Question

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

• A. $\frac{72}{7}\frac{r\rho c}{\sigma}$
• B. $\frac{7}{72}\frac{r\rho c}{\sigma}$
• C. $\frac{27}{7}\frac{r\rho c}{\sigma}$
• D. $\frac{7}{27}\frac{r\rho c}{\sigma}$

Answer 1

Answer: (B)

$\frac{7}{72}\frac{r\rho c}{\sigma}$

Answer 2

$\frac{dT}{dt} = \frac{\sigma A}{mcJ}(T^{4} - T_{0}^{4})$ [In the given problem fall in

temperature of body $dT = (200 - 100) = 100K$

Temperature of surrounding T₀ = 0K, Initial temperature of body T = 200K]

$\frac{100}{dt} = \frac{\sigma 4\pi r^{2}}{\frac{4}{3}\pi r^{3}\rho cJ}(200^{4} - 0^{4})$

⇒$dt = \frac{r\rho cJ}{48\sigma} \times 10^{- 6}s = \frac{r\rho c}{\sigma}.\frac{4.2}{48} \times 10^{- 6} = \frac{7}{80}\frac{r\rho c}{\sigma}\mu s\widetilde{–}\frac{7}{72}\frac{r\rho c}{\sigma}\mu s$ [As J = 4.2]