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The rate of solar energy striking Earth averages 168 watts per square meter. The rate of energy radiated from Earth's surface averages 390 watts per square meter. Comparing these numbers, one might expect that the planet would cool quickly, yet it does not. Why not?

Short Answer

Expert verified
In summary, the Earth does not cool down quickly despite radiating more energy than it receives from the sun because factors like the greenhouse effect, albedo, and the role of the atmosphere help maintain its energy balance. The greenhouse effect helps retain heat within the Earth's atmosphere, preventing the planet from cooling down too quickly. The Earth’s albedo, or reflectivity, affects its ability to absorb solar energy and maintain a balanced temperature. The Earth's atmosphere also plays a critical role, acting as a blanket to retain energy and warming the lower layers. These factors combine to maintain the Earth's temperature despite the difference in incoming and outgoing energy rates.

Step by step solution

01

Understand the Earth's energy balance

Earth's energy balance is mainly maintained by the absorption and reflection of solar radiation and the radiation of energy back into space. The amount of solar energy absorbed by Earth, mainly from shortwave radiation, is equal to the amount of energy radiated back to space in the form of longwave radiation. This balance is crucial for maintaining Earth's temperature.
02

Consider the greenhouse effect

The greenhouse effect plays an essential role in maintaining Earth's temperature. Greenhouse gases, such as water vapor, carbon dioxide, and methane, absorb part of the outgoing longwave radiation radiated by Earth's surface. These gases then reradiate the energy in all directions, with some energy going back to Earth's surface. The greenhouse effect helps retain heat within the Earth's atmosphere, preventing the planet from cooling down too quickly.
03

Reflect on albedo

Albedo is a measure of the Earth's reflectivity, which affects its ability to absorb solar energy. A higher albedo means more sunlight is reflected back into space, while a lower albedo means more sunlight is absorbed by the Earth. The Earth's albedo is determined by various factors, such as the presence of clouds, ice, snow, and vegetation. Earth's average albedo is around 30%, meaning 30% of incoming solar radiation is reflected back into space, and about 70% is absorbed by the Earth.
04

Discuss the atmosphere's role

The Earth's atmosphere plays a vital role in the energy balance. It absorbs and scatters incoming solar radiation and radiates energy back into space. The atmosphere also acts as a blanket, helping to retain some of the energy radiated by the Earth's surface within the atmospheric layers. This results in the warming of the Earth's lower atmosphere, known as the troposphere.
05

Conclusion

In summary, the Earth does not cool down quickly despite radiating more energy than it receives from the sun because of factors such as the greenhouse effect, albedo, and the role of the atmosphere. The Earth's energy balance and its interaction with the atmosphere maintain the Earth's temperature and keep it from cooling too quickly.

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Most popular questions from this chapter

The solar power striking Earth every day averages 168 watts per square meter. The peak electrical power usage in New York City is 12,000 megawatts. Considering that present technology for solar energy conversion is only about \(10 \%\) efficient, from how many square meters of land must sunlight be collected in order to provide this peak power? (For comparison, the total area of the city is \(830 \mathrm{~km}^{2}\).)

What properties make a substance a good coagulant for water purification?

Phosphorus is present in seawater to the extent of 0.07 ppm by mass. If the phosphorus is present as phosphate, \(\mathrm{PO}_{4}{ }^{3-}\), calculate the corresponding molar concentration of phosphate in seawater.

The enthalpy of evaporation of water is \(40.67 \mathrm{~kJ} / \mathrm{mol}\). Sunlight striking Earth's surface supplies \(168 \mathrm{~W}\) per square meter \((1 \mathrm{~W}=1 \mathrm{watt}=1 \mathrm{~J} / \mathrm{s}) .\) (a) Assuming that evaporation of water is only due to energy input from the Sun, calculate how many grams of water could be evaporated from a 1.00 square meter patch of ocean over a 12 -hour day. (b) The specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\). If the initial temperature of a 1.00 square meter patch of ocean is \(26^{\circ} \mathrm{C},\) what is its final temperature after being in sunlight for 12 hours, assuming no phase changes and assuming that sunlight penetrates uniformly to depth of \(10.0 \mathrm{~cm}\) ?

(a) What are trihalomethanes (THMs)? (b) Draw the Lewis structures of two example THMs.

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