The wavelength of maximum intensity of the sun’s radiation is observed to be near 500 nm. Assume the sun to be a blackbody and calculate (a) the sun’s surface temperature, (b) the power per unit area R(T) emitted from the sun’s surface, and (c) the energy received by the Earth each day from the

Question

The wavelength of maximum intensity of the sun’s radiation is observed to be near 500 nm. Assume the sun to be a blackbody and calculate (a) the sun’s surface temperature, (b) the power per unit area R(T) emitted from the sun’s surface, and (c) the energy received by the Earth each day from the sun’s radiation.

Strategy (a) We use Equation (3.14) with [latex]\lambda_{\max }[/latex] to determine the sun’s surface temperature. (b) We assume the sun is a blackbody. We use the temperature T with Equation (3.16) to determine the power per unit area R(T). (c) Because we know R(T), we can determine the amount of the sun’s energy intercepted by the Earth each day.

[latex]\lambda_{\max } T=2.898 imes 10^{-3} m \cdot K[/latex] (3.14)

[latex]R(T)=\epsilon \sigma T^{4}[/latex] (3.16)

Accepted Answer

(a) From Equation (3.14) we calculate the sun’s surface temperature with [latex]\lambda_{\max }=500[/latex] nm.

[latex](500 nm ) T_{ ext {sun }}=2.898 imes 10^{-3} m \cdot K \frac{10^{9} nm }{ m }[/latex]

[latex]T_{ sun }=\frac{2.898 imes 10^{6}}{500} K =5800 K[/latex] (3.17)

(b) The power per unit area R(T) radiated by the sun is

[latex]R(T)=\sigma T^{4}=5.67 imes 10^{-8} \frac{ W }{ m ^{2} \cdot K ^{4}}(5800 K )^{4}[/latex]

[latex]=6.42 imes 10^{7} W / m ^{2}[/latex] (3.18)

(c) Because this is the power per unit surface area, we need to multiply it by [latex]4 \pi r^{2}[/latex], the surface area of the sun. The radius of the sun is [latex]6.96 imes 10^{5}[/latex] km.

Surface area [latex](\operatorname{sun})=4 \pi\left(6.96 imes 10^{8} m ight)^{2}=6.09 imes 10^{18} m ^{2}[/latex]

Thus the total power, [latex]P_{ ext {sun }}[/latex], radiated from the sun’s surface is

[latex]P_{ ext {sun }}=6.42 imes 10^{7} \frac{ W }{ m ^{2}}\left(6.09 imes 10^{18} m ^{2} ight)[/latex]

[latex]=3.91 imes 10^{26} W[/latex] (3.19)

The fraction F of the sun’s radiation received by Earth is given by the fraction of the total area over which the radiation is spread.

[latex]F=\frac{\pi r_{E}^{2}}{4 \pi R_{E s}^{2}}[/latex]

where [latex]r_{E}=[/latex] radius of Earth [latex]=6.37 imes 10^{6} m , ext { and } R_{E s}=[/latex] mean Earth-sun distance [latex]=1.49 imes 10^{11}[/latex] m. Then

[latex]F=\frac{\pi r_{E}^{2}}{4 \pi R_{E s}^{2}}=\frac{\left(6.37 imes 10^{6} m ight)^{2}}{4\left(1.49 imes 10^{11} m ight)^{2}}=4.57 imes 10^{-10}[/latex]

Thus the radiation received by the Earth from the sun is

[latex]\begin{aligned}P_{ ext {Earth }}( ext { received }) &=\left(4.57 imes 10^{-10} ight)\left(3.91 imes 10^{26} W ight) \&=1.79 imes 10^{17} W\end{aligned}[/latex]

and in one day the Earth receives

[latex]U_{ ext {Earth }}=1.79 imes 10^{17} \frac{ J }{ s } \frac{60 s }{\min } \frac{60 min }{ h } \frac{24 h }{ ext { day }}[/latex]

[latex]=1.55 imes 10^{22} J[/latex] (3.20)

The power received by the Earth per unit of exposed area is

[latex]R_{ ext {Earth }}=\frac{1.79 imes 10^{17} W }{\pi\left(6.37 imes 10^{6} m ight)^{2}}=1400 W / m ^{2}[/latex] (3.21)

This is the source of most of our energy on Earth. Measurements of the sun’s radiation outside the Earth’s atmosphere give a value near 1400 [latex]W / m ^{2}[/latex], so our calculation is fairly accurate. Apparently the sun does act as a blackbody, and the energy received by the Earth comes primarily from the surface of the sun.

Question 3.5: The wavelength of maximum intensity of the sun’s radiation i...

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