This very crude analysis shows a 0.01 decrease in earth’s albedo causes a 0.92C increase in the annual global average temperature.

Earth’s Energy Balance Overall:
Light from the sun striking the earth  = Δheat radiated from earth’s surface to space.
This relation is not instantaneous, but represents a quasi-equilibrium condition, and is virtually true when averaged over surface of the earth and over one year.

(The largest exception is heat absorbed and retained by the oceans, which will be shown to have an effect lowering earth’s equilibrium temperature by 0.16C.)

From the figure above, notice that sunlight strikes the earth outer atmosphere as a parallel (collimated) beam, therefore the amount of sunlight striking the earth at any time is equivalent to light striking a disk perpendicular to the radiation, at earth’s distance from the sun, the radius “r” of which is earth’s radius:

 I_o (watts/m²) * π r²

_Io varies very slightly around 1367 watts/m²

Much light is reflected off the earth back to space.   This reflective property is known as “Albedo”, symbolized herein by “A”.   For earth, A is typically accepted as 0.30.  Therefore earth’s surface receives 1.00 – 0.30 = 70% of radiation incident at the top of atmosphere.

Compute Heat In = Heat Out

So net heat (as mostly visible light) striking the earth’s surface is

(1-A)*I_o (watts/m²) * πr² where A is earth albedo = 0.3 or 30%

but per unit area of earth’s spherical surface (= 4π r²) the average absorbed sunlight, seeing that πr² cancels out,

(1 – A)*I_o /4 (watts/m²) = 0.7 * 1367 / 4 watts/m² = 239 watts/m²

Compute Earth Surface Temperature with no Greenhouse Effect

Please notice that earth radiates its heat (as infrared) to space in all directions.  The equation for radiation would follow the Stefan-Boltzmann Law if there were no Greenhouse Effect:

Q = heat loss to space (watts/m²) = σT_a⁴   = 239 w/m²  

Where surface area of a sphere = 4πr²
ε = emissivity of earth, approximately 0.97
σ = Stefan-Boltzmann constant = 5.67E-8 watts/m²/deg-K⁴
T_a= equivalent annual average earth surface temperature, deg K.

Heat (as infrared light) is radiated in different amounts from the earth’s surface, clouds, and atmosphere.  Therefore, the energy balance for earth, averaged over all its surface for a year, if there were no Greenhouse Effect is approximately described by:

0.70*I_o * πr² = 4πr² εσT_a⁴ or

Ta = ((.7*1367)/(4*.97*5.67E-08))^0.25 solving for temperature:         equation (1)
T_a = 257 K = -16C = 3F

Earth’s actual annual average temperature is 15.5C, meaning this calculation is off by 31.5C!!!  But wait!  the current absolute temperature is 288.5K, which in relation to the calculated 257K is under by only 11%.

Let’s now see what happens when albedo A is decreased by 0.01 to 0.29:

In equation (1) substituting 1- 0.29 = 0.71 creates a T_a increase of 0.92C or 0.92K per 0.01 decrease in earth’s albedo.