Question

A detective finds a victim of murder at 9 am. At that time, the temperature of the body is 90.3 degrees F. One hour later, the body temperature has cooled to 89 degrees F. _________ is the temperature of the room in which the body was found maintains a constant temperature of 68 degrees F.

Answer 1

Answer:

Step-by-step explanation:

Given

at 9 am temperature of the body is $90.3^{\circ}F$

1 hr later temperature $89^{\circ}F$

Temperature of room $T=68^{\circ}F$

According to newtons law

$\frac{\mathrm{d} T}{\mathrm{d} t}=-k\left ( T-T_{room}\right )$

$\frac{\mathrm{d} T}{\mathrm{d} t}=-k\left ( T-68\right )$

$\frac{dT}{T-68}=-kdt$

Integrating

$=\int \frac{dT}{T-68}=-kdt$

$\ln |T-68|=-kt+c$

$T-68=Ae^{-kt}$

Now at 9 am i.e. t=0,

$90.3=A+68$

$A=22.3$

At 10 am i.e. t=1 hr T=89

$89-68=22.3e^{-k}$

$e^{-k}=21$

$k=0.06006$

Temperature of Normal human body is $T=98.6 ^{\circ}F$

$98.6-68=e^{-0.06006t}$

$30.6=e^{-0.06006t}$

Taking log both sides

$t=-5.27$

i.e. 5 hr and  16.2 min

i.e. victim murdered at 3 am and 43.8 min