1. Apr 5, 2006

### scorpio_wan1945

If the amount of solar energy striking the outer edge of the earth’s atmosphere is approximately 1.4 kW/m2 and the amount of the earth’s surface area exposed to the sunlight is 1.6 x 10E13 m2, how much energy does the earth lose through radiation within each 24 hours? The average earth atmosphere temperature increases within the same period is 10-6 oC, the average specific heat of earth is 100 J/kg-oC, and the earth’s mass is 6 x 10E24 kg.

im stuck over here, what equation that we need to use to solve this problem?

2. Apr 5, 2006

Hi! Here's what I think...

Basically, when solar energy strikes the outer edge of the Earth's atmosphere, the Earth will gain heat energy (first set of data) and its temperature will increase. However, this increase is not as large as it should be (second set of data), because the Earth also loses heat energy through radiation.

Now, do you understand how the problem should be solved?

By the way, what does the "10-6" in 10-6 oC mean?

Last edited: Apr 6, 2006
3. Apr 5, 2006

### Astronuc

Staff Emeritus
Basically the earth's atmosphere is not heating up or storing energy. The light travels through the atmosphere strikes the surface and is radiated outward to space, again through the atmosphere.

There is a power flux (kW/m2) and a surface area (m2) which yields power. Now if the atmosphere does not store energy and the earth's surface does not increase in temperature, then balance of energy means that most energy is re-radiated to space. One can assume that the atmosphere is relatively thin compared to the radius of the earth.

The 1.4 kW/m2 should be based on flux normal to surface, so that one should use the projected area of the earth (with respect to the sun), not the surface area of a hemisphere.

4. Apr 5, 2006