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(A) Consider a concentric spherical shell of radius $r$ and thickness $dr$ as shown in the figure.In a steady state,the radial rate of flow of heat $H

Vedclass · Accepted Answer

$\frac{{{r_1}\,{r_2}}}{{({r_2} - {r_1})}}$

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(A) Consider a concentric spherical shell of radius $r$ and thickness $dr$ as shown in the figure.In a steady state,the radial rate of flow of heat $H$ through this shell is given by Fourier's law of heat conduction:$H = \frac{{dQ}}{{dt}} = - KA\frac{{dT}}{{dr}}$Since the area of the spherical shell is $A = 4\pi {r^2}$,we have:$H = - K(4\pi {r^2})\frac{{dT}}{{dr}}$Rearranging the terms to integrate:$\frac{{dr}}{{{r^2}}} = - \frac{{4\pi K}}{H}dT$Integrating from $r_1$ to $r_2$ and $T_1$ to $T_2$:$\int_{{r_1}}^{{r_2}} \frac{{dr}}{{{r^2}}} = - \frac{{4\pi K}}{H} \int_{{T_1}}^{{T_2}} dT$$\left[ - \frac{1}{r} \right]_{{r_1}}^{{r_2}} = - \frac{{4\pi K}}{H} (T_2 - T_1)$$\left( \frac{1}{{{r_1}}} - \frac{1}{{{r_2}}} \right) = \frac{{4\pi K}}{H} (T_1 - T_2)$$\frac{{{r_2} - {r_1}}}{{{r_1}{r_2}}} = \frac{{4\pi K}}{H} (T_1 - T_2)$Solving for $H$:$H = \frac{{4\pi K{r_1}{r_2}({T_1} - {T_2})}}{{{r_2} - {r_1}}}$Thus,the rate of heat flow is proportional to $\frac{{{r_1}{r_2}}}{{{r_2} - {r_1}}}$.

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

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