\[
27.3 = 33.6 e^{-0.2086t}
\]

Divide both sides by 33.6:
\[
\frac{27.3}{33.6} = e^{-0.2086t}
\]
\[
0.8125 = e^{-0.2086t}
\]

Take the natural logarithm of both sides:
\[
\ln(0.8125) = -0.2086t
\]
\[
-0.2077 = -0.2086t
\]

Solve for \( t \):
\[
t = \frac{0.2077}{0.2086} = 0.996 \, \text{hours}
\]

### Step 4: Calculate the time of death.

Since this is approximately 1 hour before noon (the time the body was found), we conclude that the time of death occurred:

\[
t \approx 1 \, \text{hour before noon}.
\]

Therefore, the time of death was around **1 hour before** the body was found at noon.

Would you like any further clarification or more details on this?

---

### Related Questions:
1. How does Newton's Law of Cooling apply to real-world forensic investigations?
2. Can environmental factors (e.g., humidity) influence the cooling constant \( k \)?
3. What are the assumptions in using Newton's Law of Cooling for estimating time of death?
4. How would a higher room temperature affect the cooling rate and the time of death calculation?
5. How do forensic scientists deal with varying temperatures in different rooms of a building?

**Tip:** For more accurate time of death estimates, combining Newton’s Law of Cooling with other forensic methods, such as rigor mortis or livor mortis analysis, can improve precision.

A body was found in the basement of the Underwater Basket Weaving Building at 12:00 noon today, where the temperature is a steady 65 degrees Fahrenheit. When found, the core temperature was 92.3 degrees Fahrenheit. Two hours later, at 2:00 PM, the core temperature had fallen to 87.1. Assuming that the body temperature was 98.6 at the time of death, use Newton's law of cooling to find the time of death. ROUND TO 2 DECIMAL PLACES. The time of death was about __ hours before the body was found.

This problem uses Newton's Law of Cooling to calculate the time of death of a body found in a 65°F environment. With a core temperature of 92.3°F when found and 87.1°F two hours later, the cooling rate is calculated, leading to the estimated time of death. Suitable for forensic mathematics and applied physics students.

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