Question

At 10:30 AM, detectives discover a dead body in a room and measure its temperature at 27°C. One hour later, the body's temperature had dropped to 24.8°C. Determine the time of death (when the body temperature was a normal 37°C), assuming that the temperature in the room was held constant at 22°C. (Round your answer to two decimal places.)

:The time of death was approximately_______ hours before 10:30 AM.

Answer 1

The time of death was approximately 31.2 minutes before 10:30 AM.

How to determine the time?

We employed Newton's Law of Cooling in order to determine time;

This law is expressed by the formula:

`\[ T(t) = T_a + (T_0 - T_a) \cdot e^(-kt) \]`

In this equation:

`\( T(t) \)` is the temperature at time
`\( t \)`,

`\( T_0 \)` is the initial temperature of the body,

`\( T_a \)` is the ambient temperature (room temperature),

`\( k \)` is a constant, and

`\( e \)` is the base of the natural logarithm.

Given the specifics:

`\( T_0 = 37 \) \textdegree C` (normal body temperature),

`\( T_a = 22 \) \textdegree C` (room temperature),

`\( T(0) = 27 \) \textdegree C` (temperature at 10:30 AM),

`\( T(1) = 24.8 \) \textdegree C` (temperature one hour later),

We aimed to find
`\( t \)`, representing the time of death when the body temperature reached 37°C.

`\[ 37 = 22 + (27 - 22) \cdot e^(-kt) \]`

`\[ 15 = 5 \cdot e^(-kt) \]`

`\[ e^(-kt) = 3 \]`

`\[ -kt = \ln(3) \]`

`\[ t = -(\ln(3))/(k) \]`

Determining the value of
`\( k \)` involved using the information from the initial temperature drop:

`\[ 24.8 = 22 + (27 - 22) \cdot e^(-k \cdot 1) \]`

`\[ 2.8 = 5 \cdot e^(-k) \]`

`\[ e^(-k) = 0.56 \]`

`\[ -k = \ln(0.56) \]`

`\[ k \approx 0.576 \]`

Substituting this
`\( k \)` value back into the equation for
`\( t \)`:

`\[ t \approx -(\ln(3))/(0.576) \]`

`\[ t \approx 0.52 \text{ hours} \]`

To convert this to minutes, multiply by 60:

`\[ t \approx 31.2 \text{ minutes} \]`

Therefore, the time of death was approximately 31.2 minutes before 10:30 AM.

Answer 2

The time of death was approximately 4 hours before 10:30 AM.

Step-by-step explanation:

Temperature of the body when discovered by detective at 10:30 AM= 27°C

After discovery, Temperature of the body at 11.30 AM= 24.8°C

Change in time = ΔT = 1 hour

Rate of change of body temperature with respect to time :

`R=(24.8^oC-27^oC)/(1 hour)=-2.2^oC/hour`

Normal body temperature = 37°C

Change in time since death = ΔT'

`R=(27^oC-37^oC)/(\Delta T')`

`-2.2^oC/hour=(27^oC-37^oC)/(\Delta T')`

`\Delta T'=(27^oC-37^oC)/(-2.2^oC/Hour)=4.54 hour\approx 4 hours`

The time of death was approximately 4 hours before 10:30 AM.