Question

At $\mathit{t}\mathbf{=}\mathbf{0}$a sealed test tube containing a chemical is immersedin a liquid bath. The initial temperature of the chemicalin the test tube is${\mathbf{80}}^{\mathbf{o}}\mathbf{F}$. The liquid bath has a controlledtemperature (measured in degrees Fahrenheit)given by${\mathit{T}}_{\mathbf{m}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathbf{-}\mathbf{40}{\mathit{e}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{t}}\mathbf{,}\mathit{t}\mathbf{>}\mathbf{0}$, whereis measured in minutes.(a) Assume that$\mathit{k}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{1}$in (2). Before solving the IVP,describe in words what you expect the temperatureof the chemical to be like in the short term. In the long term.(b) Solve the initial-value problem. Use a graphing utility to plot the graph ofon time intervals of various lengths.Do the graphs agree with your predictions in part (a)?

Short answer

(a) The chemical decreases in a short term and increases in a long term.

(b) The equation is $T\left(t\right)=100-(4t+20)e^{-0.1t}$.

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

Solve for temperature of the chemical.

Let the temperature in a test tube immersed in a liquid path be,

$T_m\left(t\right)=100-40e^{-0.1t},t⩾0$… (1)

And with the conditions $T(t=0)={80}^0\mathrm{F}$.

(a)

Substitute the value$t=0$in the equation (1) to obtain the initial path.

$\begin{array}{rcl}{T}_m& =& 100-40e^0\\ & =& 100-40\\ & =& {60}^0\mathrm{F}\end{array}$

Thus, the temperature of the chemical should decrease.

Substitute the value$t=\infty$in the equation (1) to obtain the initial path.

$$\begin{gathered} T_m=100-40e^{-\infty } \\ =100-40\times 0 \\ ={100}^0\hspace{0.17em}\mathrm{F} \end{gathered}$$

Thus, the temperature of the chemical should increase.

Solve for first order equation.

(b)

Let the temperature of the liquid and chemical be,

$$\begin{gathered} \frac{dT}{dt}=-0.1\left(T-T_m\right) \\ \frac{dT}{dt}=-0.1\left(T-100+40e^{-0.1t}\right) \\ \frac{dT}{dt}=-0.1T+10-4e^{-0.1t} \end{gathered}$$

$\frac{dT}{dt}+0.1T=10-4e^{-0.1t}$… (2)

As (2) is linear, find the integrating factor and integrate the function. Let the integrating factor be,

role="math" localid="1663841814038" $\begin{array}{rcl}I.F& =& e^{\int p\left(t\right)dt}\\ & =& e^{\int 0.1dt}\\ & =& e^{0.1t}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}{e}^{0.1t}\frac{dT}{dt}+0.1e^{0.1t}T& =& 10e^{0.1t}-4\\ \frac{d}{dt}\left(e^{0.1t}T\right)& =& 10e^{0.1t}-4\\ \int d\left(e^{0.1t}T\right)& =& \int \left(10e^{0.1t}-4\right)dt\\ e^{0.1t}T& =& 10\int e^{0.1t}dt-4\int dt\\ e^{0.1t}T& =& 10\times \frac{1}{0.1}{e}^{0.1t}-4t+c\\ e^{0.1t}T& =& 100e^{0.1t}-4t+c\end{array}$

$T\left(t\right)=100-4t{e}^{-0.1t}+c{e}^{-0.1t}$… (3)

Obtain the values of constants

To find the values of constants, apply the point$(T,t)=\left({80}^0\mathrm{F},0\mathrm{minutes}\right)$in the equation (3), then

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

Substitute the value of c in the equation (3).

$$\begin{gathered} T\left(t\right)=100-4t{e}^{-0.1t}-20e^{-0.1t} \\ T\left(t\right)=100-(4t+20)e^{-0.1t} \end{gathered}$$

Obtain the values of constants.

Let the graph of the prediction in part (a) be,