Hi,

Nearly all of the energy used on earth comes originally from the sun, and that solar radiation is intercepted at 1370 Wm^-^2 at the top of the atomsphere, and that 49%of this radiation is absorbed by the earths surface.
If a shallow dish containing ice or warte, placed on the earth surface, absorbs the same the same fraction of solar radiation as the surface itself, at what rate is energy absorbed from the sun by such a container of surface 0.039m^2 ?
Can you show me how thhis is worked out please. Thanks Carl.

.49*1370 = watts per square meter absorbed

Multiply that by the dish area.

Is the answer 67130 Wm^2

Sure, Carl! To calculate the rate at which energy is absorbed by a container, we need to consider the area of the container and the fraction of solar radiation absorbed by it.

Given:
Solar radiation intercepted at the top of the atmosphere (I) = 1370 Wm^-2
Fraction of solar radiation absorbed by the Earth's surface (A) = 49%
Surface area of the container (A_c) = 0.039m^2

First, we need to find the amount of solar radiation absorbed by the Earth's surface. This can be calculated by multiplying the intercepted solar radiation with the absorption fraction:
Energy absorbed by the Earth's surface (E_s) = I * A

Next, we can calculate the energy absorbed by the container by multiplying the energy absorbed by the Earth's surface with the ratio of the container's surface area to the Earth's surface area:
Energy absorbed by the container (E_c) = E_s * (A_c / 1m^2)

Let's plug in the given values and calculate the result:

Energy absorbed by the Earth's surface (E_s) = 1370 Wm^-2 * 0.49 = 671.3 Wm^-2

Energy absorbed by the container (E_c) = 671.3 Wm^-2 * (0.039m^2 / 1m^2) = 26.16 W

Therefore, the rate at which energy is absorbed from the sun by the container is approximately 26.16 watts.

I hope this helps! Let me know if you have any further questions.