Q 3.3-5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

It was noon on a cold December day in Tampa: 16°C. Detective Taylor arrived at the crime scene to find the sergeant leaning over the body. The sergeant said there were several suspects. If they knew the exact time of death, then they could narrow the list. Detective Taylor took out a thermometer and measured the temperature of the body: 34.5°C. He then left for lunch. Upon returning at 1:00 p.m., he found the body temperature to be 33.7°C. When did the murder occur? [Hint: Normal body temperature is 37°C.]

Step-by-Step Solution

Answer

The murder was occurred at 9:08 a.m.

Step 1: Analyzing the given statement

It was noon on a cold December day in Tampa: 16°C. The initial temperature of the body measured by detective Taylor in the noon is 34.5°C. At 1:00 p.m., the temperature of the body was found to be 33.7°C, which means the temperature of the body after 1 hour was 33.7°C. By using Newton’s law of cooling, we have to determine the time at which the murder occurred, i.e., the time at which the temperature of the body was 37°C.

Newton’s Law of Cooling is,

 $\mathit{T}\left(t\right)\mathbf{=}{\mathit{M}}_{\mathbf{0}}\mathbf{+}\left(T_0-M_0\right){\mathit{e}}^{\mathbf{-}\mathbf{k}\mathbf{t}}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(1\right)$      

Here, we will take the values as,

Initial temperature of the body,${\mathit{T}}_{\mathbf{0}}\mathbf{=}\mathbf{34}\mathbf{.}{\mathbf{5}}^{\mathbf{o}}\mathit{C}$, 

Temperature in Tampa in the noon, ${\mathit{M}}_{\mathbf{0}}\mathbf{=}{\mathbf{16}}^{\mathbf{o}}\mathit{C}$

Temperature of the body after 60 min,$\mathit{T}\left(60\right)\mathbf{=}\mathbf{33}\mathbf{.}{\mathbf{7}}^{\mathbf{o}}\mathit{C}$

Step 2: To find the value of k in the formula of Newton’s Law of cooling

 Using the given values in equation (1), to find the value of k,

$$\begin{gathered} T\left(60\right)=16+\left(34.5-16\right)e^{-60k} \\ 33.7=16+\left(18.5\right)e^{-60k} \\ 33.7-16=\left(18.5\right)e^{-60k} \\ 17.7=\left(18.5\right)e^{-60k} \\ e^{60k}=\frac{18.5}{17.7} \\ e^{60k}=1.0452 \\ 60k=ln\left(1.0452\right) \\ k=\frac{ln\left(1.0452\right)}{60} \\ k=0.00074 \end{gathered}$$

We will use this value of k in next step to find the time at which the temperature of the body was 37°C.

Step 3: To determine the time at which the murder occurred, i.e., the time at which the temperature of the body was 37°C

Substituting $T\left(t\right)={37}^{o}C$in equation (1),

 $$\begin{gathered} T\left(t\right)=16+\left(34.5-16\right)e^{-\left(0.00074\right)t} \\ 37=16+\left(18.5\right)e^{-\left(0.00074\right)t} \\ 37-16=\left(18.5\right)e^{-\left(0.00074\right)t} \\ 21=\left(18.5\right)e^{-\left(0.00074\right)t} \\ \frac{21}{18.5}=e^{-\left(0.00074\right)t} \\ -\left(0.00074\right)t=ln\left(1.135\right) \\ t=-\frac{ln\left(1.135\right)}{0.00074} \\ t=-171min \end{gathered}$$$$\begin{gathered} T\left(t\right)=16+\left(34.5-16\right)e^{-\left(0.00074\right)t} \\ 37=16+\left(18.5\right)e^{-\left(0.00074\right)t} \\ 37-16=\left(18.5\right)e^{-\left(0.00074\right)t} \\ 21=\left(18.5\right)e^{-\left(0.00074\right)t} \\ \frac{21}{ 18.5}=e^{-\left(0.00074\right)t} \\ -\left(0.00074\right)t=ln\left(1.135\right) \\ t=-\frac{ln\left(1.135\right)}{0.00074} \\ t=-171min \end{gathered}$$

Here, negative sign indicates that the temperature of the body was 37°C, 171 minutes before the noon. 

Hence, the murder was occurred at 9:08 a.m.