Question

A small metal bar, whose initial temperature was $\mathbf{20}\mathbf{°}\mathbf{C}$, is dropped into a large container of boiling water. How long will it take the bar to reach$\mathbf{90}\mathbf{°}\mathbf{C}$if it is known that its temperature increasesrole="math" localid="1663918838903" $\mathbf{2}\mathbf{°}$in 1second? How long will it take the bar to reach$\mathbf{98}\mathbf{°}\mathbf{C}$?

Short answer

The time to take the bar to reach$T={90}^o$ is 82.1 seconds, and to reach $T={98}^o$is 145.7.

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,

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

And with the conditions,$T(t=0\mathrm{seconds})={20}^{\mathrm{o}}\mathrm{F}$and $T(t=1\mathrm{seconds})={22}^{\mathrm{o}}\mathrm{F}$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$\begin{array}{rcl}\frac{dT}{T-T_{out}}& =& kdt\\ \int \frac{1}{T-{100}^{\mathrm{o}}\mathrm{F}}dT& =& k\int dt\\ ln\left(T-{100}^{\mathrm{o}}\mathrm{F}\right)& =& kt+c_1\\ e^{ln\left(T-{100}^{\mathrm{o}}\mathrm{F}\right)}& =& e^{\left(kt+c_1\right)}\\ T-{100}^{\mathrm{o}}\mathrm{F}& =& e^{kt}{e}^{c_1}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}T& =& e^{kt}{e}^{c_1}+100\\ & =& c{e}^{kt}+100\end{array}$… (2)

Obtain the values of constants.

To find the values of constants, apply the point$(T,t)=\left({20}^0\mathrm{C},1\right.\mathrm{seconds})$in the equation (2), then

$\begin{array}{rcl}{20}^0& =& c{e}^0+{100}^0\\ c{e}^0& =& 20-100\\ c& =& -80\end{array}$

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

$T=-80e^{kt}+100$… (3)

Again, apply the other point$(T,t)=\left({22}^0\mathrm{C},1\right.\mathrm{seconds})$in the equation (3).

role="math" localid="1663919366690" $\begin{array}{rcl}22& =& -80e^k+{100}^0\\ 22-100& =& -80e^k\\ -78& =& -80e^k\\ \frac{-78}{-80}& =& e^k\\ 0.975& =& e^k\\ ln(0.975)& =& k\\ k& =& -0.0253\end{array}$

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

$T=-80e^{-0.0253t}+{100}^{\mathrm{o}}\mathrm{C}$… (4)

Obtain the initial temperature of the inside room.

Substitute the value$T={90}^o$ minuteinto the equation (4).

$\begin{array}{rcl}{90}^{\mathrm{o}}& =& -80e^{-0.0253t}+{100}^{\mathrm{o}}\\ 90-100& =& -80e^{-0.0253t}\\ -10& =& -80e^{-0.0253t}\\ \frac{10}{80}& =& e^{-0.0253t}\\ ln(0.125)& =& -0.0253t\\ t& =& \frac{ln(0.125)}{-0.0253}\\ & =& \frac{-2.08}{-0.0253}\\ & \approx & 82.1\mathrm{seconds}\end{array}$

Substitute the value $T={98}^o$minute into the equation (4).

$\begin{array}{rcl}{98}^{\mathrm{o}}& =& -80e^{-0.0253t}+{100}^{\mathrm{o}}\\ 98-100& =& -80e^{-0.0253t}\\ -2& =& -80e^{-0.0253t}\\ \frac{2}{80}& =& e^{-0.0253t}\\ ln(0.025)& =& -0.0253t\\ t& =& \frac{ln(0.025)}{-0.0253}\\ & =& \frac{-3.689}{-0.0253}\\ & \approx & 145.7\mathrm{seconds}\\ & & \end{array}$