## 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).

## Verified Solution

$\begin{matrix} E \ = \ \sigma A T^{4} & = \ \left(5.67 \ \times \ 10^{-8} \ {W}/{m^{2}} \ \cdot \ K^{4}\right) \ \times \ \left(5.1 \ \times 10^{14} \ m^{2}\right) \ \times \ \left(15 \ + \ 273 \ K\right)^{4} \\ & = \ 2.0 \ \times \ 10^{17} \ W \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \end{matrix}$
$\text{Initial (simple) rate of return} \ = \ \frac{\text{Annual savings} \ S \ \left({\}/{yr}\right)}{\text{Extra first cost} \ \Delta P\left(\\right)}$ (5.3)
$\lambda _{max} \left(\text{earth}\right) \ = \ \frac{2898}{T \ \left(K\right)} \ = \ \frac{2898}{288} \ = \ 10.1 \ \mu m$
$\lambda _{max} \left(\text{sun}\right) \ = \ \frac{2898}{5800} \ = \ 0.5 \ \mu m$