Any object that is warmer than absolute zero will radiate energy (usually as infrared waves, but as visible light if the object is hot enough). The more surface area the object has, the more energy it will radiate. Also, the hotter the object is, the more energy it will radiate as well. The famous (“black-body radiation”) equation that predicts the power radiated to an object’s temperature and surface area is:
,
where 
and T is temperature in degrees Kelvin. (0° C is 273° K).
At the earth’s distance from the sun, the power received is 1350 watts per square meter. (Less solar power filters down to ground level).
Helpful formulas:
Area of disk: 
Surface area of sphere: 
You may have wondered why the earth’s temperature is what it is, and how much influence the sun has. Well, wonder no more, because you’re about to figure it out!
Assume that the only incoming source of heat for the earth is the power from sunlight (1350 W/m2).
1a. The earth’s radius is 6371 km. Calculate the earth’s cross-sectional area in m2.
1b. Using this area, calculate the amount of power from sunlight (in watts) that the earth intercepts. This is the power received by the earth from sunlight.
2. Next, although you don’t know the resulting earth temperature, just let it be “T” for now. Because the earth is at this temperature T, it will radiate (infrared) energy into space following the black-body equation.
2a. Calculate the total surface area of the earth, in m2.
2b. Work out an expression for the power that the earth radiates, which will of course depend on T. This energy will be radiated away into space as infrared waves.
2c. Assume that this power radiated away by the earth exactly balances the power from sunlight received by the earth. (Why? Otherwise the earth would keep warming up or cooling off until it reached equilibrium.) This, finally, allows you to solve for the temperature, T of this hypothetical earth.
3. T will be in degrees-Kelvin. Convert to your favorite temperature scale (e.g., Celsius, Fahrenheit). Does this temperature seem reasonable? Remember, it will represent some kind of average over the whole earth (averaging the temperatures of Antarctica with Tahiti, etc.).