Question

A thermometer is removed from a room where the temperature is $\mathbf{70}\mathbf{°}\mathbf{F}$and is taken outside, where the air temperature is $\mathbf{10}\mathbf{°}\mathbf{F}$. After one-half minute the thermometer reads $\mathbf{50}\mathbf{°}\mathbf{F}$. What is the reading of the thermometer at $\mathit{t}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{min}$? How long will it takefor the thermometer to reach$\mathbf{15}\mathbf{°}\mathbf{F}$?

Short answer

The reading of the thermometer at $t=1$is $T\approx 36.7°\mathrm{F}$, and the time to reach the thermometer $T={15}^{\mathrm{o}}\mathrm{F}$ is $t=3.064\mathrm{minutes}$.

Step-by-step solution

Step 1:Define growth and decay.

The mathematical formulation of Newton’s empirical law of cooling/warming ofan object is given by the linear first-order differential equation, $\frac{\mathbf{d}\mathbf{T}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{k}\left(T-T_m\right)$ 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 temperature that is, the temperature of the medium around the object.

Solve for first order growth and decay equation.

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

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

And with the conditions,role="math" localid="1663915760669" $T(t=0\mathrm{mins})={70}^{\mathrm{o}}\mathrm{F}$and $T(t=0.5\mathrm{mins})={50}^{\mathrm{o}}\mathrm{F}$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$\begin{array}{rcl}\frac{dT}{T-T_{out}}& =& kdt\\ \int \frac{1}{T-{10}^{\mathrm{o}}}dT& =& k\int dt\\ ln\left(T-{10}^{\mathrm{o}}\right)& =& kt+c_1\\ e^{ln\left(T-{10}^{\mathrm{o}}\right)}& =& e^{\left(kt+c_1\right)}\\ T-{10}^{\mathrm{o}}& =& e^{kt}{e}^{c_1}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}T& =& e^{kt}{e}^{c_1}+10\\ & =& c{e}^{kt}+10\end{array}$… (2)

Obtain the values of constants.

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

$\begin{array}{rcl}70& =& c{e}^0+10\\ c& =& 70-10\\ & =& 60\end{array}$

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

$T=60e^{kt}+10$… (3)

Again, apply the other point$(T,t)=\left({50}^{\mathrm{o}}\mathrm{F},0.5\right.\mathrm{minutes})$in the equation (3).

$\begin{array}{rcl}50& =& 60e^{0.5k}+10\\ 50-10& =& 60e^{0.5k}\\ 40& =& 60e^{0.5k}\\ 0.667& =& e^{0.5k}\\ ln(0.667)& =& 0.5k\\ k& =& \frac{ln(0.667)}{0.5}\\ & =& \frac{-0.405}{0.5}\\ k& =& -0.811\end{array}$

Substitute the value of$k$ in the equation (3).

$T=60e^{-0.811t}+{10}^{\mathrm{o}}$… (4)

Obtain the reading of the thermometer at the given time.

Substitute the value $t=1$minute into the equation (4).

$\begin{array}{rcl}T& =& 60e^{-0.811\times 1}+{10}^{\mathrm{o}}\mathrm{F}\\ & =& 60e^{-0.811}+{10}^{\mathrm{o}}\mathrm{F}\\ & =& 60\times \frac{4}{9}+{10}^{\mathrm{o}}\mathrm{F}\\ & =& \left(\frac{80}{3}\right)F+{10}^{\mathrm{o}}\mathrm{F}\\ T& \approx & 36.7^{\mathrm{o}}\mathrm{F}\end{array}$

Obtain the time to reach the given thermometer.

Substitute the valueminuteinto the equation (4).

$T={15}^{\mathrm{o}}\mathrm{F}$

role="math" localid="1663917307184" $\begin{array}{rcl}{15}^{\mathrm{o}}\mathrm{F}& =& 60e^{-0.811t}+10\\ 15-10& =& 60e^{-0.811t}\\ 5& =& 60e^{-0.811t}\\ 0.8333& =& e^{-0.811t}\\ ln(0.833)& =& -0.811t\\ t& =& \frac{ln(0.833)}{-0.811}\\ & =& \frac{-2.485}{-0.811}\\ & =& 3.064\mathrm{minutes}\end{array}$