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(C) The amount of radiation emitted by a body is given by the Stefan-Boltzmann law: $Q = \sigma A t (T^4 - T_0^4)$.Since the temperature $T$,surroundi

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$(\frac{\pi}{6})^{1/3} : 1$

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(C) The amount of radiation emitted by a body is given by the Stefan-Boltzmann law: $Q = \sigma A t (T^4 - T_0^4)$.Since the temperature $T$,surroundings temperature $T_0$,emissivity $\sigma$,and time $t$ are the same for both bodies,the ratio of heat emitted is proportional to the surface area: $\frac{Q_{sphere}}{Q_{cube}} = \frac{A_{sphere}}{A_{cube}} = \frac{4\pi r^2}{6a^2} \dots (i)$.Given that the volumes are equal: $\frac{4}{3}\pi r^3 = a^3$,which implies $a = (\frac{4}{3}\pi)^{1/3} r$.Substituting $a$ into equation $(i)$:$\frac{Q_{sphere}}{Q_{cube}} = \frac{4\pi r^2}{6((\frac{4}{3}\pi)^{1/3} r)^2} = \frac{4\pi r^2}{6(\frac{4}{3}\pi)^{2/3} r^2} = \frac{4\pi}{6 \cdot \frac{4^{2/3}}{3^{2/3}} \cdot \pi^{2/3}} = \frac{4^{1/3} \pi^{1/3}}{6 \cdot 3^{-2/3}} = \frac{4^{1/3} \pi^{1/3}}{2 \cdot 3 \cdot 3^{-2/3}} = \frac{4^{1/3} \pi^{1/3}}{2 \cdot 3^{1/3}} = (\frac{4\pi}{2^3 \cdot 3})^{1/3} = (\frac{4\pi}{24})^{1/3} = (\frac{\pi}{6})^{1/3}$.Thus,the ratio is $(\frac{\pi}{6})^{1/3} : 1$.

$A$ sphere and a cube of the same material and same volume are heated up to the same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiation emitted will be

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