Question

A thermometer reading $\mathbf{70}\mathbf{°}\mathbf{F}$is placed in an oven preheated to aconstant temperature. Through a glass window in the oven door,an observer records that the thermometer reads$\mathbf{110}\mathbf{°}\mathbf{F}$after$\frac{\mathbf{1}}{\mathbf{2}}$minute and$\mathbf{145}\mathbf{°}\mathbf{F}$after 1minute. How hot is the oven?

Short answer

The temperature of the oven is $T_{ove}={390}^{\mathrm{o}}\mathrm{C}$.

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, $\frac{\mathit{d}\mathit{T}}{\mathit{d}\mathit{t}}\mathbf{=}\mathit{k}\mathbf{(}\mathbf{T}\mathbf{-}{\mathbf{T}}_{\mathbf{m}}\mathbf{)}$where $k$is a constant of proportionality,$T\left(t\right)$ is the temperature of the object for $t>0$, and$T_m$ 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,

$\frac{dT}{dt}=k\left(T-T_{ove}\right)$… (1)

And with the conditions,$T(t=0\mathrm{mins})={70}^{\mathrm{o}}\mathrm{F},T(t=0.5\mathrm{mins})={110}^{\mathrm{o}}\mathrm{F}$ and $T(t=1\mathrm{mins})={145}^{\mathrm{o}}\mathrm{F}$.

For container A:

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

$\begin{array}{rcl}\frac{dT}{T-T_{ove}}& =& kdt\\ \int \frac{1}{T-T_{ove}}dT& =& k\int dt\\ ln\left(T-T_{ove}\right)& =& kt+c_1\\ e^{ln\left(T-T_{owe}\right)}& =& e^{\left(kt+c_1\right)}\\ T-T_{ove}& =& e^{kt}{e}^{c_1}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}T& =& e^{kt}{e}^{c_1}+T_{ove}\\ & =& c{e}^{kt}+T_{ove}\end{array}$… (2)

Obtain the values of constants.

To find the values of constants, apply the point$(T,t)=\left({70}^0\mathrm{F},0\right.\mathrm{minutes})$ in the equation (2), then

$\begin{array}{rcl}70& =& c{e}^0+T_{ove}\\ c& =& 70-T_{ove}\end{array}$

Substitute the value of$c$ in the equation (2).

$T=\left(70-T_{ove}\right)e^{kt}+T_{ove}$… (3)

Again, apply the other point$(T,t)=\left({110}^0\mathrm{F},0.5\right.\mathrm{minutes})$ in the equation (3).

$\begin{array}{rcl}110& =& \left(70-T_{ove}\right)e^{0.5k}+T_{ove}\\ 110-T_{ove}& =& \left(70-T_{ove}\right)e^{0.5k}\end{array}$

$e^{0.5k}=\frac{110-T_{ove}}{70-T_{ove}}$… (a)

Again, apply the other point $(T,t)=\left({145}^{\mathrm{o}}\mathrm{F},1\right.\mathrm{minute})$in the equation (3).

$\begin{array}{rcl}145& =& \left(70-T_{ove}\right)e^k+T_{ove}\\ 145-T_{ove}& =& \left(70-T_{ove}\right)e^{0.5k}\end{array}$

$e^k=\frac{145-T_{ove}}{70-T_{ove}}$… (b)

Obtain the temperature of the oven.

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

$\begin{array}{rcl}{e}^k& =& e^{{\left(k\right)}^2}\\ \frac{145-T_{ove}}{70-T_{ove}}& =& {\left(\frac{110-T_{ove}}{70-T_{ove}}\right)}^2\\ {\left(110-T_{ove}\right)}^2& =& {\left(70-T_{ove}\right)}^2\times \frac{145-T_{ove}}{70-T_{ove}}\\ {\left(110-T_{ove}\right)}^2& =& \left(70-T_{ove}\right)\times \left(145-T_{ove}\right)\end{array}$

$\begin{array}{rcl}12,100-220T_{ove}+T_{ove}^2& =& 10,150-70T_{ove}-145T_{ove}+T_{ove}^2\\ 5T_{ove}& =& 1,950\\ T_{ove}& =& \frac{1,950}{5}\\ T_{ove}& =& {390}^{\mathrm{o}}\mathrm{C}\end{array}$