Understanding the Hydrologic Cycle: Key Concepts and Applications

Lab 2: Hydrologic Cycle Adapted for UMD GEOG 211 by Amanda Hoffman-Hall Sources: Living Physical Geography (Theodore I. Erski), NCDC, USGS, John Wiley and Sons, Thompson Higher Education In this lab we will delve deeper into certain parts of the Hydrosphere. The Hydrosphere is the combined mass of water found on, under, and above the surface of the planet. We’ll start with an interesting property of water, which may seem minor and small, but when applied to the ocean can have huge impacts! Then we’ll move to atmospheric moisture, skip surface water for now, and finish with groundwater. 1) Unique Property of Water - Specific Heat Specific heat is the amount of energy required to raise 1 gram of a substance 1°C. Different substances have different specific heat values, as seen in the table below. Water is unique in that it has a very high specific heat! When you know the specific heat of a substance you can figure out how many calories are needed to change the temperature of the substance. The relationship is expressed as follows: Q=cmΔT Q = calories added c = specific heat of substance m = mass of substance (in grams) ΔT = change in temperature

For example, how many calories must be added to 1 gram of granite to increase its temperature 5°C? Q=cmΔT Q = (0.19 calories/gram) x (1 gram) x (5°C) Q = 0.95 calories Another example, how many calories must be added to 1 gram of water to increase its temperature by 5°C? Q=cmΔT Q = (1 calorie/gram) x (1 gram) x (5°C) Q = 5 calories Notice it takes many more calories to increase the temperature of water 5°C than it takes for granite. You could test this by holding a granite rock in one hand and an equal amount of water in the other while standing outside in the sun. You will notice that the granite rock will begin to feel warm much more quickly than the water. If we need to determine temperature change as a result of adding calories, we simply rearrange the above equation as follows: ΔT = Q/cm For example, how would we calculate the temperature change of 1 gram of granite if we add 0.95 calories to it? ΔT = Q/cm ΔT = (0.95 calories)/((0.19 calories/gram) x (1 gram)) ΔT = 5°C Another example, how would we calculate the temperature change of 1 gram of wet mud if we add 3 calories to it? ΔT = Q/cm ΔT = (3 calories)/((0.60 calories/gram) x (1 gram)) ΔT = 5°C

Using the previous two equations, complete the following table. Hint: 1 kg = 1000 g . Pay attention to the mass unit! (1A, 1B, 1C each is worth 4 pts) The graphic below shows annual monthly mean temperatures of four different Southern African cities. While many factors are responsible for temperature, some differences are more obvious than others. Mombassa is generally the warmest city shown thanks to being closer to the Equator than the others. Notice though, that the annual lines for Gaborone and Johannesburg shift widely throughout the year, whereas Mombassa and Walvis Bay only experience slight seasonal differences in temperatures. Gaborone and Walvis Bay are situated at similar latitudes, but Gaborone experiences hot summers and cold winters , while Walvis Bay experiences only minor seasonal differences in temperature .

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1D) Despite being located at similar latitudes, Gaborone and Johannesburg exhibit more pronounced seasonal changes than Walvis Bay and Mombasa due to their [location1]. This absence of the [effect1], which involves the [property1] of water, allows for more temperature extremes. Conversely, the consistent temperatures of Walvis Bay and Mombasa can be attributed to their [location2], which allows for the [effect2] to leverage the [property1] to moderate seasonal variations. (5 pts) 2) Humidity There are three different terms commonly used to describe humidity. Specific Humidity, Maximum Humidity, and Relative Humidity. Specific Humidity is the easiest to describe. Specific Humidity refers to the amount of water vapor that is actually IN a parcel of air. Typically expressed as grams of water vapor per kilogram of air (g/kg). In contrast to Specific Humidity, Maximum Humidity refers to how much water vapor COULD BE in a parcel of air. You can also think of this as the CAPACITY of a parcel of air to hold water vapor. The maximum humidity of a parcel of air is temperature dependent. The graph and table below show this relationship.

***keep this table handy – you’ll be using it for the rest of the lab! 2A) Raising a kilogram’s air temperature from -10°C to 0°C increases its maximum humidity (capacity) by how many g/kg? (3 pts)

2B) Raising a kilogram’s air temperature from 20°C to 35°C increases its maximum humidity (capacity) by how many g/kg? (3 pts) Knowing Specific and Maximum Humidity can be interesting, but often doesn’t give the general public a good idea of what it will actually feel like outside. For example, a TV meteorologist could tell their viewers that the specific humidity is 5 g of water per every 1 kg of air. But what would that really mean to the public? 5 g of water is no big deal within a kilogram of 40°C air – however, 5 g of water would feel very wet indeed within a kilogram of 10°C air. Since the 10°C parcel can only hold 7 grams total of water vapor, it will be nearly full! Whereas the 40°C parcel still has lots of room for more vapor until it reaches saturation. Therefore, meteorologists tend to use Relative Humidity to describe humidity conditions. Relative Humidity refers to what percentage of the parcel of air contains water vapor (i.e. the ratio of Specific Humidity to Maximum Humidity!) Relative Humidity (%) = (Specific Humidity/Maximum Humidity) x 100 For example, what is the relative humidity of a parcel of air that has a specific humidity of 10 g and a maximum humidity of 20 g/kg? Relative Humidity = (10/20) x 100 Relative Humidity = 0.5 x 100 Relative Humidity = 50% Another example, what is the relative humidity of a parcel of air that has a specific humidity of 10 g and a temperature of 25°C? Relative Humidity = (10/20 * ) x 100 Relative Humidity = 0.5 x 100 Relative Humidity = 50% *Note: to find the Max Humidity of a 25°C parcel, use the Maximum Humidity table shown earlier. 2C) What is the relative humidity of a parcel of air that has a specific humidity of 5 g/kg and a maximum humidity of 25 g/kg? (4 pts) 2D) What is the relative humidity of a parcel of air that has a specific humidity of 5 g/kg and a temperature of 15°C? (4 pts)

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2E) Assuming specific humidity remains constant – decreasing the temperature of a parcel of air increases the _______ humidity. (4 pts) 2F) Assuming specific humidity remains constant – decreasing the temperature of a parcel of air decreases the _______ humidity. (4 pts) 3) Clouds The last two questions from Section 2 revealed how decreasing the temperature of a parcel of air can alter the humidity. But how does a parcel of air decrease in temperature? We can’t just stick the parcel into a refrigerator! Recall from lecture the graph to the right. Beneath the Tropopause, air decreases in temperature as it climbs in elevation! The amount of temperature lost per increase in elevation is known as the Environmental Lapse Rate (ELR) . If a parcel of air containing water vapor is warmer than the air surrounding it, it will rise at a rate different from the ELR known as the Dry Adiabatic Lapse Rate (DAR). There is also a WET Adiabatic Lapse Rate. This rate applies after a parcel reaches saturation, i.e. 100% Relative Humidity. Before that point, even up to 99.9% Relative Humidity, the DRY rate applies.

The point at which a parcel becomes saturated (reaches 100% Relative Humidity) is known as the Condensation Level . You can also think of this level as when a parcel will become a cloud! To calculate the Condensation Level of a parcel of air you need to know the following: the parcel’s temperature, the dew point temperature, and the dry adiabatic rate. CL = (T ap – T dp ) / DAR T ap = temperature of air parcel T dp = dew-point temperature DAR = dry adiabatic rate (constant of 1°C/100m) For example, a parcel of air has a specific humidity of 7 g/kg and a temperature of 15°C . Find the Condensation Level. Step 1: Find T ap T ap is given in the question as 15°C. Step 2: Find T dp The dew point temperature is the temperature at which the parcel would reach saturation, i.e. 100% Relative Humidity. From what we’ve learned about RH, we know that 100% RH is achieved when Specific Humidity is equal to Maximum Humidity . Using the Maximum Humidity table from Section 2 of this lab, we look up whatever temperature corresponds to a Maximum Humidity equal to 7g/kg (the Specific Humidity given in the question). That temperature, according to the table, is 10°C. Step 3: Calculate Condensation Level (CL) We need to find how high this particular parcel will need to rise to cool from its original temperature of 15°C to its dew point temperature of 10°C. To do this we divide the necessary change in temperature by the rate at which the parcel will cool (the Dry Adiabatic Lapse Rate of 1°C/100m):

3A) A parcel of air has a specific humidity of 3.5 g/kg and a temperature of 10°C. What is the condensation level? (4 pts) 3B) A parcel of air has a specific humidity of 10 g/kg and a temperature of 35°C. What is the condensation level? (4 pts) Unfortunately calculating the condensation level doesn’t tell us if the parcel will ever actually condense to form a cloud . This occurring depends on the Environmental Lapse Rate (ELR) of the surrounding air. If the Adiabatic Lapse Rate is greater than the ELR , air is stable and the parcel will not rise. If the Adiabatic Lapse Rate is less than the ELR , air is unstable and the parcel will continue to rise until it reaches equilibrium with the surrounding air.

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How does this work? Imagine an asphalt parking lot, much warmer than the surrounding grassy areas (or a ground fire, like in the image to the right – see the pyro-cumulus cloud!). A bubble of warm air is created just above the parking lot, compared to the cooler still air around it. As this bubble rises, it cools down according to the dry adiabatic lapse rate (10°C per km), while the still air around it cools according to the environmental lapse rate, which is 12°C per km in this case (the ELR is dependent on many things, beyond the scope of this introductory course). Therefore, the original parking lot air bubble cools down slower than the surrounding air. This means it is always slightly warmer than the surrounding air and continues to rise past its Condensation Level, which causes a cloud! This is known as unstable air and results in many different types of clouds . Stable air on the other hand, is when an air parcel like our parking lot bubble, cools down as fast (or faster) than the surrounding air, causing it to sink back down again. No clouds, no precipitation . Two parcels of air are formed, one over a parking lot in College Park with a specific humidity of 5g/kg and temperature of 15°C. The other is formed over a metal roof in Baltimore with a specific humidity of 14g/kg and temperature of 25°C. In College Park, the Environmental Lapse Rate is 13˚C/km, in Baltimore it is 7˚C/km. 3C) What is the condensation level of the College Park parcel? Report in kilometers. (3 pts) 3D) What is the condensation level of the Baltimore parcel? Report in kilometers. (3 pts) 3E) Which of these two parcels will actually form a cloud? (3 pts)

Differences in air stability can create many different types of clouds. While cloud formation in detail is beyond the scope of this class, with your knowledge of stable and unstable air, as well as the graphic below describing cloud heights and types, you can make some inferences into what types of clouds may result from different scenarios . Utilizing your critical thinking skills, answer the following question. 3F) Consider an area where still air has an ELR of 15°C/km, up to 8km above sea level. The dry adiabatic lapse rate is 10°C/km, and the wet adiabatic lapse rate is 5°C/km. A particular parcel has a condensation level of 500m. Which of the following clouds is most likely to form? (5 pts) NASA's Worldview tool provides the ability to interactively browse global, full- resolution satellite imagery and then download the underlying data. You can view and analyze various aspects of our planet's ecosystem, including cloud formations, which are a vital part of the hydrologic cycle. By using the Worldview tool, you can observe cloud cover on a global scale or zoom in to see detailed cloud formations over specific regions. The large datasets

provide the opportunity to track and study weather patterns, analyze the different types of cloud formations. Step 1: Visit the Worldview website https://worldview.earthdata.nasa.gov/ Step 2: Choose the clouds dataset Step 3: Navigate the worldview tool! And read the notes on the right bottom during the steps.

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3G) In the Pacific Northwest, USA, marine stratocumulus clouds often form due to a process known as coastal upwelling. Where are marine stratocumulus clouds most likely to form? (5 pts) 3H) Examine the screenshot from the NASA Worldview tool provided. It shows a particular cloud formation pattern over the ocean. What is this pattern primarily called? (5 pts)

4) Groundwater Shifting from our detailed discussion of atmospheric water, we’re now going to move to another part of the hydrologic cycle. Skipping over surface water which we’ll learn about later in this course (rivers, lakes, etc.), we’ll move on to another type of water even more difficult to see than atmospheric water – groundwater! Read pp.196-203 in your textbook, and answer the following questions. 4A) Groundwater provides about ____ % of the world’s irrigation water for agriculture and nearly half the world’s drinking water. (3 pts) 4B) From 1950 to 2000, how much did groundwater withdrawal increase in the U.S. and Canada? (3 pts) 4C) Which continent does not have major aquifers? (3 pts)

4D) True/False: Over-pumping of groundwater can lead to a phenomenon known as a "cone of depression". (3 pts) 4E) The groundwater issue in the High Plains Aquifer is primarily related to: (3 pts) 4F) A town uses water from a local river to cool its power plant and then returns that water back to the river. The same town also uses river water for agricultural irrigation where much of the water is lost to evapotranspiration. In this scenario, the power plant's use of water is considered ___________, and the use of water for irrigation is considered ___________. (3 pts) In the United States, one of the most used (and consequently, threatened) groundwater supplies is the Ogallala Aquifer . Download OgallalaAquifer.kmz from ELMS and open in Google Earth. This map shows water-level changes in the Ogallala Aquifer from pre-development to the year 2011. Predevelopment is defined as before 1950, except in Texas, where it is defined as before 2000. 4G) Which of the following states are NOT over the Ogallala Aquifer? (3 pts) 4H) What does the red color symbolize in the Ogallala’s legend? (3 pts) 4I) Based on your observations from the Ogallala Aquifer data in Google Earth, which of the following statements best describes the overall trend in groundwater levels over time? (5 pts) You have now completed Lab 2. Make sure you submit all your answers before 11:59 pm on the due date!

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Lab 2 Hydrologic Cycle

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