Question

A dead body was found within a closed room of a house where the temperature was a constant ${\mathbf{70}}^{\mathbf{o}}\mathbf{F}$. At the time of discovery the core temperature of the body was determined to be${\mathbf{85}}^{\mathbf{o}}\mathbf{F}$. One hour later a second measurement showed that the core temperature of the body was${\mathbf{80}}^{\mathbf{o}}\mathbf{F}$. Assume that the time of death corresponds to$\mathbf{t}\mathbf{=}\mathbf{0}$and that the core temperature at that time was$\mathbf{98}\mathbf{.}{\mathbf{6}}^{\mathbf{o}}\mathbf{F}$.Determine how many hours elapsed before the body was found.[Hint: Let${\mathbf{t}}_{\mathbf{1}}\mathbf{>}\mathbf{0}$denote the time that the body was discovered.]

Short answer

The time elapsed before the body was found is ${\mathrm{t}}_1=1.59\mathrm{hours}$.

Step-by-step solution

Define Newton’s law of cooling/heating.

The mathematical formulation of Newton’s empirical law of cooling/warming ofan object is given by the linear first-order differential equation, $\frac{\mathbf{d}\mathbf{T}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{k}\mathbf{(}\mathbf{T}\mathbf{-}{\mathbf{T}}_{\mathbf{m}}\mathbf{)}$ where $\mathrm{k}$is a constant of proportionality, $\mathrm{T}\left(\mathrm{t}\right)$is the temperature of the object for $\mathrm{t}>0$,and ${\mathrm{T}}_{\mathrm{m}}$is the ambient temperaturethat is, the temperature of the medium aroundthe object.

Solve for first order equation.

Let the difference between the thermometer temperature and outside temperature be,

$\frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{k}\left(\mathrm{T}-{\mathrm{T}}_{\mathrm{m}}\right)$… (1)

And with the conditions,$\mathrm{T}(\mathrm{t}=0\mathrm{hours})=98.6^{\mathrm{o}}\mathrm{F},\mathrm{T}\left(\mathrm{t}={\mathrm{t}}_1\mathrm{hours}\right)={85}^{\mathrm{o}}\mathrm{F}$ and$\mathrm{T}\left(\mathrm{t}={\mathrm{t}}_1+1\mathrm{hours}\right)={80}^{\mathrm{o}}\mathrm{F}$.

As the equation (1) is linear and separable, so integrate the equation and separate the variables.

role="math" localid="1664093602525" $\begin{array}{rcl}\frac{\mathrm{dT}}{\mathrm{T}-{\mathrm{T}}_{\mathrm{m}}}& =& \mathrm{kdt}\\ \int \frac{1}{\mathrm{T}-{70}^{\mathrm{o}}}\mathrm{dT}& =& \mathrm{k}\int \mathrm{dt}\\ \ln \left(\mathrm{T}-{70}^{\mathrm{o}}\right)& =& \mathrm{kt}+{\mathrm{c}}_1\\ {\mathrm{e}}^{\ln \left(\mathrm{T}-{70}^{\mathrm{o}}\right)}& =& {\mathrm{e}}^{\left(\mathrm{kt}+{\mathrm{c}}_1\right)}\\ \mathrm{T}-{70}^{\mathrm{o}}& =& {\mathrm{e}}^{\mathrm{kt}}{\mathrm{e}}^{{\mathrm{c}}_1}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}\mathrm{T}& =& {\mathrm{e}}^{\mathrm{kt}}{\mathrm{e}}^{{\mathrm{c}}_1}+70\\ & =& {\mathrm{ce}}^{\mathrm{kt}}+70\end{array}$… (2)

Obtain the values of constants.

To find the values of constants, apply the point $\mathrm{t}(\mathrm{T},\mathrm{t})=(98.6\mathrm{F},0\mathrm{hours})$in the equation (2), then

$\begin{array}{rcl}98.6^{\mathrm{o}}& =& {\mathrm{ce}}^0+{70}^{\mathrm{o}}\\ {\mathrm{ce}}^0& =& 98.6-70\\ \mathrm{c}& =& 28.6\end{array}$

Substitute the value of in the equation (2).

role="math" localid="1664093730286" $\mathrm{T}=28.6{\mathrm{e}}^{\mathrm{kt}}+{70}^{\mathrm{o}}$… (3)

Again, apply the other point$(\mathrm{T},\mathrm{t})=\left({85}^{\mathrm{o}}\mathrm{F},{\mathrm{t}}_1\right.\mathrm{hours})$in the equation (3).

$\begin{array}{rcl}{85}^{\mathrm{o}}& =& 28.6{\mathrm{e}}^{{\mathrm{kt}}_1}+{70}^{\mathrm{o}}\\ 85-70& =& 28.6{\mathrm{e}}^{{\mathrm{kt}}_1}\\ 15& =& 28.6{\mathrm{e}}^{{\mathrm{kt}}_1}\end{array}$

${\mathrm{e}}^{{\mathrm{kt}}_1}=\frac{15}{28.6}$

${\mathrm{kt}}_1=\ln \left(\frac{15}{28.6}\right)$… (a)

Again, apply the other point $(\mathrm{T},\mathrm{t})=\left({80}^{\mathrm{o}}\mathrm{F},{\mathrm{t}}_1+1\right.\mathrm{hours})$in the equation (3).

$\begin{array}{rcl}{80}^{\mathrm{o}}& =& 28.6{\mathrm{e}}^{\mathrm{k}\left({\mathrm{t}}_1+1\right)}+{70}^{\mathrm{o}}\\ 80-70& =& 28.6{\mathrm{e}}^{\mathrm{k}\left({\mathrm{t}}_1+1\right)}\\ 10& =& 28.6{\mathrm{e}}^{{\mathrm{kt}}_1}\\ {\mathrm{e}}^{\mathrm{k}\left({\mathrm{t}}_1+1\right)}& =& \frac{10}{28.6}\end{array}$

$\mathrm{k}\left({\mathrm{t}}_1+1\right)=\ln \left(\frac{15}{28.6}\right)$… (b)

Obtain the temperature of the oven.

By comparing the equation (a) and (b), the temperature of the oven is,

$\begin{array}{rcl}\mathrm{k}\left({\mathrm{t}}_1+1\right)-{\mathrm{kt}}_1& =& \ln \left(\frac{10}{28.6}\right)-\ln \left(\frac{15}{28.6}\right)\\ {\mathrm{kt}}_1+\mathrm{k}-{\mathrm{kt}}_1& =& \ln \left(\frac{10}{28.6}\right)-\ln \left(\frac{15}{28.6}\right)\\ \mathrm{k}& =& \ln \left(\frac{10}{28.6}\right)-\ln \left(\frac{15}{28.6}\right)\end{array}$

$\begin{array}{rcl}\mathrm{k}& =& -1.0508-(-0.64535)\\ & =& -1.0508+0.64535\\ \mathrm{k}& =& -0.405\end{array}$

Substitute the value$\mathrm{k}=-0.405$ in the equation (a).

$\begin{array}{rcl}-0.405{\mathrm{t}}_1& =& \ln \left(\frac{15}{28.6}\right)\\ {\mathrm{t}}_1& =& \frac{\ln \left(\frac{15}{28.6}\right)}{-0.405}\\ {\mathrm{t}}_1& =& 1.59\mathrm{hours}\end{array}$