First slide

Radiation

Question

A sphere and a cube of same material and same total surface area are placed in the same evacuated space turn by turn after they are heated to the same temperature. Find the ratio of their initial rates of cooling in the enclosure.

Moderate

A

$\sqrt{\frac{π}{6}} : 1$
B

$\sqrt{\frac{π}{3}} : 1$
C

$\frac{π}{\sqrt{6}} : 1$
D

$\frac{π}{\sqrt{3}} : 1$

Solution

Rate of emission of energy = ${σT}^{4} s$
Let $m_{1}$ be the mass of sphere, C is specific heat and ( $\frac{dθ}{dt}$ ), the rate of cooling.
For sphere
${σT}^{4} S = m_{1} C {(\frac{dθ}{dt})}_{s}$ ------------(1)
Let $m_{2}$ be the mass of cube, C its specific heat and $(\frac{dθ}{dt})$ , the rate of cooling.
For cube ${σT}^{4} S = m_{2} C {(\frac{dθ}{dt})}_{c}$ .----------(2)
From equations (l) and (2)
$\frac{{(\frac{dθ}{dt})}_{s}}{{(\frac{dθ}{dt})}_{c}} = \frac{m_{2}}{m_{1}} = \frac{R_{s}}{R_{c}}$
$or \frac{a^{3} ρ}{(\frac{4}{3}) {πr}^{2} ρ} = \frac{R_{s}}{R_{c}}$
where a is the side of cube and r is the radius of sphere, $ρ$ is the density.
$∴ \frac{R_{s}}{R_{c}} = \frac{3 a^{3}}{4 {πr}^{3}}$
But since S is the same, so
$6 a^{2} = 4 {πr}^{2}$
or $a^{2} = (\frac{2}{3}) {πr}^{2}$
$∴ \frac{R_{s}}{R_{c}} = \frac{3 {(\frac{2 {πr}^{2}}{3})}^{\frac{3}{2}}}{4 {πr}^{3}} = \frac{2 π \sqrt{2 π}}{\sqrt{3} (4 π)}$
= $\sqrt{\frac{2 π}{12}} = \sqrt{\frac{π}{6}}$

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