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.

• A. $\sqrt{\frac{\pi }{6}} \text{:} 1$
• B. $\sqrt{\frac{\pi }{3}} \text{:} 1$
• C. $\frac{\pi }{\sqrt{6}} \text{:} 1$
• D. $\frac{\pi }{\sqrt{3}} \text{:} 1$

Answer 1

Answer: (A)

Rate of emission of energy $= \sigma \text{T}^{4} \text{S}$
Let m₁ be the mass of sphere, C is specific heat and $\left(\text{d} \theta / \text{dt}\right)$ , the rate of cooling.
For sphere
$\sigma \left(\text{T}\right)^{4} \text{S} = \left(\text{m}\right)_{1} \text{C} \left(\frac{\text{d} \theta }{\text{dt}}\right)_{\text{S}}$ ...(i)
Let m₂ be the mass of cube, C its specific heat and $\left(\text{d} \theta / \text{dt}\right)$ , the rate of cooling
For cube
$\sigma \left(\text{T}\right)^{4} \text{S} = \left(\text{m}\right)_{2} \text{C} \left(\frac{\text{d} \theta }{\text{dt}}\right)_{\text{C}}$ ...(ii)
From Eqs.(i) and (ii)
$\frac{\left(\text{d} \theta / \text{dt}\right)_{\text{s}}}{\left(\text{d} \theta / \text{dt}\right)_{\text{c}}} = \frac{\left(\text{m}\right)_{2}}{\left(\text{m}\right)_{1}}$
$=\frac{\left(\text{a}\right)^{3} \rho }{\left(4 / 3\right) \pi \left(\text{r}\right)^{2} \rho }$
where a is the side of cube and r is the radius of sphere, $\rho $ is the density.
Required ratio $ \frac{\text{R}_{\text{s}}}{\text{R}_{\text{c}}} = \frac{3 \text{a}^{3}}{4 \pi \text{r}^{3}}$
But since $\text{S }$ (surface area) is the same,
$6 \text{a}^{2} = 4 \pi \text{r}^{2}$
or $\left(\text{a}\right)^{2} = \left(2 / 3\right) \pi \left(\text{r}\right)^{2}$
$∴ \, \, \, \frac{\left(\text{R}\right)_{\text{s}}}{\left(\text{R}\right)_{\text{c}}} = \frac{3 \left(2 \pi \left(\text{r}\right)^{2} / 3\right)^{3 / 2}}{4 \pi \left(\text{r}\right)^{3}} = \frac{2 \pi \sqrt{2 \pi }}{\sqrt{3} \left(4 \pi \right)}$
$= \sqrt{\frac{2 \pi }{1 2}} = \sqrt{\frac{\pi }{6}}$