Chapter 7.3, Problem 42E

To determine

To Find: The expression for the temperature $T\left(r\right)$ between the spheres.

Answer to Problem 42E

The required function is $T\left(r\right)=\frac{20}{r}+35$ .

Explanation of Solution

Given:

The given equation is,

  $\frac{d^{2}t}{d{r}^2}+\frac{2}{r}\frac{dT}{dr}=0$

The sphere with the radius 1 m has the temperature of 15 degree Celsius and it lies inside the concentric sphere that has the radius of 2 m and the temperature of $25°\text{C}$ .

The expression for the $S=\frac{dT}{dr}$ .

Calculation:

Consider the differential equation is,

  $\frac{d^{2}t}{d{r}^2}+\frac{2}{r}\frac{dT}{dr}=0$

Consider that T is the temperature of the sphere and the radius is r .

Then, let us consider $u=\frac{dT}{dr}$ then, $\frac{du}{dr}=\frac{d^{2}T}{d{r}^2}$

Thus, the equation is,

  $\begin{array}{l}\frac{du}{dr}+\frac{2}{r}u=0\\ \frac{du}{dr}=\frac{-2u}{r}\\ \frac{1}{u}du=\frac{-2}{r}dr\end{array}$

Then,

  $\begin{array}{l}{\displaystyle \int \frac{1}{u}du}={\displaystyle \int \frac{-2}{r}dr}\\ \ln u=-2\ln r+C_1\\ u=r^{-2}{e}^{c_1}\\ \frac{dT}{dr}=r^{-2}{e}^{c_1}\end{array}$

Consider $e^{c_1}=k_1$ .

Then,

  $\begin{array}{l}\frac{dt}{dr}=k_1{r}^{-2}\\ {\displaystyle \int T}=-k_1{r}^{-1}+k_2\\ T=-\frac{k_1}{r}+k_2\end{array}$

Consider for the radius of 1 m and the temperature of $T=15°\text{C}$ .

  $\begin{array}{l}15=-\frac{k_1}{1}+k_2\\ k_2-k_1=15\end{array}$

Consider for the radius of 2 m and the temperature of $T=25°\text{C}$ .

  $\begin{array}{l}25=-\frac{k_1}{1}+k_2\\ 2k_2-k_1=50\end{array}$

Then, from the above equations,

  $\begin{array}{l}{k}_1=20\\ k_2=35\end{array}$

Thus, the required temperature function is,

  $T\left(r\right)=\frac{20}{r}+35$