Question

Suppose that as a body cools, the temperature of the surrounding medium increases because it completely absorbs the heat being lost by the body. Let $\mathit{T}\mathbf{\left(}\mathit{t}\mathbf{\right)}$and${\mathit{T}}_{\mathit{m}}\left(\mathit{t}\right)$be the temperatures of the body and the medium at time $t$, respectively.

If the initial temperature of the body is${\mathit{T}}_1$and the initial temperature of the medium is ${\mathit{T}}_2$, then it can be shown in this case that Newton’s law of cooling is$\frac{\mathit{d}\mathit{T}}{\mathit{d}\mathit{t}}=\mathit{k}(\mathit{T}-{\mathit{T}}_{\mathit{m}}),\mathit{k}<0,$where${\mathit{T}}_{\mathbf{m}}\mathbf{=}{\mathit{T}}_{\mathbf{2}}\mathbf{+}\mathit{B}\mathbf{(}{\mathit{T}}_{\mathbf{1}}\mathbf{-}\mathit{T}\mathbf{)}\mathbf{,}\mathit{B}\mathbf{>}\mathbf{0}$is a constant.

(a) The foregoing DE is autonomous. Use the phase portrait concept of Section 2.1 to determine the limiting value of the temperature$\mathit{T}\left(\mathit{t}\right)$as $t\to \infty .$. What is the limiting value of ${\mathit{T}}_{\mathit{m}}\left(\mathit{t}\right)$as $t\to \infty .$?

(b) Verify your answers in part (a) by actually solving the differential equation.

(c) Discuss a physical interpretation of your answers in part (a).

Short answer

a)$li{m}_{t\to \infty }T\left(t\right)=li{m}_{t\to \infty }{T}_m\left(t\right)=\frac{B{T}_1+T_2}{1+B}$

b)$li{m}_{t\to \infty }\frac{B{T}_1+T_2}{1+B}+\frac{T_1-T_2}{1+B}{e}^{k(1+B)t}-\frac{B{T}_1-T_2}{1+B}$

c) Temperatures equalize as equilibrium occurs in the system.

Step-by-step solution

Definition of differential equation

A differential equation is an equation containing the derivatives or differentials of one or more dependent variables, with respect to one or more independent variables.

Find the value of limt→∞⁡Tm(t)

$li{m}_{t\to \infty }=\frac{B{T}_1+T_2}{1+B}$The equation is autonomous since depends only on T. Take the derivative on both sides of T.

$\begin{array}{c}\frac{dT}{dt}=k(T-T_m)\\ =k(T-T_2-B(T_1-T\left)\right)\\ =k\left(\right(1+B)T-(B{T}_1+T_2\left)\right)\\ =k(1+B)\left(T-\frac{B{T}_1+T_2}{1+B}\right)\end{array}$

Since$B>0$and$k<0$, the differential equation has the single critical point $\frac{B{T}_1+T_2}{1+B}$. Now use the fact that$k<0$and conclude that the sign of $\frac{dT}{dt}$is negative above the critical point, and positive below it. Therefore, the critical point is attractor. We can now conclude that the limiting point$T\left(t\right)$of is actually that critical point., that is,

$li{m}_{t\to \infty }=\frac{B{T}_1+T_2}{1+B}$

From this result, we can obtain the value of $li{m}_{t\to \infty }{T}_m\left(t\right)$

Solve the differential equation

Solve the differential equation to obtain the same results.

$\begin{array}{c}\frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{k}(\mathrm{T}-{\mathrm{T}}_{\mathrm{m}})\\ \frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{k}(\mathrm{T}-{\mathrm{T}}_2-{\mathrm{BT}}_1+\mathrm{B}(\mathrm{T}\left)\right)\\ \frac{\mathrm{dT}}{\mathrm{dt}}-\mathrm{k}(1+\mathrm{B})=-\mathrm{k}({\mathrm{T}}_2+{\mathrm{BT}}_1)\end{array}$

This is linear DE and has the integrating factor $e^{-\int k(1+B)dt}-e^{-k(1+B)t}$,so, we can solve it as follows:

$\begin{array}{c}\frac{d}{dt}\left[e^{-k(1+B)t}T\right]=-k(B{T}_1+T_2)e^{-k(1+B)t}\\ e^{-k(1+B)t}T=\frac{B{T}_1+T_2}{1+B}{e}^{-k(1+B)t}+C\\ T\left(t\right)=\frac{B{T}_1+T_2}{1+B}+C{e}^{-k(1+B)t}\end{array}$

Use initial condition$T\left(0\right)=T_1$to calculate constant C

$\begin{array}{c}T\left(0\right)=\frac{B{T}_1+T_2}{1+B}+C{e}^{-k(1+B)\left(0\right)}\\ T_1=\frac{B{T}_1+T_2}{1+B}+C\\ C=T_1-\frac{B{T}_1+T_2}{1+B}\\ =\frac{T_1-T_2}{1+B}\end{array}$

Thus the final solution is$T\left(t\right)=\frac{B{T}_1+T_2}{1+B}+\frac{T_1-T_2}{1+B}{e}^{k(1+B)t}$

Find the limit$\left(lim\right)┬(t\to \infty )T\left(t\right).$.

localid="1668437450622" $\underset{t\to \infty }{lim}\frac{B{T}_1+T_2}{1+B}+\frac{T_1-T_2}{1+B}{e}^{k(1+B)t}=\frac{B{T}_1+T_2}{1+B}$

Therefore we obtain the same results in both steps.

Discussing a physical interpretation

Notice that both$T\left(t\right)$ and $T_m\left(t\right)$are the same after a long period of time. Intuitively, that

is completely reasonable because energy won’t escape outside of the system, since it

will be evenly distributed in the media.