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Numerical · Q21

Q.Treating the Sun as a black-body sphere of radius R=6.96×108 mR = 6.96\times10^{8}\ \text{m} with a surface temperature of 5800 K5800\ \text{K}, use Stefan's law (E=σT4E = \sigma T^4, applied over the Sun's full surface area 4πR24\pi R^2) to estimate the total power radiated by the Sun. Take σ=5.67×10−8 W m−2K−4\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}.

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Given: R=6.96×108 mR = 6.96\times10^{8}\ \text{m}, T=5800 KT = 5800\ \text{K}, σ=5.67×10−8 W m−2K−4\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}.

Energy radiated per unit area (Stefan's law):

E=σT4=5.67×10−8×(5800)4E = \sigma T^4 = 5.67\times10^{-8} \times (5800)^4

Computing (5800)4(5800)^4: 58002=3.364×1075800^2 = 3.364\times10^7, so (5800)4=(3.364×107)2=1.132×1015(5800)^4 = (3.364\times10^7)^2 = 1.132\times10^{15}.

E=5.67×10−8×1.132×1015=6.42×107 W m−2E = 5.67\times10^{-8} \times 1.132\times10^{15} = 6.42\times10^{7}\ \text{W m}^{-2}

Surface area of the Sun:

4πR2=4π (6.96×108)2=4π×4.844×1017=6.09×1018 m24\pi R^2 = 4\pi\,(6.96\times10^{8})^2 = 4\pi \times 4.844\times10^{17} = 6.09\times10^{18}\ \text{m}^2

Total power radiated: …

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