Question

Two large containers A and B of the same size are filled with different fluids. The fluids in containers A and B are maintained at $\mathbf{0}\mathbf{°}\mathbf{C}$and$\mathbf{100}\mathbf{°}\mathbf{C}$, respectively. A small metal bar, whose initial temperature is$\mathbf{100}\mathbf{°}\mathbf{C}$, is lowered into container A. After 1minute the temperature of the bar is$\mathbf{90}\mathbf{°}\mathbf{C}$. After 2minutes the bar is removed and instantly transferred to the other container. After 1minute in container B the temperature of the bar rises$\mathbf{10}\mathbf{°}$. How long, measured from the start of the entire process, will it take the bar to reach$\mathbf{99}\mathbf{.}\mathbf{9}\mathbf{°}\mathbf{C}$?

Short answer

The total time at which the metal bar reaches$T=99°$ from its initial condition is$9.02$ minutes.

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, $\frac{\mathbf{d}\mathbf{T}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{k}\left(T-T_m\right)$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 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,$T=T_0$ for container A $T(t=0\mathrm{minutes})={100}^0\mathrm{F}$, $T(t=1\mathrm{minutes})={90}^{\mathrm{o}}\mathrm{F}$and$T=T_0$ for container B atrole="math" localid="1663920982907" $(\mathrm{t}=2\mathrm{minutes}\mathrm{in}\mathrm{container}A)$ and$T=\left(T_0+10\right)$ for container B at role="math" localid="1663920992736" $(\mathrm{t}=1\mathrm{minutes}\mathrm{in}\mathrm{container}A)$.

For container A:

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

$\begin{array}{rcl}\frac{dT}{dt}& =& k_A(T-0)\\ \frac{dT}{T}& =& k_{A}dt\\ \int \frac{1}{T}dT& =& k_A\int dt\\ ln\left(T\right)& =& k_{A}t+c_1\\ e^{ln\left(T\right)}& =& e^{\left(k_{1}t+c_1\right)}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}T& =& e^{k_{A}t}{e}^{c_1}\\ & =& c{e}^{k_{A}t}\end{array}$… (2)

Obtain the values of constants.

To find the values of constants, apply the point$(T,t)=\left({100}^{\mathrm{o}}\mathrm{F},0\mathrm{minutes}\right.)$ in the equation (2), then

$\begin{array}{rcl}{100}^{\mathrm{o}}& =& c{e}^{\mathrm{o}}\\ c& =& 100\end{array}$

Substitute the value of in the equation (2).

$T=100e^{k_{A}t}$… (3)

Again, apply the other point $(T,t)=\left({90}^0\mathrm{F},1\mathrm{minutes}\right.)$in the equation (3).

$\begin{array}{rcl}{90}^{*}& =& 100e^{k_A\left(1\right)}\\ e^{k_A}& =& \frac{90}{100}\\ k_A& =& ln(0.9)\\ & =& -0.105\end{array}$

Substitute the value of in the equation (3).

$T=100e^{-0.105t}$… (4)

Solve for first order equation.

For container B:

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

$\begin{array}{rcl}\frac{dT}{dt}& =& k_B(T-100)\\ \frac{dT}{T-100}& =& k_{B}dt\\ \int \frac{1}{T-100}dT& =& k_B\int dt\\ ln(T-100)& =& k_{B}t+h_1\\ e^{ln(T-100)}& =& e^{\left(k_{B}t+h_1\right)}\\ T-100& =& e^{\left(k_{B}t+h_1\right)}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}T& =& e^{k_{B}t}{e}^{h_1}+{100}^o\mathrm{C}\\ & =& h{e}^{k_{B}t}+{100}^o\mathrm{C}\end{array}$… (5)

Obtain the values of constants.

To find the values of constants, apply the point$(T,t)=\left({81}^{\mathrm{o}}\mathrm{F},0\mathrm{minutes}\right.)$ in the equation (5), then

$\begin{array}{rcl}{81}^{\mathrm{o}}& =& h{e}^{\mathrm{o}}+100\\ h& =& 81-100\\ & =& -19\end{array}$

Substitute the value of in the equation (5).

$T=-19e^{kBt}+{100}^0$… (6)

Again, apply the other point$(T,t)=\left({(81+10)}^{\mathrm{o}}\mathrm{F},1\right.\mathrm{minutes})$ in the equation (6).

$\begin{array}{rcl}{91}^{\mathrm{o}}& =& -19e^{k_B\left(1\right)}+{100}^{\mathrm{o}}\\ 91-100& =& -19e^{k_B}\\ -9& =& -19e^{k_B}\\ e^{k_B}& =& \frac{9}{19}\\ k_B& =& ln\left(\frac{9}{19}\right)\\ & =& -0.747\end{array}$

Substitute the value of in the equation (6).

$T=-19e^{-0.747t}+100$… (7)

Obtain the initial temperature.

Substitute the valueinto the equation (7).

$\begin{array}{rcl}99.9^{\mathrm{o}}& =& -19e^{-0.747t}+{100}^{\mathrm{o}}\\ 99.9-100& =& -19e^{-0.747t}\\ -0.1& =& -19e^{-0.747t}\\ e^{-0.747t}& =& \frac{1}{190}\\ -0.747t& =& ln\left(\frac{1}{190}\right)\\ & =& \frac{ln\left(\frac{1}{190}\right)}{-0.747}\\ & =& \frac{-5.247}{-0.747}\\ & =& 7.02\mathrm{minutes}\end{array}$

The total time at which the metal bar reaches from its initial condition is,

$\begin{array}{rcl}\mathrm{Total}\mathrm{time}& =& \mathrm{Time}\mathrm{in}\mathrm{container}\mathrm{A}+\mathrm{Time}\mathrm{in}\mathrm{container}\mathrm{B}\\ & =& 2\mathrm{minutes}+7.02\mathrm{minutes}\\ & =& 9.02\mathrm{minutes}\end{array}$