Question

A solid copper sphere(density $\rho$ and specific heat c) of radius r at an initial temperature $200$K is suspended inside a chamber whose walls are at almost $0$ K. The time required to the temperature of sphere to drop to $100$ K is _________?

• A. $\frac{9r\rho c}{72\times {10}^6\sigma }$sec.
• B. $\frac{7r\rho c}{72\times {10}^6\sigma }$sec.
• C. $\frac{7r\rho c}{82\times {10}^6\sigma }$sec.
• D. $\frac{19r\rho c}{72\times {10}^7\sigma }$sec.

Answer 1

Answer: (B)

$\frac{7r\rho c}{72\times {10}^6\sigma }$sec.

Given : $\rho$ = density , $C=$ specific capacity , initial temperature = 200$K$

solution :

we have to calculate heat loss due to radiation:

$mC\frac{dT}{dt}=\sigma A{T}^4$

$\left(\frac{4}{3}\pi r^3\right)\left(\rho \right)C\frac{dT}{dt}=\sigma \left(4\pi r^2\right)T^4$

$\frac{r\rho C}{3\sigma }{\int }_{100}^{200}\frac{dT}{T^4}={\int }_0^{t}dt$

$\frac{r\rho C}{9\sigma X{10}^6}(1-\frac{1}{8})=t$

$t=\frac{7r\rho C}{72X{10}^6\sigma }$

hence the correct opt : B