Question

At 10:30 AM, detectives discover a dead body in a room and measure its temperature at 31∘C. One hour later, the body’s temperature had dropped to 24.8∘C. Determine how many hours had passed since the death occured (when the body temperature was a normal 37∘C), by the time the body was discovered, assuming that the temperature in the room was held constant at 20∘C. (Use decimal notation. Give your answer to one decimal place. Round any intermediate calculations, if needed, to no less than three decimal places.)

Answer 1

The reference time is 10:30 AM,

The body temperature is 31 ° C,

The room temperature ($T_A$) is 20 ° c,

Newton and the law of cooling are useful in this situation, the equation says,

$\frac{dT}{dt} = -k(T-T_A)$

Where k is proportionality constant

t= time

T= Temperature of Body,

Thus,

$\frac{dT}{dt} = - k(T-20)$

$\frac{1}{T-20}dT = -kdt$

We can integrate here,

$\int \frac{1}{T-20}dT = -k\int dt$

$ln(T-20) = -kt+c$

$T-20=e^{-kt+c}$

We have our first equation,

$T=20+e^{-kt}e^c$

When the measurement t = 0 was made, the temperature T = 31 ° c

$31=20+e^{-k(0)}e^c$

11=e^c

We have know that $e^c = 11$

Our equation now is,

$T=20+11e^{-kt}$

When t=1hour then T=24.8, then

$24.8=20+11e^{-k(1)}$

$24.8-20=11e^{-k}$

$\frac{4.8}{11}=e^{-k}$

$ln(\frac{4.8}{11})=-k$

$-k=-0.83$

Then our equation is now,

$T=20+11e^{0.83t}$

Finally, we know can find the time when T=37, so

$37=20+11e^{-0.83t}$

$e^{-0.83t}=\frac{17}{11}$

-0.83t=ln\frac({17}{11})

$t=-0.52$

$t=31min$

Therefore the time is 10:30-31min, we have that at 9:59AM the body was still alive