# Applications of First Order Differential Equations: A dead body was found within a closed room of a

Carla Murphy 2022-01-15 Answered
Applications of First Order Differential Equations: A dead body was found within a closed room of a house where the temperature was a constant ${24}^{\circ }C$. At the time of discovery, the core temperature of the body was determined to be ${28}^{\circ }C$. One hour later, a second measurement showed that the core temperature of the body was ${26}^{\circ }C$. The core temperature of the body at the time of death is ${37}^{\circ }C$. A. Determine how many hours elapsed before the body was found. B. What is the temperature of the body 5 hours from the death of the victim?
You can still ask an expert for help

## Want to know more about Laplace transform?

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Bernard Lacey
GaceCoect5v
${T}^{\prime }=k\left(T-{T}_{m}\right)$ (Newton's Law of Cooling)
$\int \frac{dT}{T-{T}_{m}}=\int kdt$
$\mathrm{ln}|T-{T}_{m}|=kt+{C}_{1}\to T\left(t\right)={T}_{m}+c{e}^{kt}$
$T\left(\text{closed room}\right)=21\to {T}_{m}=21$

Substitute in $T\left(t\right)={T}_{m}+c{e}^{kt}$
$37=21+c{e}^{0}\to c=16$
$T=21+16{e}^{kt}$ (1)
Where t is the time at discovery
The temperature was ${27}^{\circ }$ at the time of discovery
$27=21+16{e}^{kt}\to 16{e}^{kt}=6$ (2)
The temperature was ${25}^{\circ }$ after one hour $\left(t+1\right)$
$25=21+16{e}^{k\left(t+1\right)}\to 16{e}^{k\left(t+1\right)}=4$ (3)
Dividing (2) by (3)
$\frac{{e}^{kt}}{{e}^{k\left(t+1\right)}}=\frac{6}{4}=\frac{3}{2}$
${e}^{kt-kt-k}=\frac{3}{2}\to {e}^{-k}=\frac{3}{2}\to -k=\mathrm{ln}\frac{3}{2}\to k=-0.41$
Substitute in (2)
$16{e}^{-0.41t}=6\to {e}^{-0.41t}=\frac{6}{16}=\frac{3}{8}$
$-0.41t=\mathrm{ln}\frac{3}{8}$
$t=\mathrm{ln}\frac{3}{8}÷\left(-0.41\right)$
$t=2.4$ hours
alenahelenash