Question

For a post-mortem report, a doctor requires to known approximate time of death of the deceased. He records the first temperature at $10.00$ a.m. to be ${93.4}^o$F. After $2$ hours the finds the temperature to be ${91.4}^o$F. If the room temperature (which is constant) is ${72}^o$F, estimate the time of death.(Assume normal temperature a human body to be ${98.6}^o$F). $[lo{g}_e\frac{19.4}{21.4}=-0.0426\times 2.303$ and $lo{g}_e\frac{26.6}{21.4}=0.0945\times 2.303]$.

Answer 1

Using Newton;s law of cooling:

$T\left(t\right)=T_s+(T_o-T_s)e^{-kt}$

where $T\left(t\right)=$ temp. at any time 't' from beginning.

$T_s=$surrounding temp.$={72}^{o}F$

$T_o=$initial temp.$={98.6}^{o}F$

$k=$ constant

$t=$time from beginning.

Let the man died 't' hours before $10.00a.m.$

Then,

$93.4=72+(98.6-72)e^{-kt}.................\left(1\right)$

and after 2 hours

$91.4=72+(98.6-72)e^{-k(t+2)}$

$\Rightarrow 91.4=72+(98.6-72)e^{-kt}{e}^{-2k}.................\left(2\right)$

From (1) and (2) we get

$19.4=21.4e^{-2k}$

$\Rightarrow k=\frac{-1}{2}\times lo{g}_e\frac{19.4}{21.4}$

$\Rightarrow k=.0213\times 2.303$

Putting above value in (1) we get

$21.4=26.6e^{-.0213\times 2.303t}$

$\Rightarrow t=4.43hours=4hours26min.\left(about\right)$

Hence the man died at about $5:34a.m.$