Question

For a post-mortem report, a doctor requires to know the approximate time of death of the deceased. He records the first temperature at 10.00 a.m. to be 93.4 F. After 2 hours he finds the temperature to be 91.4 F. If the room temperature (which is constant) is 72 F, estimate the time of the death. Assume the temperature of the human body to be 98.6 F.

Answer 1

Hint: Here the first set of data is recorded and then after some time another set is recorded. It is observed that as time increases the temperature decreases. This problem can be solved by using Newton’s law of cooling. We are given two conditions in which a body cools down from a certain temperature to another We need to determine the time of the death.

Complete step by step solution:
From the law: \[T(t)=T(s)+\{{{T}_{0}}-T(s)\}{{e}^{-kt}}\]
Where \[T(t)\] is the temperature at time, t. \[T(s)\] is the temperature of the surrounding, \[{{T}_{0}}\] is the initial temperature of the body.
Assuming the man had died before 10 am, \[93.4=72+\{98.6-72\}{{e}^{-kt}}\]
$ 93.4=72+\{98.6-72\}{{e}^{-kt}} \\
\Rightarrow 93.4-72=26.6{{e}^{-kt}} \\
\Rightarrow 21.4=26.6{{e}^{-kt}} \\
\Rightarrow {{e}^{-kt}}=0.8 \\
\Rightarrow -kt=\ln (0.8) \\
\Rightarrow -kt=-0.22 \\
\Rightarrow kt=0.22 \\$
So, \[kt=0.22\]--(1)
After 2 hours, \[91.4=72+\{98.6-72\}{{e}^{-k(t+2)}}\]
$\Rightarrow 91.4=72+\{98.6-72\}{{e}^{-k(t+2)}} \\
\Rightarrow 19.4=26.6{{e}^{-k(t+2)}} \\
\Rightarrow {{e}^{-k(t+2)}}=0.72 \\
\Rightarrow -k(t+2)=\ln (0.72) \\
\Rightarrow -k(t+2)=-0.32 \\
\Rightarrow kt+2k=0.32 $
From (1),
$\Rightarrow 0.22+2k=0.32 \\
\Rightarrow 2k=0.10 \\
\therefore k=0.05 $
Putting the value of k in (1), we get t=4.4 h

Thus, the man died 4h 24 mins earlier that is at 5:36 am.

Note: Newton’s law of cooling tells us the rate at which a body exposed to atmosphere changes temperature through by loss of heat to the surrounding which is approximately proportional to the difference between the object’s temperature and its surroundings, provided the difference is small. As per this law rate of cooling is dependent upon the temperature, suppose we want a hot cup of tea to cool down, a cup of hot coffee will cool more quickly if we put it in the refrigerator.