Two spherical black bodies of radius r_{1} an | vedclass

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Question

(A) According to the Stefan-Boltzmann Law,the power $P$ radiated by a spherical black body of radius $r$ and temperature $T$ is given by $P = \sigma A

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$\left(\frac{T_{2}}{T_{1}}\right)^{2}$

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(A) According to the Stefan-Boltzmann Law,the power $P$ radiated by a spherical black body of radius $r$ and temperature $T$ is given by $P = \sigma A T^{4}$,where $A = 4 \pi r^{2}$ is the surface area.Since both bodies radiate the same power,we have $P_{1} = P_{2}$.Substituting the formula,we get $4 \pi r_{1}^{2} \sigma T_{1}^{4} = 4 \pi r_{2}^{2} \sigma T_{2}^{4}$.Canceling the common terms $4 \pi \sigma$,we get $r_{1}^{2} T_{1}^{4} = r_{2}^{2} T_{2}^{4}$.Rearranging the terms to find the ratio $\frac{r_{1}^{2}}{r_{2}^{2}}$,we get $\frac{r_{1}^{2}}{r_{2}^{2}} = \frac{T_{2}^{4}}{T_{1}^{4}}$.Taking the square root on both sides,we get $\frac{r_{1}}{r_{2}} = \frac{T_{2}^{2}}{T_{1}^{2}} = \left(\frac{T_{2}}{T_{1}}\right)^{2}$.

Two spherical black bodies of radius $r_{1}$ and $r_{2}$ with surface temperatures $T_{1}$ and $T_{2}$ respectively,radiate the same power. Then the ratio $r_{1}: r_{2}$ is:

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