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

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} (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} (dT)/(dt)=-k(T-T_0) \\ (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)(dT)/(dt)+e^(kt)kT=e^(kt)kT_0`

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

`(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)=(40)/(58.6) \\ ln((40)/(58.6))=-kt_f \\ t_f=-(ln((40)/(58.6)))/(k) \\ t_f=(ln((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