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(C) The rate of heat emission by radiation is given by the Stefan-Boltzmann law: $\Delta E = e A \sigma (T^4 - T_0^4)$.Since the material,temperature,

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

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(C) The rate of heat emission by radiation is given by the Stefan-Boltzmann law: $\Delta E = e A \sigma (T^4 - T_0^4)$.Since the material,temperature,and surroundings are the same,the ratio of radiation emitted depends only on the surface area $A$: $\frac{(\Delta E)_{	ext{sphere}}}{(\Delta E)_{	ext{cube}}} = \frac{A_{	ext{sphere}}}{A_{	ext{cube}}} = \frac{4 \pi R^2}{6 a^2}$.Given that the volumes are equal: $\frac{4}{3} \pi R^3 = a^3$,which implies $a = R \left( \frac{4\pi}{3} ight)^{1/3}$.Substituting $a$ into the area ratio: $\frac{4 \pi R^2}{6 [R (4\pi/3)^{1/3}]^2} = \frac{4 \pi}{6 (4\pi/3)^{2/3}} = \frac{4 \pi}{6} \cdot \frac{3^{2/3}}{(4\pi)^{2/3}} = \frac{(4\pi)^{1/3}}{6 \cdot 3^{-2/3}} = \frac{(4\pi)^{1/3}}{6^{1/3} \cdot 6^{2/3} \cdot 3^{-2/3}} = \left( \frac{4\pi}{6 \cdot (4/3)^{2/3}} ight) = \left( \frac{\pi}{6} ight)^{1/3}$.Thus,the ratio is $\left( \frac{\pi}{6} ight)^{1/3} : 1$.

$A$ 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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