Question

A body was found at 10 a.m. in a warehouse where the temperature was40°F. The medical examiner found the temperature of the body to be 80°F.What was the approximate time of death? Use Newton's law of cooling, with k= 01947.

Answer 1

SOLUTION

Newton's law of cooling states that the rate of change of temperature is proportional to the difference to the ambient temperature.

So, from here we have

$\begin{gathered} \frac{dT}{dt}\propto T-T_0 \\ If\text{ T}>T_0 \end{gathered}$

then the body should cool so the derivative should be negative, hence we insert the proportionality constant and arrive at

$\begin{gathered} \frac{dT}{dt}=-k(T-T_0) \\ \frac{dT}{dt}+kT=kT_0 \end{gathered}$

Can now use the integrating factor method of solving ODEs.

$I(x)=e^{\int kdt}=e^{kt}$

Multiplying both sides by I(x) we get

$e^{kt}\frac{dT}{dt}+e^{kt}kT=e^{kt}kT_0$

Notice that by using the product rule we can rewrite the LHS, leaving:

$\frac{d}{dt}[Te^{kt}]=e^{kt}kT_0$

Integrate both sides with respect to t, we have

$\begin{gathered} Te^{kt}=kT_0\int e^{kt}dt \\ Te^{kt}=T_0e^{kt}+C \\ divide\text{ by }e^{kt} \\ T(t)=T_0+Ce^{-kt} \end{gathered}$

Average human body temperature is 98.6 degree fahrenheit, we have

$\begin{gathered} T(0)=98.6 \\ 98.6=40+Ce^0 \\ C=58.6 \\ Let\text{ t}_f\text{ be the time at which body is found} \end{gathered}$

we have

$\begin{gathered} T(t_f)=80 \\ 80=40+58.6e^{-kt_f} \\ e^{-kt_f}=\frac{40}{58.6} \\ ln(\frac{40}{58.6})=-kt_f \\ t_f=-\frac{ln(\frac{40}{58.6})}{k} \\ t_f=\frac{ln(\frac{40}{58.6})}{0.1947} \\ t_f=1.96hr \end{gathered}$

So from time of death, assuming body immediately started to cool, it took 1.96 hours to reach 80°F at which point it was found.

1.96 hr = 117,6 mins subtracting from 10 a.m, approximate time of death is 8 : 02 : 24 am

Hence the answer is

8 : 02 : 24 am