Chapter 11, Problem 67AP

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Chapter
Section

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

# Earth’s surface absorbs an average of about 960. W/m2 from the Sun’s irradiance. The power absorbed is Pabs = (960. W/m2) (Adisc), where A disc  =  π R E 2 is Earth’s projected area. An equal amount of power is radiated so that Earth remains in thermal equilibrium with its environment at nearly 0 K. Estimate Earth’s surface temperature by setting the radiated power from Stefan’s law equal to the absorbed power and solving for the temperature in Kelvin. In Stefan’s law, assume e = 1 and take the area to be A  = 4 π R E 2 the surface area of a spherical Earth. (Note: Earth’s atmosphere acts like a blanket and warms the planet to a global average about 30 K above the value calculated here.)

To determine
The Earth’s surface temperature.

Explanation

Given Info: The power absorbed is 960W/m2(Adisc) , emissivity e is 1 and Earth is assumed to completely spherical.

Formula to calculate the  power absorbed is,

• Pabsorbed is the power absorbed,
• Adisc is the area of the disc,

Formula to the area of the disc is,

• RE is the radius of the Earth,

Use πRE2 for Adisc in the above expression to rewrite Pabsorbed .

Pabsorbed=960W/m2(πRE2)

The formula for the power radiated is,

• σ is the Stefan – Boltzmann constant,
• e is the emissivity,
• T is the temperature,

Conclusion:

The power absorbed by the Earth is radiated so that it remains in thermal equilibrium with its environment at nearly 0 K that is Pabsorbed=Pradiated

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