91Ó°ÊÓ

• The amount of solar flux density per unit area is termed as solar constant. and it is given as

${\mathrm{σ}}_{\text{solar}}=\frac{P}{4\mathrm{π}{R}^2}$

• $$\begin{gathered} \text{From the Stefan's law of radiationwehavethat} \\ \text{amount of heat radiation is as follows:} \end{gathered}$$ $H=\mathrm{σ}e\left(\mathrm{Ï€}{r}^2\right)T^4$

The above two equations will be used to get the desired result.

Solar constant is given as :

${\mathrm{σ}}_{\text{solar}}=\frac{P}{4\mathrm{π}{R}^2}$

Here,$P$is the solar power and data-custom-editor="chemistry" $\mathrm{R}${ is the mean distance between the sun and Venus. }

The mean distance between the sun and the Venus is $70\%$of the distance between the earth and the sun.

$R=70\%\left(150×{10}^9\mathrm{m}\right)$

$=1.05×{10}^{11}\mathrm{m}$

Putting $3.9×{10}^{26}\mathrm{W}\text{for}P$and $1.05×{10}^{11}\mathrm{m}$$\text{for}R\text{in the equation}{\mathrm{σ}}_{\text{solar}}=\frac{P}{4\mathrm{π}{R}^2}$

${\mathrm{σ}}_{\text{solar}}=\frac{\left(3.9×{10}^{26}\mathrm{W}\right)}{4\mathrm{π}{\left[1.05×{10}^{11}\mathrm{m}\right]}^2}$

$=2.8×{10}^3\mathrm{W}/{\mathrm{m}}^2$

$\text{Hence, the solar constant at the location of Venus is}2.8×{10}^3\mathrm{W}/{\mathrm{m}}^2$.

The amount of heat absorbed by planet is

$H={\mathrm{σ}}_{\text{solar}}A$

$={\mathrm{σ}}_{\text{solar}}\left(4\mathrm{π}{r}^2\right)$

$\text{From the Stefan's law of radiation, the amount of heat radiation is as follows}$

$H=\mathrm{σ}e\left(\mathrm{π}{r}^2\right)T^4$
role="math" localid="1647761137436" $\mathrm{σ}\text{is the Stefan's constant and}T\text{is the absolute temperature.}$

$\text{Putting}{\mathrm{σ}}_{\text{solar}}\left(4\mathrm{π}{r}^2\right)\text{for}H$

${\mathrm{σ}}_{\text{Solar}}\left(4\mathrm{π}{r}^2\right)=\mathrm{σ}e\left(\mathrm{π}{r}^2\right)T^4$

$T={\left[\frac{{\mathrm{σ}}_{\text{Solar}}}{4\mathrm{σ}}\right]}^{\frac{1}{4}}$

Putting , $2800$$w/m^2$for ${\mathrm{σ}}_{\text{Solar}}$$\text{and}5.6×{10}^{-8}\mathrm{W}/{\mathrm{m}}^2·{\mathrm{K}}^4\text{for}\mathrm{σ}$.

$T={\left[\frac{\left(2800\mathrm{W}/{\mathrm{m}}^2\right)}{4\left(5.67×{10}^{-8}\mathrm{W}/{\mathrm{m}}^2·{\mathrm{K}}^4\right)}\right]}^{\frac{1}{4}}$

$T=333K$

As the clouds reflect $77\%$incoming sunlight, then the remaining $23\%$will be absorbed Hence, the new solar constant will be

${\mathrm{σ}}_{\text{solar}}{}^{\text{'}}=(23\%){\mathrm{σ}}_{\text{solar}}$

$=(0.23)\left(2.8×{10}^3\mathrm{W}/{\mathrm{m}}^2\right)$

$=644\mathrm{W}/{\mathrm{m}}^2$

$\text{The new surface temperature of the Venusis}$

$T^{\text{'}}={\left[\frac{{\mathrm{σ}}_{\text{Solar}}}{4\mathrm{σ}}\right]}^{{]}^{\frac{1}{4}}}$

Putting the value of $644\mathrm{W}/{\mathrm{m}}^2\text{for}{\mathrm{σ}}_{\text{solar}}\text{'and}5.6×{10}^{-8}\mathrm{W}/{\mathrm{m}}^2·{\mathrm{K}}^4$for $\mathrm{σ}$.

$T^{\text{'}}={\left[\frac{644\mathrm{W}/{\mathrm{m}}^2}{4\left(5.67×{10}^{-8}\mathrm{W}/{\mathrm{m}}^2·{\mathrm{K}}^4\right)}\right]}^{\frac{1}{4}}$

$=231\mathrm{K}$

$\text{Hence, the surface temperature of Venus, when reflectivity is taking into account, is}231\mathrm{K}\text{.}$

The upper blanket send as much energy downward as it sends upward, so its total emission for each unit of sunlight is two units.

The equilibrium requires that it also absorb two units of infrared radiation and these two units must come from the lower blanket. Since the lower blanket is radiating twice as much as energy upward as the upper blanket, its temperature must be greater by a factor of $2^{\frac{1}{4}}$.

In the same way, the lower blanket must also send as much energy down as it sends up - in this case, two units, since it emits a total of four units of which one comes from the upper blanket. The other three must come from the ground, as a check note that the ground is absorbed two units from the lower blanket and one from the sun so it must emit three units. In order for the ground to emit three times as much as energy upward as the upper blanket, its temperature must be greater by a factor of $3^{\frac{1}{4}}$.

Hence, the ${70}^{\text{th}}$blanket is warmer than first blanket by the order of $(70{)}^{1/4}$. So, the ground is warmer than the order of $(71{)}^{1/4}$.$\text{The temperature of the Venusisgivenas}$

$$\begin{gathered} T_{\text{venus}}=(71{)}^{\frac{1}{4}}·T^{\text{'}} \\ {T}_{\text{venus}}=(71{)}^{\frac{1}{4}}(231\mathrm{K}) \\ =670\mathrm{K} \end{gathered}$$

$\text{Hence, the ground temperature of the Venus is}670\mathrm{K}$