Question

Solar radiation at the top of the atmosphere is 1340 W/m2, whereas at the surface this number is reduced to about 700 W/m2. This means that about 640 W/m2 is absorbed by the atmosphere. If the atmosphere extends 80 km above the surface, and the radius of the Earth is 3959 miles, determine how much the temperature would increase in the atmosphere by the absorption of the solar radiation. To find the mass of the atmosphere, use the density of air as 1.225 kg/m3. The specific heat of air is 1000 J/kg oC.The temperature change of a substance can be determined by dT =(rate of energy flow) *(# seconds per day)*(area of earth's surface)/(mass of air*specific heat of air).This calculation assumes that the atmosphere is uniform vertically.

Answer 1

Given

Solar radiation at the top of the atmosphere,

$Q=1340\text{ W/m}^2$

Solar radiation at the bottom of the atmosphere,

$Q^{\prime}=700\text{ W/m}^2$

Solar radiation absorbed by the atmosphere is

$\begin{gathered} \\ P=640\text{ W/m}^2 \end{gathered}$

The time, t=24 hr=24x60x60 s

Specific heat,

$s=\frac{1000J}{kg^oC}$

The radius of the earth,

$R_1=3959\text{ miles=6371392.89 metres}$

The length with the atmosphere,

$R_2=6371392.89+80000=6451392.896m$

To find

Determine how much the temperature would increase in the atmosphere by the absorption of the solar radiation

Explanation

Volume of the atmosphere,

$\begin{gathered} V=\frac{4}{3}\pi(R_2^3-R_1^3) \\ \Rightarrow V=\frac{4}{3}\pi(6451392.896^3-6371392.89^3) \\ \Rightarrow V=4.13\times10^{19}m^3 \end{gathered}$

Now,

Mass

$m=1.225\times1000=5.06\times10^{19}=5.06\times10^{19}kg$

Thus the change in temperature is

$dT=\frac{640\times24\times60\times60\times4\pi(6.37\times10^6)^2}{5.06\times10^{19}\times1000}=0.57^oC$

Conclusion

The change in temperature is

$0.57^oC$