Question

Newton’s Law of Cooling/Warming As shown in Figure 3.3.12, a small metal bar is placed inside container A, and container A then is placed within a much larger container B. As the metal bar cools, the ambient temperature ${\mathit{T}}_{\mathit{A}}\left(\mathit{t}\right)$of the medium within container A changes according to Newton’s law of cooling. As container A cools, the temperature of the medium inside container B does not change significantly and can be considered to be a constant TB. Construct a mathematical model for the temperatures $\mathit{T}\left(\mathit{t}\right)$and ${\mathit{T}}_{\mathit{A}}\left(\mathit{t}\right)$, where $\mathit{T}\left(\mathit{t}\right)$is the temperature of the metal bar inside container A. As in Problems 1, 5, and 20, this model can be solved by using prior knowledge. Find a solution of the system subject to the initial conditions $\mathit{T}\left(0\right)={\mathit{T}}_0,T_A\left(0\right)=T_1.$

Short answer

The solution is

$$\begin{gathered} T\left(t\right)=T_A+\left(T_0-T_A\right)e^{-kt} \\ {T}_A\left(t\right)=\left(T_1-\frac{1}{2}T-\frac{1}{2}{T}_B\right)e^{-kt}+\frac{1}{2}T+\frac{1}{2}{T}_B \end{gathered}$$

Step-by-step solution

Given information

Initial conditions $T\left(0\right)=T_0,T_A\left(0\right)=T_1$.

To solve the differential equation

We must create a mathematical model based on differential equations to describe the temperature of the metal bar T and the temperature of the container A.

$$\begin{gathered} \frac{dT}{dt}=-k\left(T-T_A\right)....\left(1\right) \\ \frac{d{T}_A}{dt}=k\left(T-T_A\right)-k\left(T_A-T_B\right) \\ \frac{d{T}_A}{dt}=-k\left(2T_A-T-T_B\right)....\left(2\right) \end{gathered}$$

The container A is heated first during the cooling of the bar where $(T_A-T_B)>(T-T_A),$with the conditions,

$$\begin{gathered} T\left(0\right)=T_0 \\ {T}_A\left(0\right)=T_1 \end{gathered}$$

To solve the differential equation

$$\begin{gathered} \int \frac{1}{T-T_A}dT=-k\int dt \\ ln\left(T-T_A\right)=-kt+c_1 \\ {e}^{ln\left(T-T_A\right)}=e^{-kt+c_1} \\ T-T_A=e^{c_1}{e}^{-kt} \end{gathered}$$

Then we have

$$\begin{gathered} T\left(t\right)=T_A+e^{c_1}{e}^{-kt} \\ =T_A+c{e}^{-kt}...\left(a\right) \end{gathered}$$

$$\begin{gathered} T\left(t\right)=T_A+e^{c_1}{e}^{-kt} \\ =T_A+c{e}^{-kt}...\left(a\right) \end{gathered}$$

To find the value of constant c , apply the point of condition $\left(T,t\right)=\left(T_0,0\right)$into equation (a)

$$\begin{gathered} T_0=T_A+c{e}^0 \\ c=T_0-T_A \end{gathered}$$

Substitute c into equation (a)$T\left(t\right)=T_A+\left(T_0-T_A\right)e^{-kt}...\left(b\right)$

Solve the first differential equation

$$\begin{gathered} \int \frac{1}{2T_A-T-T_B}dT=-k\int dt \\ \frac{1}{2}\int \frac{1}{2T_A-T-T_B}dT=-k\int dt \\ ln\left(2T_A-T-T_B\right)=-2kt+h_1 \end{gathered}$$

$$\begin{gathered} e^{ln\left(2T_A-T-T_B\right)}=e^{h_1}{e}^{-2kt} \\ 2T_A=h{e}^{-2kt}+T+T_B \\ {T}_A\left(t\right)=\frac{1}{2}h{e}^{-2kt}+\frac{1}{2}T+\frac{1}{2}{T}_B...\left(c\right) \end{gathered}$$

Find the value of constant h, to apply the point of condition$\left(T_A,t\right)=\left(T_1,0\right)$

$$\begin{gathered} T_1=\frac{1}{2}h{e}^0+\frac{1}{2}T+\frac{1}{2}{T}_B \\ h=2T_1-T-T_B \end{gathered}$$

Substitute with h into equation (h)

$T_A\left(t\right)=\left(T_1-\frac{1}{2}T-\frac{1}{2}{T}_B\right)e^{-kt}+\frac{1}{2}T+\frac{1}{2}{T}_B$

Temperature of container A at time t

Conclusion

The solution is$$\begin{gathered} T\left(t\right)=T_A+\left(T_0-T_A\right)e^{-kt} \\ {T}_A\left(t\right)=\left(T_1-\frac{1}{2}T-\frac{1}{2}{T}_B\right)e^{-kt}+\frac{1}{2}T+\frac{1}{2}{T}_B \end{gathered}$$