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

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 )`

`(dT)/(T-68)=-kdt`

Integrating

`=\int (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