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A solid copper sphere (density $\rho $ and specific heat capacity $c$ ) of radius $r$ at an initial temperature $200K$ is suspended inside a chamber whose walls are at almost $0K$ . The time required (in $\mu s$) for the temperature of the sphere to drop to $100\, K$ is
$\frac{{72}}{7}\frac{{r\rho c}}{\sigma }$
$\frac{7}{{72}}\frac{{r\rho c}}{\sigma }$
$\frac{{27}}{7}\frac{{r\rho c}}{\sigma }$
$\frac{7}{{27}}\frac{{r\rho c}}{\sigma }$
Solution
(b) $\frac{{dT}}{{dt}} = \frac{{\sigma \,A}}{{mcJ}}\,\,({T^4} – T_0^4)$ [In the given problem fall in temperature of body $dT = (200 – 100) = 100K$, temp. of surrounding $T0 = 0K$ , Initial temperature of body $T = 200K].$
$\frac{{100}}{{dt}} = \frac{{\sigma 4\pi {r^2}}}{{\frac{4}{3}\pi {r^3}\rho \,c\,J}}({200^4} – {0^4})$
==> $dt = \frac{{r\rho \,c\,J}}{{48\sigma }} \times {10^{ – 6}}s = \frac{{r\rho \,c}}{\sigma }.\frac{{4.2}}{{48}} \times {10^{ – 6}}$
$ = \frac{7}{{80}}\frac{{r\rho \,c}}{\sigma }\mu \,s\tilde –\frac{7}{{72}}\frac{{r\rho \,c}}{\sigma }\mu \,s$ [As $J = 4.2$ ]
