Chapter 9, Problem 8Q

(a)

To determine

The amount of sunlight, in terms of power (in watt), that will be reflected back into space due to Earth’s albedo. Given that the total power of the sunlight that reaches the top of the atmosphere is $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$.

Answer to Problem 8Q

Solution:

$\text{5}\text{.4}\times {\text{10}}^{\text{14}}\text{W}$

Explanation of Solution

Given data:

The total power of the sunlight that reaches the top of the atmosphere is $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$.

Formula used:

The expression for the total power of reflected light is,

$\text{Power}\text{of}\text{reflected}\text{light}=\left(\text{Albedo}\right)\left(\text{total}\text{insolation}\right)$

Explanation:

The amount of sunlight that will be reflcted from the Earth’s surface can be calculated using the following formula.

$\text{Power}\text{of}\text{reflected}\text{light}=\left(\text{Albedo}\right)\left(\text{total}\text{insolation}\right)$

Substitute 0.31 for Albedo and $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$ for total insolation.

$\begin{array}{c}\text{Power}\text{of}\text{reflected}\text{light}=\text{0}\text{.31}\left(\text{1}\text{.75}\times {\text{10}}^{\text{17}}\text{ W}\right)\\ =\text{5}\text{.4}\times {\text{10}}^{\text{16}}\text{ W}\end{array}$

Conclusion:

Therefore, due to Earth’s albedo, $\text{5}\text{.4}\times {\text{10}}^{\text{14}}\text{W}$ of incoming solar radiation will be reflected back into the space.

(b)

To determine

The amount of radiation (in terms of power) that would be emitted by the Earth in the absence of atmosphere, given that the total power of the sunlight that reaches the top of the atmosphere is $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$.

Answer to Problem 8Q

Solution:

$1.21\times {10}^{17}W$

Explanation of Solution

Given data:

The total power of the sunlight that reaches the top of the atmosphere is $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$.

Formula used:

The expression for the total power of radiated light is,

$\text{Power}\text{of}\text{radiated}\text{light}=\text{Amount of radiation absorbed or radiated}\times \text{total}\text{insolation}$

Explanation:

As the amount of albedo is 0.31, the amount of absorbed radiation will be $1-0.31=0.69$. In the absence of atmosphere, the amount of radiation that will be absorbed by the Earth’s surface will be radiated back into space in its entirety.

The expression for radiated light is,

$\text{Power}\text{of}\text{radiated}\text{light}=\text{Amount of radiation absorbed or radiated}\times \text{total}\text{insolation}$

Substitute 0.69 for amount of radiation absorbed or radiated and $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$ for total insolation.

$\begin{array}{c}\text{Power}\text{of}\text{radiated}\text{light}=\text{0}\text{.69}\times \left(\text{1}{\text{.75×10}}^{\text{17}}\text{ W}\right)\\ =1.21\times {\text{10}}^{\text{17}}\text{ W}\end{array}$

Conclusion:

Therefore, the amount of radiation that would be emited by the Earth in the absence of atmosphere will be $1.21\times {10}^{17}W$.

(c)

To determine

The amount of radiation (in terms of power) that would be radiated by one square meter of the Earth’s surface in the absence of atmosphere, given that the total power of the sunlight that reaches the top of the atmosphere is $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$.

Answer to Problem 8Q

Solution:

$237{\text{ W/m}}^{\text{2}}$

Explanation of Solution

Given data:

The total power of the sunlight that reaches the top of the atmosphere is $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$.

Formula used:

In order to calculate the amount of radiations that would be radiated by one square meter of the Earth’s surface or flux, the following formula will be used:

$\text{Flux}=\frac{P_{\text{re-radiated}}}{\text{Surface}\text{area}\text{of}\text{the}\text{Earth}}$

Here, P stands for power of radiation.

The surface area of a spherical body is calculated using the following formula:

$A=4\pi r^2$

Here, r is radius.

Explanation:

From part (b), the re-radiated radiation from the entire surface or power of radiation (P) of the Earth in the absence of atmosphere was calculated to be $1.21\times {10}^{17}\text{ W}$.

Consider Earth to be a perfect sphere and recall the formula for the surface area of a sphere (Earth).

$A=4\pi r^2$

Substitute $\text{6}\text{.38}\times {\text{10}}^{\text{6}}\text{ m}$ for r (radius of Earth).

$\begin{array}{c}A=\text{4}\pi \left(\text{6}\text{.38}\times {\text{10}}^{\text{6}}\text{ m}\right)\\ =5.1\text{1}\times {\text{10}}^{\text{14}}{\text{ m}}^{\text{2}}\end{array}$

Recall the expression for calculating flux.

$\text{Flux}=\frac{P_{\text{re-radiated}}}{\text{Surface}\text{area}\text{of}\text{the}\text{Earth}}$

Substitute $1.21\times {10}^{17}\text{ W}$ for $P_{\text{re-radiated}}$ and $5.1\text{1}\times {\text{10}}^{\text{14}}{\text{ m}}^{\text{2}}$ for A.

$\begin{array}{c}\text{Flux}=\frac{\text{1}\text{.21}\times {\text{10}}^{\text{17}}\text{ W}}{\text{5}\text{.11}\times {\text{10}}^{\text{14}}{\text{m}}^{\text{2}}}\\ ={\text{237 W/m}}^{\text{2}}\end{array}$

Conclusion:

The power of radiation that would be emitted by one square meter of the Earth’s surface in the absence of atmosphere will be $237{\text{ W/m}}^{\text{2}}$.

(d)

To determine

The average temperature of the surface of the Earth in both Kelvin (K) as well as in degree Celsius (°C), given that the total power of the sunlight that reaches the top of the atmosphere is $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$.

Answer to Problem 8Q

Solution:

$\text{254 K}\text{and }-\text{19}°\text{C}$

Explanation of Solution

Given data:

The total power of the sunlight that reaches the top of the atmosphere is $\text{1}{\text{.75×10}}^{\text{17}}\text{ W}$.

Formula used:

For calculating the temperature of the Earth’s surface in Kelvin, the Stefan-Boltzmann formula can be used.

$\begin{array}{c}F=\sigma T^4\\ T=\sqrt[4]{\frac{F}{\sigma }}\end{array}$

Here, T is the temperature, $\sigma$ is the Stefan-Boltzmann constant $\left(\text{5}{\text{.67×10}}^{\text{-8}}{\text{ W/m}}^{\text{2}}{\text{K}}^{\text{4}}\right)$ and F is flux.

Temperature in °C can be calculated using the following formula:

$\text{Temperature in }°\text{C}=\text{temperature in K}-\text{273}\text{.15}$

Introduction:

For calculating the average temperature of the Earth’s surface, the Stefan-Boltzmann law is applied. As per this law, the higher the temperature of a surface, the more energy it radiates. This radiated energy is no dobt less than that radiated by the Sun.

Explanation:

The value of flux (F) is $237{\text{ W/m}}^{\text{2}}$ as calculated in part (c).

Write the expression for the Stefan-Boltzomann forlmula in terms of T.

$T=\sqrt[4]{\frac{F}{\sigma }}$

Substitute ${\text{237 W/m}}^{\text{2}}$ for F and $5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}{\text{K}}^{\text{4}}$ for $\sigma$.

$\begin{array}{c}T=\sqrt[4]{\frac{237{\text{ W/m}}^{\text{2}}}{5.67\times {10}^{-8}{\text{ W/m}}^{\text{2}}{\text{K}}^{\text{4}}}}\\ =254\text{ K}\end{array}$

Recall the formula to calculate tmerperature in degree Celsius.

$\text{Temperature in }°\text{C}=\text{temperature in K}-\text{273}\text{.15}$

Substitute 254 K for temperature in K.

$\begin{array}{c}\text{Temperature in }°\text{C}=\text{254 K}-\text{273}\text{.15}\\ \text{=}-\text{19}°\text{C}\end{array}$

Conclusion:

Therefore, the average surface temperature of the Earth in Kelvin and degree Celsius is $\text{254 K}\text{and}-\text{19}°\text{C}$, respectively.

(e)

To determine

The reason that Earth’s actual temperature is higher than the one calculated in part (d).

Answer to Problem 8Q

Solution:

The Earth’s actual surface temperature is higher than the one calculated in part (d) because of greenhouse effect.

Explanation of Solution

Introduction:

The values calculated in parts (b), (c), and (d) have been obtained by assuming that there is no atmosphere present on the Earth. However, this is not the case as the atmosphere of the Earth is made up of abundant gases, like water vapor, carbon dioxide, oxygen, nitrogen, methane, and other trace gases.

Explanation:

The atmosphere of Earth is made of gases called “greenhouse” gases that trap the outgoing long infrared radiations after sunset. These radiations, thus trapped, maintain the temperature of the atmosphere and prevent the Earth from turning ice cold at night.

In case the concentration of greenhouse gases like carbon dioxide, methane, vapor, and nitrogen oxide, increases in the atmosphere, the greenhouse effect will escalate. This raises the temperature of the Earth’s surface, which is more than what was calculated in part (d) of the question.

Conclusion:

Therefore, the average temperature of the Earth’s surface is more than that calculated in part (d) because of the presence of greenhouse gases in the atmosphere, which prevents the temperature from dipping below freezing point at night.