91Ó°ÊÓ

The sun's intensity at the distance of the earth $=1370W/m^2$

Energy reflected by water and clouds $=30\%$

Absorbed energy$=70\%$

The actual average temperature of the earth$={15}^{∘}C$

For this problem we need to consider that the Earth will radiate from its entire surface, that is, $E=4{\mathrm{Ï€¸é}}^2$On the other hand, the Earth will be lit on half of its surface area. However, the intensity will be the given one only at one point. The power given to the earth will be in total the given intensity times an "effective" area of$A=\mathrm{Ï€}{R}^2$. One can reach this conclusion by integrating the intensity times the cosine of the angle it makes with the normal to the surface, or, as we will do, following another logic: the power given to the Earth will be equal to the intensity times the area of the shadow created by the Earth. It is clear that this shadow is a sphere of radius $R.$Therefore, our "effective" area is, as we expected, role="math" localid="1648124670922" $A={\mathrm{Ï€¸é}}^2$

Having this in mind, since we have thermal equilibrium , the incoming power is equal to the power that is radiated following the Stefan-Boltzmann law. This means that we can write\

$IA=A_E\mathrm{σ}{T}^4$

The emissivity has been omitted since we are given it is very close to one.

From this, we can express the temperature as

$T=\sqrt[4]{\frac{I}{\mathrm{σ}}\frac{A}{A_E}}=\sqrt[4]{\frac{I·\mathrm{π}{R}^2}{\mathrm{σ}·4\mathrm{π}{R}^2}}$

Simplifying, we get

$T=\sqrt[4]{\frac{I}{4\mathrm{σ}}}$

For the numerical calculations, we have

$T=\sqrt[4]{\frac{1370·0.7}{4·5.67·{10}^{-8}}}=255\mathrm{K}$

Please note that the intensity was multiplied by $0.7$since only $70\%$of the energy is absorbed, according to the text. The absorption was not included in the parametrical solution to simplify it.

Converting our temperature to Celsius, we have

data-custom-editor="chemistry" $T=-18°\mathrm{C}$

The earth's average temperature is $-{18}^{∘}C$if the earth had no atmosphere.