# Suppose the temperature of a body when discovered is 85°F.Two hours laters, the temperature is 74°F and room temperature is 68°F.Find the time when the body was discovered after death(assume the body temperature to be 98.6°F at time of death.)?

Jul 4, 2018

$\approx 1 \text{ hour " 8 " mins}$

#### Explanation:

Using Newton's Law of cooling :

• $\dot{T} = - k \left(T - {T}_{a}\right) \setminus \square q \quad \text{where " T_a = 68 " is ambient}$

There are 3 conditions:

• $\left\{\begin{matrix}T \left(0\right) = {T}_{o} = 98.6 \\ T \left(\tau\right) = 85 q \quad \leftarrow \text{ body discovered } \\ T \left(\tau + 2\right) = 74\end{matrix}\right.$

$\square$ separates for integration:

$\frac{\mathrm{dT}}{T - {T}_{a}} = - k \setminus \mathrm{dt}$

${\int}_{{T}_{o}}^{T \left(t\right)} \frac{\mathrm{dT}}{T - {T}_{a}} = - k t$

Because $T > {T}_{o}$:

${\left[\ln \left(T - {T}_{a}\right)\right]}_{{T}_{o}}^{T \left(t\right)} = - k t$

$\frac{T - {T}_{a}}{{T}_{o} - {T}_{a}} = {e}^{- k t}$

$\therefore \boldsymbol{T = {T}_{a} + \left({T}_{o} - {T}_{a}\right) {e}^{- k t}}$

The 3 conditions translate as:

• $\left\{\begin{matrix}{T}_{o} = 98.6 \\ 85 = 68 + \left(98.6 - 68\right) {e}^{- k \tau} q \quad q \quad q \quad \mathbb{\left(A\right)} \\ 74 = 68 + \left(98.6 - 68\right) {e}^{- k \left(\tau + 2\right)} q \quad \mathbb{\left(B\right)}\end{matrix}\right.$

$\frac{\mathbb{\left(B\right)}}{\mathbb{\left(A\right)}} \implies {e}^{- 2 k} = \frac{74 - 68}{85 - 68}$

$\implies k = \frac{1}{2} \ln \left(\frac{17}{6}\right)$

From $\mathbb{\left(A\right)}$:

$- k \tau = \ln \left(\frac{85 - 68}{98.6 - 68}\right)$

$\tau = \frac{\ln \left(\frac{98.6 - 68}{85 - 68}\right)}{\frac{1}{2} \ln \left(\frac{17}{6}\right)} \approx 1.129 \text{ hrs}$