Answer to Question #222692 in Differential Equations for handy

Question

Detectives have discovered the murder victim in juja at 6:30 am and the body temperature of the victim was then 26⁰celcius after 30minutes the police surgeon arrived and found the body temperature to be 23⁰ Celsius. If the air temperature was 16⁰ Celsius throughout and the normal body temperature is 37⁰ Celsius at what time did the police surgeon estimate that the crime had occurred(victim murdered)?

Answer 1

$\text{At 6:30 a.m, t = 0 and the temperature $T_0=26^0C$} \\\text{After 30 minutes t = 0.5hr and the temperature $T_0=23^0C$ and the room temperature}\\\text{The change in temperature is given by }\\\frac{dT}{dt}=-k(T-T_c)=-k(T-16)\\\text{Integrating, we have that}\\In(T-16)=-kt+c\\\text{Taking exponents, we have}\\T-16=Ae^{-kt}\text{ where, $A=e^c$, therefore}\\T=Ae^{-kt}+16 \\\text{when t=0 and $T_0=26^0C$, we have that A = 10}\\\implies T=Ae^{-kt}+16 \\\text{when t=0.5 and $T_0=23^0C$, we have that k = 0.7133}\\\implies T=10e^{-0.7133t}+16 \\\text{To calculate the time of death, we take T as normal body temperature and we }\\\text{compute t}\\37=10e^{-0.7133t}+16 \\\text{Hence, $t=-1.040hr$, therefore the death occured about 1 hour before 6:30 a.m}$