Answer to Question #199545 in Differential Equations for Raj kumar

Question

Suppose the temperature of a body when discovered is 85^o F. Two hours later, the 

temperature is 74^o F and the room temperature is 68^o F. Find the time when the body was discovered after death (assume the body temperature to be 98.6^o F at the time of death.)

Answer 1

Newton's law of cooling states that the temperature of a body changes at a rate proportional to the difference in temperature between the body and its surroundings.

Let $T_0$ be the initial temperature of a body and $T$ be the temperature of the body at time $t$ . If $T_s$ be the temperature of the surroundings, then Newton's law of cooling states that

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

where $k$ is a positive constant.

$\Rightarrow \frac{dT}{ T-T_s}=-kdt$

Integrating both sides,

$\int_{T_0}^T \frac{dT}{ T-T_s}=-k\int _0^tdt$

$\Rightarrow[ ln(T-T_s)]_{T_0}^T=-kt\\ \Rightarrow ln(\frac{T-T_s}{T_0-T_s})=-kt\\ \Rightarrow \frac{T-T_s}{T_0-T_s}=e^{-kt}\ ...............(1)$

In the given problem, initial temperature $T_0=98.6\degree F$ ( temperature of the body at the time of death $t=0)$

and surrounding temperature $T_s=68\degree F$ ( room temperature).

Let at $t=t_1$ the body was discovered and its temperature was $T_1=85\degree F$

From Eq.(1),

$\frac{T_1-T_s}{T_0-T_s}=e^{-kt_1}\ ...............(2)$

After two hours at $t=t_1+2$ the temperature of the body is $T_2=74\degree F$

From Eq.(1),

$\frac{T_2-T_s}{T_0-T_s}=e^{-k(t_1+2)}\ ...............(3)$

Dividing Eq.(2) by Eq.(3),

$\frac{T_1-T_s}{T_2-T_s}=\frac{e^{-kt_1}}{e^{-k(t_1+2)}}\\ \Rightarrow \frac{T_1-T_s}{T_2-T_s}=e^{2k}\\ \Rightarrow k=\frac{1}{2}ln(\frac{T_1-T_s}{T_2-T_s})\\ \Rightarrow k=\frac{1}{2}ln(\frac{85-68}{74-68})\\ \Rightarrow k=0.52$

Substitute the values of $k, T_1, T_s, T_0$ in Eq.(2), we will get $t_1$ .

$\therefore \frac{85-68}{98.6-68}=e^{-0.52t_1}$

On simplification, $t_1=1.13\ hours$

Therefore, the body was discovered 1.13 hours after death.