- Home
- Standard 11
- Physics
The figure shows a system of two concentric spheres of radii $r_1$ and $r_2$ and 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
$\frac{{{r_1}\,{r_2}}}{{({r_1} - {r_2})}}$
$({r_2} - {r_1})$
$({r_2} - {r_1})({r_1}\,{r_2})$
$In \left( {\frac{{{r_2}}}{{{r_1}}}} \right)$
Solution
(a) Consider a concentric spherical shell of radius $r$ and thickness $dr$ as shown in fig.
The radial rate of flow of heat through this shell in steady state will be $H = \frac{{dQ}}{{dt}} = – KA\frac{{dT}}{{dr}} = – K\,(4\pi {r^2})\frac{{dT}}{{dr}}$
==> $\int_{\,{r_1}}^{\,{r_2}} {\frac{{dr}}{{{r^2}}} = – \frac{{4\pi K}}{H}\int_{\,{T_1}}^{{T_1}} {dT} } $
Which on integration and simplification gives
$H = \frac{{dQ}}{{dt}} = \frac{{4\pi K{r_1}{r_2}({T_1} – {T_2})}}{{{r_2} – {r_1}}}$
==> $\frac{{dQ}}{{dt}} \propto \frac{{{r_1}{r_2}}}{{({r_2} – {r_1})}}$
