Chapter 7
Q. 7.1
The Earth’s Spectrum. Consider the earth to be a blackbody with average surface temperature 15ºC and area equal to 5.1 × 10^{14} m². Find the rate at which energy is radiated by the earth and the wavelength at which maximum power is radiated. Compare this peak wavelength with that for a 5800 K blackbody (the sun).
Step-by-Step
Verified Solution
Using (7.2), the earth radiates:
\begin{matrix} E \ = \ \sigma A T^{4} & = \ \left(5.67 \ \times \ 10^{-8} \ {W}/{m^{2}} \ \cdot \ K^{4}\right) \ \times \ \left(5.1 \ \times 10^{14} \ m^{2}\right) \ \times \ \left(15 \ + \ 273 \ K\right)^{4} \\ & = \ 2.0 \ \times \ 10^{17} \ W \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \end{matrix}
The wavelength at which the maximum power is emitted is given by (5.3):
\text{Initial (simple) rate of return} \ = \ \frac{\text{Annual savings} \ S \ \left({\$}/{yr}\right)}{\text{Extra first cost} \ \Delta P\left(\$\right)} (5.3)
\lambda _{max} \left(\text{earth}\right) \ = \ \frac{2898}{T \ \left(K\right)} \ = \ \frac{2898}{288} \ = \ 10.1 \ \mu m
For the 5800 K sun,
\lambda _{max} \left(\text{sun}\right) \ = \ \frac{2898}{5800} \ = \ 0.5 \ \mu mIt is worth noting that earth’s atmosphere reacts very differently to the much longer wavelengths emitted by the earth’s surface (Fig. 7.1) than it does to the short wavelengths arriving from the sun (Fig. 7.2). This difference is the fundamental factor responsible for the greenhouse effect.
