Question

The figure shows a system of two concentric sphere of radii $r_1$ and $r_2$ and kept at temperature $T_1$ and $T_2$, respectively. The radial rate of flow of heat in a substance between the two concentric spheres, is proportional to:

• A. $\frac{(r_2-r_1)}{\left(r_1{r}_2\right)}$
• B. $ln\left(\frac{r_2}{r_1}\right)$
• C. $\frac{r_1{r}_2}{(r_2-r_1)}$
• D. $(r_2-r_1)$

Answer 1

The correct option is C $\frac{r_1{r}_2}{(r_2-r_1)}$

Let us draw two spherical shells of radii $x$ and $x+dx$ concentric with the given system.Let the temperatures of these shells be $\theta$ and $\theta +d\theta$ respectively.The amount of heat flowing radially inward through the material between $x$ and $x+dx$ is

$\frac{\delta Q}{\delta t}=\frac{K.4\pi x^{2}d\theta }{dx}$

Thus, $K.4\pi {\int }_{{\theta }_1}^{{\theta }_2}d\theta =\frac{\delta Q}{\delta t}{\int }_{r_1}^{r_2}\frac{dx}{x^2}$

or $K.4\pi ({\theta }_2-{\theta }_1)=\frac{\delta Q}{\delta t}(\frac{1}{r_1}-\frac{1}{r_2})$

or $\frac{\delta Q}{\delta t}=\frac{4\pi K{r}_1{r}_2({\theta }_2-{\theta }_1)}{r_2-r_1}$

$\frac{\delta Q}{\delta t}\propto \frac{r_1{r}_2}{(r_2-r_1))}$