Question

The rate at which a body cools also depends on its exposed surface area S. If Sis a constant, then a modification of (2) is$\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{kS}\mathbf{(}\mathbf{T}\mathbf{-}{\mathbf{T}}_{\mathbf{m}}\mathbf{)}$,where$\mathbf{k}\mathbf{<}\mathbf{0}$and${\mathbf{T}}_{\mathbf{m}}$is a constant. Suppose that two cups A and B are filled with coffee at the same time. Initially, the temperature of the coffee is${\mathbf{150}}^{\mathbf{o}}\mathbf{F}$. The exposed surface area of the coffee in cup B is twice the surface area of the coffee in cup A. After 30min the temperature of the coffee in cup A is${\mathbf{100}}^{\mathbf{o}}\mathbf{F}$. If${\mathbf{T}}_{\mathbf{m}}\mathbf{=}{\mathbf{70}}^{\mathbf{o}}\mathbf{F}$, then what is the temperature of the coffee in cup B after 30 min?

Short answer

The temperature of the coffee in cup B is ${\mathrm{T}}_{\mathrm{B}}=81.{25}^{\mathrm{o}}$.

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, $\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{k}\mathbf{(}\mathbf{T}\mathbf{-}{\mathbf{T}}_{\mathbf{m}}\mathbf{)}$where $\mathrm{k}$is a constant of proportionality, $\mathrm{T}\left(\mathrm{t}\right)$is the temperature of the object for $\mathrm{t}>0$,and ${\mathrm{T}}_{\mathrm{m}}$is the ambient temperature that is, the temperature of the medium aroundthe object.

Solve for first order equation.

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

$\frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{kS}\left(\mathrm{T}-{\mathrm{T}}_{\mathrm{m}}\right)$… (1)

And with the conditions,$\mathrm{T}={\mathrm{T}}_{\mathrm{A}}={\mathrm{T}}_{\mathrm{B}}(\mathrm{t}=0\mathrm{minutes})={150}^{\mathrm{o}}\mathrm{F},{\mathrm{T}}_{\mathrm{A}}(\mathrm{t}=30\mathrm{minutes})={100}^{\mathrm{o}}\mathrm{F}$and${\mathrm{S}}_{\mathrm{B}}=2{\mathrm{S}}_{\mathrm{A}}$.

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

localid="1664096827788" $\begin{array}{rcl}\frac{\mathrm{dT}}{\mathrm{T}-{\mathrm{T}}_{\mathrm{m}}}& =& \mathrm{kSdt}\\ \int \frac{1}{\mathrm{T}-{70}^{\mathrm{o}}}\mathrm{dT}& =& \mathrm{kS}\int \mathrm{dt}\\ \ln \left(\mathrm{T}-{70}^{\mathrm{o}}\right)& =& \mathrm{kSt}+{\mathrm{c}}_1\\ {\mathrm{e}}^{\ln \left(\mathrm{T}-{70}^{\mathrm{o}}\right)}& =& {\mathrm{e}}^{\left(\mathrm{kSt}+{\mathrm{c}}_1\right)}\\ \mathrm{T}-{70}^{\mathrm{o}}& =& {\mathrm{e}}^{\mathrm{kSt}}{\mathrm{e}}^{{\mathrm{c}}_1}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}\mathrm{T}& =& {\mathrm{e}}^{\mathrm{kSt}}{\mathrm{e}}^{\mathrm{c}1}+70\\ & =& {\mathrm{ce}}^{\mathrm{kSt}}+{70}^{\mathrm{o}}\end{array}$… (2)

Obtain the values of constants.

To find the values of constants, apply the point$(\mathrm{T},\mathrm{t})=\left({150}^{\mathrm{o}}\mathrm{F},0\right.\mathrm{minutes})$in the equation (2), then

$\begin{array}{rcl}{150}^{\mathrm{o}}& =& c{e}^0+{70}^{\mathrm{o}}\\ c{e}^0& =& 150-70\\ c& =& 80\end{array}$

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

$\mathrm{T}=80{\mathrm{e}}^{\mathrm{kSt}}+70$… (3)

Again, apply the other point$\left({\mathrm{T}}_{\mathrm{A}},\mathrm{t}\right)=\left({100}^{\mathrm{o}}\mathrm{F},30\right.\mathrm{minutes})$in the equation (3).

$\begin{array}{rcl}100& =& 80{\mathrm{e}}^{{\mathrm{kS}}_{\mathrm{A}}\left(30\right)}+70\\ 100-70& =& 80{\mathrm{e}}^{30{\mathrm{kS}}_{\mathrm{A}}}\\ 30& =& 80{\mathrm{e}}^{30{\mathrm{ks}}_{\mathrm{A}}}\\ {\mathrm{e}}^{30{\mathrm{kS}}_{\mathrm{A}}}& =& \frac{30}{80}\end{array}$

$30{\mathrm{kS}}_{\mathrm{A}}=\ln \left(\frac{30}{80}\right)$… (a)

Again, apply the other point $\mathrm{t}=30$and ${\mathrm{S}}_{\mathrm{B}}=2{\mathrm{S}}_{\mathrm{A}}$in the equation (3).

$\begin{array}{rcl}{\mathrm{T}}_{\mathrm{B}}& =& 80{\mathrm{e}}^{{\mathrm{kS}}_{\mathrm{B}}\left(30\right)}+70\\ {\mathrm{T}}_{\mathrm{B}}& =& 80{\mathrm{e}}^{2{\mathrm{kS}}_{\mathrm{A}}\left(30\right)}+70\\ {\mathrm{T}}_{\mathrm{B}}-70& =& 80{\mathrm{e}}^{60{\mathrm{kS}}_{\mathrm{A}}}\\ {\mathrm{e}}^{60{\mathrm{kS}}_{\mathrm{A}}}& =& \frac{{\mathrm{T}}_{\mathrm{B}}-70}{80}\\ 60{\mathrm{kS}}_{\mathrm{A}}& =& \ln \left(\frac{{\mathrm{T}}_{\mathrm{B}}-70}{80}\right)\end{array}$

$30{\mathrm{kS}}_{\mathrm{A}}=\frac{1}{2}\ln \left(\frac{{\mathrm{T}}_{\mathrm{B}}-70}{80}\right)$… (b)

Obtain the temperature of the oven.

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

$\begin{array}{rcl}\left(\frac{30}{80}\right)& =& \frac{1}{2}\ln \left(\frac{{\mathrm{T}}_{\mathrm{B}}-70}{80}\right)\\ 2\times (-0.9808)& =& \ln \left(\frac{{\mathrm{T}}_{\mathrm{B}}-70}{80}\right)\\ \ln \left(\frac{{\mathrm{T}}_{\mathrm{B}}-70}{80}\right)& =& -1.9616\\ \frac{{\mathrm{T}}_{\mathrm{B}}-70}{80}& =& {\mathrm{e}}^{-1.9616}\\ {\mathrm{T}}_{\mathrm{B}}-70& =& 80{\mathrm{e}}^{-1.9616}\\ {\mathrm{T}}_{\mathrm{B}}& =& 80{\mathrm{e}}^{-1.9616}+70\\ & =& \frac{80\times 9}{64}+70\\ & =& 11.25+70\\ {\mathrm{T}}_{\mathrm{B}}& =& 81.{25}^{\mathrm{o}}\end{array}$