Question

A thermometer is taken from an inside room to the outside,where the air temperature is $\mathbf{5}\mathbf{°}\mathbf{F}$. After 1minute the thermometer reads$\mathbf{55}\mathbf{°}\mathbf{F}$, and after 5minutes it reads$\mathbf{30}\mathbf{°}\mathbf{F}$.What is the initial temperature of the inside room?

Short answer

The initial temperature of the inside room is $T\approx 64.5^0\mathrm{F}$.

Step-by-step solution

DefineNewton’s law of cooling/heating.

The mathematical formulation of Newton’s empirical law of cooling/warming of an object is given by the linear first-order differential equation, $\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathit{k}\mathbf{(}\mathbf{T}\mathbf{-}{\mathbf{T}}_{\mathbf{m}}\mathbf{)}$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 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,$T(t=1\mathrm{mins})={55}^{\mathrm{o}}\mathrm{F}$and $T(t=5\mathrm{mins})={30}^{\mathrm{o}}\mathrm{F}$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$\begin{array}{rcl}\frac{dT}{T-T_{out}}& =& kdt\\ \int \frac{1}{T-5^{\mathrm{o}}\mathrm{F}}dT& =& k\int dt\\ ln\left(T-5^{\mathrm{o}}\mathrm{F}\right)& =& kt+c_1\\ e^{ln\left(T-5^{\mathrm{o}}\mathrm{F}\right)}& =& e^{\left(kt+c_1\right)}\\ T-5^{\mathrm{o}}\mathrm{F}& =& e^{kt}{e}^{c_1}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}T& =& e^{kt}{e}^{c_1}+5\\ & =& c{e}^{kt}+5\end{array}$… (2)

Obtain the values of constants.

To find the values of constants, apply the point$(T,t)=\left({55}^{\mathrm{o}}\mathrm{F},1\right.\mathrm{minutes})$in the equation (2), then

$\begin{array}{rcl}55& =& c{e}^k+5\\ c{e}^k& =& 50\\ c& =& \frac{50}{e^k}\end{array}$

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

role="math" localid="1663918067854" $\begin{array}{rcl}T& =& \frac{50}{e^k}{e}^{kt}+5^{\mathrm{o}}\\ & =& 50e^{kt}{e}^{-k}+5^{\mathrm{o}}\\ & =& 50e^{kt-k}+5^o\end{array}$

$T=50e^{k(t-1)}+5$… (3)

Again, apply the other point$(T,t)=\left({30}^{\mathrm{o}}\mathrm{F},5\right.\mathrm{minutes})$in the equation (3).

$\begin{array}{rcl}30& =& 50e^{k(t-1)}+5\\ 30& =& 50e^{k(5-1)}+5\\ 30-5& =& 50e^{4k}\\ 25& =& 50e^{5k}\\ 0.5& =& e^{4k}\\ ln(0.5)& =& 4k\\ k& =& \frac{ln(0.5)}{4}\\ & =& \frac{-0.693}{5}\\ & =& -0.173\end{array}$

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

$T=50e^{-0.173(t-1)}+5$

Obtain the initial temperature of the inside room.

Substitute the value $t=0$minute into the equation (4).

$\begin{array}{rcl}T& =& 50e^{-0.173(0-1)}+5^{\mathrm{o}}\mathrm{F}\\ & =& 50e^{0.173}+5^{\mathrm{o}}\mathrm{F}\\ & =& 50\times 1.189+5^{\mathrm{o}}\mathrm{F}\\ & =& 59.5^{\mathrm{o}}\mathrm{F}+5^{\mathrm{o}}\mathrm{F}\\ T& \approx & 64.5^{\mathrm{o}}\mathrm{F}\end{array}$