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Chapter 11: Problem 17

Fig. \(10.1\) (p. 167 ) indicated that about 44,000 TW globally goes into evaporating water. We can turn this into an estimate of how much rain we expect per year, on average. The simplest way to do this is to think of a single square meter of ocean surface, receiving an average evaporation input power of \(120 \mathrm{~W} .^{44}\) Each millimeter of of water depth across our square meter has a volume of \(1 \mathrm{~L}\), or a mass of \(1 \mathrm{~kg}\). At a steady input of \(120 \mathrm{~W}\), \(^{45}\) how many millimeters of water are drawn off in a year? That same amount will come back down somewhere as precipitation.

Short Answer

Expert verified
**Question:** Based on the given evaporation input power of 120 W per square meter and the specific latent heat of vaporization, estimate the average amount of rainfall per year in millimeters. **Answer:** Using the calculations provided in the step-by-step solution, we find that the average amount of rainfall per year is equal to the depth (in millimeters) calculated in Step 5, which represents the amount of water evaporated in a year. To find the exact number, follow the steps and use the given values.

Step by step solution

01

Understand the given information

We are given the following information: - Globally, there is about 44,000 TW evaporating water. - Each square meter of ocean surface receives an average evaporation input power of 120 W. - Each millimeter of water depth across a square meter has a volume of 1 L and a mass of 1 kg.
02

Calculate the total energy input in a year

We are given that the steady input power received by each square meter of ocean surface is 120 W. To find out the total energy input in a year, we need to multiply the power by the total seconds in a year: Total energy/year = Energy input power * total seconds/year Total energy/year = 120 W * (365 * 24 * 3600) s/year
03

Find the energy required to evaporate 1 L (1 kg) of water

To calculate the energy required to evaporate 1 L (1 kg) of water, we use the specific latent heat of vaporization, which is approximately 2260 kJ/kg for water at room temperature. Since we are giving the energy in joules, we need to convert kJ to J: Energy required to evaporate 1 kg of water: 2260 kJ/kg * (1000 J/1 kJ) = 2260000 J/kg
04

Calculate the mass of water evaporated in a year

Now, we will find the mass of water evaporated in a year using the total energy per year calculated in Step 2 and the energy required to evaporate 1 kg of water from Step 3: Mass of water evaporated/year = Total energy/year / Energy required to evaporate 1 kg of water Mass of water evaporated/year = (120 * (365 * 24 * 3600) * J) / 2260000 J/kg
05

Convert mass to depth (millimeters)

We know that each millimeter of water depth across a square meter has a mass of 1 kg. To find out how many millimeters of water are drawn off in a year, we simply set the mass of water evaporated per year equal to the depth in millimeters: Depth (mm) = Mass of water evaporated/year
06

Conclusion

The depth (in millimeters) calculated in Step 5 represents the amount of water evaporated in a year, which can be considered as the average amount of rainfall per year as the same amount will come back as precipitation.

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Most popular questions from this chapter

A gallon of gasoline contains about \(130 \mathrm{MJ}\) of chemical energy at a mass of around \(3 \mathrm{~kg}\). How high would you have to lift the gallon of gasoline to get the same amount of gravitational potential energy? Compare the result to the radius of the earth.

While the Chief Joseph Dam on the Columbia River can generate as much as \(2.62 \mathrm{GW}\left(2.62 \times 10^{9} \mathrm{~W}\right)\) of power at full flow, the river does not always run at full flow. The average annual production is 10.7 TWh per year \(\left(10.7 \times 10^{12} \mathrm{Wh} / \mathrm{yr}\right)\). What is the capacity factor of the dam: the amount delivered vs. the amount if running at \(100 \%\) capacity the whole year?

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