Chapter 7.4, Problem 14E

To determine

To calculate : at what time the murder takes place

Answer to Problem 14E

The murder takes place at $11:55am$

Explanation of Solution

Given information : The question is

In a murder investigation, the temperature of the corpse was ${32.5}^{\circ }C$ at $1:30pm$ and ${30.3}^{\circ }C$ an hour later. Normal body temperature is ${37.0}^{\circ }C$ and the temperature of the surrounding was ${20.0}^{\circ }C$

Calculation : for the time of at which murder takes place

  $\begin{array}{l}\frac{dT}{dt}=k\left(T-T_a\right)\\ {\displaystyle \int \frac{1}{T-T_a}dT={\displaystyle \int kdt}}\\ u=T-T_a\\ du=dT\\ {\displaystyle \int u^{-1}du=\ln \left|u\right|}\\ \ln \left|T-T_a\right|=kt+C\\ \left|T-T_a\right|=e^{kt+C}=C{e}^{kt}\\ when\text{ T}\ge {\text{T}}_a\\ T\left(t\right)=C{e}^{kt}+{\text{T}}_a\\ when\text{ T}\le {\text{T}}_a\\ T-T_a=C{e}^{kt}\\ T\left(t\right)=-C{e}^{kt}+T_a\\ T\left(t\right)=C{e}^{kt}+{\text{T}}_a\left\{\begin{array}{l}{\text{where T}}_a={20}^{\circ }C;t=60\text{minutes ;}T\left(t\right)={30.3}^{\circ }C\\ T\left(0\right)={32.5}^{\circ }C\end{array}\right\}\\ 30.3=12.5e^{k60}+20\\ \frac{30.3-20}{12.5}=e^{k60}\\ \frac{1}{60}\ln \frac{10.3}{12.5}=k\\ k\approx -0.0032\\ T\left(t\right)\approx 12.5e^{-0.0032}+20\\ 37=12.5e^{-0.0032}+20\\ \frac{37-20}{12.5}=e^{-0.0032}\\ \frac{1}{-0.0032}\ln \frac{17}{12.5}=t\\ t\approx -95.3024\min utes\\ T\left(t\right)={37}^{\circ }C\\ \to 95\text{ minutes before }1:30pm\text{ is }11:55am\end{array}$