The reference time is 10:30 AM,
The body temperature is 31 ° C,
The room temperature ([tex]T_A[/tex]) is 20 ° c,
Newton and the law of cooling are useful in this situation, the equation says,
[tex]\frac{dT}{dt} = -k(T-T_A)[/tex]
Where k is proportionality constant
t= time
T= Temperature of Body,
Thus,
[tex]\frac{dT}{dt} = - k(T-20)[/tex]
[tex]\frac{1}{T-20}dT = -kdt[/tex]
We can integrate here,
[tex]\int \frac{1}{T-20}dT = -k\int dt[/tex]
[tex]ln(T-20) = -kt+c[/tex]
[tex]T-20=e^{-kt+c}[/tex]
We have our first equation,
[tex]T=20+e^{-kt}e^c[/tex]
When the measurement t = 0 was made, the temperature T = 31 ° c
[tex]31=20+e^{-k(0)}e^c[/tex]
11=e^c
We have know that [tex]e^c = 11[/tex]
Our equation now is,
[tex]T=20+11e^{-kt}[/tex]
When t=1hour then T=24.8, then
[tex]24.8=20+11e^{-k(1)}[/tex]
[tex]24.8-20=11e^{-k}[/tex]
[tex]\frac{4.8}{11}=e^{-k}[/tex]
[tex]ln(\frac{4.8}{11})=-k[/tex]
[tex]-k=-0.83[/tex]
Then our equation is now,
[tex]T=20+11e^{0.83t}[/tex]
Finally, we know can find the time when T=37, so
[tex]37=20+11e^{-0.83t}[/tex]
[tex]e^{-0.83t}=\frac{17}{11}[/tex]
-0.83t=ln\frac({17}{11})
[tex]t=-0.52[/tex]
[tex]t=31min[/tex]
Therefore the time is 10:30-31min, we have that at 9:59AM the body was still alive
