Use the following data for this problem: the total power radiated by the Sun is

Question

Use the following data for this problem: the total power radiated by the Sun is 3.84 times 1026 W; the distance between Earth and the Sun is 1.50 times 10^11 m; Earth's radius is 6.38 times 10^6 m. How much energy per second from the Sun is intercepted by Earth's cross-sectional area? Measurements show that clouds reflect 30 percent of the energy Earth receives from the Sun. How much energy per second from the Sun reaches the ground? To a good approximation, Earth re-radiates the solar energy that reaches the ground as blackbody radiation. The emissivity of the ground Ls e - 1. Assuming that the temperature is uniform over Earth's surface, calculate Earth's surface temperature. Is your answer to part (c) above or below the freezing point of water? Fortunately the carbon dioxide and water vapor in Earth's atmosphere provide a gentle (so far) greenhouse effect that raises Earth s surface temperature above the freezing point of water.

Explanation / Answer

a) Intensity of sun at the location of earth,

I = Power/Area

= 3.84*10^26/(4*pi*(1.5*10^11)^2)

= 1358 W/m^2

Energy intercepted by earth in one second = I*Area of semi sphere of earth

= 1358*2*pi*Re^2

= 1358*2*pi*(6.38*10^6)^2

= 3.47*10^17 J

b) Energy reached to earth = 0.7*3.47*10^17

= 2.429*10^17 J

c) Energy radiated by earth per second per m^2 = 2.429*10^17/(4*pi*(6.37*10^6)^2)

sigma*T^4 = 476.4

T^4 = 476.4/sigma

T = (476.6/sigma)^1/4

= (476.6/(5.67*10^-8))^0.25

= 302.8 K <<<<<<<<<<<<<<<<<----------------Answer

d) above freezing point of water.

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