A sphere and a cube of same material and same total surface area...

Rate of emission of energy $= \sigma T^{4} \text{S}$

Let m₁ be the mass of sphere, C is specific heat and ~\left(\text{d} \theta / \text{dt}\right)$,~the rate of cooling.<br /><br />For sphere<br /><br />$\sigma T^{4} \text{S} = {m}_{1} \text{C} {\left(\dfrac{\text{d} \theta}{\text{dt}}\right)}_{S}$...(i)<br /><br />Let m₂ be the mass of cube, C its specific heat and$\left(\text{d} \theta / \text{dt}\right)$,~the rate of cooling<br /><br />For cube<br /><br />$\sigma T^{4} \text{S} = {m}_{2} \text{C} {\left(\dfrac{\text{d} \theta}{\text{dt}}\right)}_{C} ...(ii)

From Eqs.(i) and (ii)

$\dfrac{{\left(\text{d} \theta / \text{dt}\right)}_{s}}{{\left(\text{d} \theta / \text{dt}\right)}_{c}} = \dfrac{{m}_{2}}{{m}_{1}}$
$= \dfrac{a^{3} ρ}{\left(4 / 3\right) \pi r^{2} ρ}$

where a is the side of cube and r is the radius of sphere, $ρ$~is the density.

Required ratio $\dfrac{{R}_{s}}{{R}_{c}} = \dfrac{3 a^{3}}{4 \pi r^{3}}$

But since $\text{S}~$(surface area) is the same,

$6 a^{2} = 4 \pi r^{2}$

or~ $a^{2} = \left(2 / 3\right) \pi r^{2}$

$∴ \dfrac{{R}_{s}}{{R}_{c}} = \dfrac{3 {\left(2 \pi r^{2} / 3\right)}^{3 / 2}}{4 \pi r^{3}} = \dfrac{2 \pi \sqrt{2 \pi}}{\sqrt{3} \left(4 \pi\right)}$

$= \sqrt{\dfrac{2 \pi}{1 2}} = \sqrt{\dfrac{\pi}{6}}$