Unit 3.4: Energy And Water Resourcesgeography

  1. Unit 3.4: Energy And Water Resources Geography Answer
  2. Unit 3.4: Energy And Water Resources Geography Textbook

3.4 Solving Energy Problems Involving Phase Changes and Temperature Changes

When a cloud drop evaporates, the energy to evaporate it must come from somewhere because energy is conserved according to the 1st Law of Thermodynamics. It can come from some external source, such as the sun, from chemical reactions, or from the air, which loses some energy and thus cools. Thus, temperature changes and phase changes are related, although we can think of phase changes as occurring at a constant temperature. The energy associated with phase changes drives much of our weather, especially our severe weather, such as hurricanes and deep convection. We can quantify the temperature changes that result from phase changes if we have a little information on the mass of the air and the mass and phases of the water.

In the previous lesson, we said that all changes of internal energy were associated with a temperature change. But the phase changes of water represent another way to change the energy of a system that contains the phase-shifter water. So often we need to consider both temperature change and phase change when we are trying to figure out what happens with heating or cooling.

For atmospheric processes, we saw that we must use the specific heat at constant pressure to figure out what the temperature change is when an air mass is heated or cooled. Thus the heating equals the temperature change times the specific heat capacity at constant pressure times the mass of the air. For dry air, we designate the specific heat at constant pressure ascpd. For water vapor, we designate the specific heat at constant pressure as cpv. So for example, the energy required to change temperature for a dry air parcel is cpd m ΔT = cpd ρV ΔT, where cpd is the specific heat capacity for dry air at constant pressure. If we have moist air, then we need to know the mass of dry air and the mass of water vapor, calculate the heat capacity of each of them, and then add those heat capacities together.

Energy must be added to increase the temperature to melt it; to evaporate it Energy is released when a substance cools condenses freezes Energy units: Ergs (1 erg = 1 g cm 2 s-2) Joules (1 J = 1 kg m 2 s-2 = 10 7 ergs) Calories (1 cal = 4.184 J) Total energy to raise temperature of 1.0 kg of water from 10°C to 110°C C p water = 4.2 x 10 3 J. In lesson 1, we will introduce the unit by discussing energy – the ability to do work. We will also be exploring the two major types of energy. Potential energy is stored energy, or energy that an object has because of its position relative to other objects. Kinetic Energy is energy that is doing work, or energy in motion.

Unit 3.4: Energy And Water Resourcesgeography

At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C. 2014 National Curriculum Resources » Geography » KS2 Geography Curriculum » Place Knowledge » Understand geographical similarities and differences through the study of human and physical geography of a region of the United Kingdom, a region in a European country, and a region within North or South America » Europe.

For liquids and solids, the specific heat at constant volume and the specific heat at constant pressure are about the same, so we have only one for liquid water (cw) and one for ice (ci).

For phase changes, there is no temperature change. Phase changes occur at a constant temperature. So to figure out the energy that must be added or removed to cause a phase change, we only need to know what the phase change is (melting/freezing, sublimating/depositing, evaporating/condensing) and the mass of water that is changing phase. So, for example, the energy needed to melt ice is lf mice.

Icicles melting. The energy for the phase change from ice to liquid water comes from the air, which must be warmer than freezing.

The following tables provide numbers and summarize all the possible processes involving dry air and water in its three forms.

Specific Heat Capacity at 0 oC (units: J kg–1 K–1)

Dry air


Water vapor


Liquid water



Latent Heat (units: J kg–1)

Vaporization @ 0 oC


Vaporization @ 100 oC


Fusion @ 0 oC


Sublimation @ 0 oC

2.501 x 1062.257 x 1060.334 x 1062.834 x 106
Temperature Change
Dry airWater vaporLiquid waterIce
cpd md ΔT = cpd ρdV ΔTcpv mv ΔT= cpv ρvV ΔTcw mliquid ΔTci mice ΔT
Phase Change
lv mvaporlv mliquidls mvaporls micelf mliquidlf mice


To solve energy problems you can generally follow these steps:

  1. Identify the energy source and write it on the left-hand side of the equation.
  2. Identify all the changes in temperature and in phase and put them on the right-hand side.
  3. You should know all of the variables in the equation except one. Rewrite the equation so that the variable of interest is on the left-hand side and all the rest are on the right-hand side.

Knowing how to perform simple energy calculations helps you to understand atmospheric processes that you are observing, and to predict future events. Why is the air chilled in the downdraft of the thunderstorm? When will the fog dissipate? When might the sun warm the surface enough to overcome a near-surface temperature inversion and lead to thunderstorms? We can see that evaporating, subliming, and melting can take up a lot of energy and that condensing, depositing, and freezing can give up a lot of energy. In fact, by playing with these numbers and equations, you will see how powerful phase changes are and what a major role they play in many processes, particularly convection.

With the elements in the tables above, you should be able to take a word problem concerning energy and construct an equation that will allow you to solve for an unknown, whether the unknown be a time or a temperature or a total mass.

In the atmosphere, these problems can be fairly complex and involve many processes. For example, when thinking about solar energy melting a frozen pond, we would need to think about not only the solar energy needed to change the pond from ice to liquid water, but we would also need to consider the warming of the land in which the pond rests and the warming of the air above the pond. Further, the land and the ice might absorb energy at different rates, so we would need to factor in the rates of energy transfer among the land and the pond and the air.

So we can make these problems quite complex, or we can greatly simplify them so that you will understand the basic concepts of energy required for temperature and phase changes. In this course, we are going to solve fairly simple problems and progress to slightly more complicated ones. Let’s look at a few examples. I will give you some examples and then you can do more for Quiz 3-3.

Example Problems

A small puddle is frozen and its temperature is 0 oC. How much solar energy is needed to melt all the ice? Assume that mice = 10.0 kg.

  1. The heating source is the sun and we are trying to calculate the total solar energy. Put this on the left-hand side.
  2. The change that we want is the melting of the ice. We know the mass and the latent heat. We write those on the right-hand side.
  3. The equation already has the unknown variable on the left-hand side.

Qdt=lfmice=(0.334×106 J kg1)(10.0 kg)=3.34×106J[email protected]@[email protected]@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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[email protected][email protected]

To put this amount of energy into perspective, this energy is equivalent to a normal person walking at about 4 mph for 2 hours (assuming the person burns 400 calories per hour, which is really 400 kilocalories per hour in scientific units).

Now let’s assume that the ice is originally at –20.0 oC. Now we have to both raise the temperature and melt the ice. If we don’t warm the ice, some of it will simply refreeze. Our equation now becomes:

Qdt=lfmice+cimiceΔT=(0.334×106 J kg1)(10.0 kg)+(2106 J kg1 K1)(10.0 kg)(0.020.0)K=3.76×106 J[email protected]@[email protected]@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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[email protected][email protected]

We see that the amount of energy required increased by about 25%. Most of the energy is still required to melt the ice, not change the temperature.

Now let’s assume that the solar heating rate is constant at 191 W m–2 and that the area of the puddle is 2.09 m2. How long does it take the sun to raise the temperature of the ice and then melt it?

Qdt=ΔQ¯ΔAAΔt=lfmice+cimiceΔTΔt=lfmice+cimiceΔTΔQ¯ΔAA=3.76×106 J(191 J m2 s1)(2.09 m2)=9.42×103s=2.6 h[email protected]@[email protected]@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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[email protected][email protected]

We could now assume that the source of heating is not the sun but instead is warm air passing over the puddle. If the temperature of the air is 20.0 oC and we assume that its temperature drops to 0.0 oC after contacting the ice, what is the mass of air that is required to warm the ice and then melt it?

Qdt=cpdmairΔTair=lfmice+cimiceΔTicemair=lfmice+cimiceΔTicecpdΔTair=3.76×106 J(1005 J kg1 K1)(20.0 K)=187kg[email protected]@[email protected]@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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[email protected][email protected]

See this video (2:28) for further explanation:

Melting Ice Unit 3.4: energy and water resources geography textbook
Click here for transcript of the Melting Ice Video.

Let's work a problem about melting a small frozen pond. When we set up the equations, we'll always put our source of heating or cooling on the left, and the things changing temperature or phase on the right. We'll start with the simplest case, and then introduce more information see how to solve the problems. The energy required to melt all the ice is simply the integral of the heating rate over time. That goes on the left side, because it's a source. The change that we're observing is the melting of the ice into liquid water. For this first part, there is no temperature change, just a phase change. We need to know the mass of the ice, and the latent heat of fusion, which tells us about the amount of energy required to convert ice to liquid water. The energy required is about three million joules. Now, let's complicate the problem a little more. Let's start with the ice at a temperature of minus 20 degrees C, but we are still interested in the energy required to melt the ice. Now, we need more energy. Well, we'll need to first raise the temperature of the ice to 0 degrees from minus 20 degrees C, and then we can melt it. So only two terms in the equation. So we add the second term, which accounts for the energy required to raise 10 kilograms of ice by 20 degrees C. Takes another million joules. Now we'll specify that the solar heating rate was 200 watts per meter squared. And now we know what the area of the puddle is. So the integral of the heating rate is just the average heating rate times time, if the heating rate is constant. And so that gives us a way to figure out how long it would take to melt all the ice. We write down the equation for heating on the left, and the changes on the right. The changes, remember, include both the melting of the ice, and the raising of the ice temperature from minus 20 degrees C to 0. Then we rearrange the equation so that only the times remain on the left. When we do this, the answer is that it takes about three hours.

Now it is your turn to solve some of these energy problems and then to take a quiz solving some more.

Quiz 3-3: Energy problems.

  1. Find Practice Quiz 3-3 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
  2. When you feel you are ready, take Quiz 3-3. You will be allowed to take this quiz only once. Good luck!

The first step in changing the way you use water in the future is by understanding how much water you use today. The best place to find this information is on your monthly water bill. Pull out your water bill and follow the steps below to learn more about it and your own water use.

On This Page:​

  • How much do you use?

Different utilities use different units for measuring water use. The most common units are centum cubic feet (CCF) and the gallon. A CCF also called an HCF (hundred cubic feet), represents one hundred cubic feet of water. The first 'C' comes from the Roman word for hundred, 'centum.” This is the most common unit used by both water and natural gas utilities. But you may be more familiar with the other unit, the gallon. One CCF is equal to 748 gallons.

What does your usage mean? The average American uses around 88 gallons per day per person in the household. That means a family of four would use around 10,500 gallons in a 30-day period. But usage varies a great deal across the country, mostly because of differences in weather patterns. For example, water use tends to be higher in drier areas of the country that rely more on irrigation for outdoor watering than in wetter parts of the country that can rely on more rainfall.

Based on information from Water Research Foundation, “Residential End Uses of Water, Version 2.” 2016; and The US Geological Survey, “Estimated Water Use in the United States.” 2010.

  • What is your usage trend?

Does your bill explain your household's usage trend? Some utilities provide graphs like the ones below that show how your water use has varied over the course of the year and previous years. This can be a helpful way of seeing when your own water use reaches its highest levels.

While using water efficiently is important throughout the year, sometimes the timing of water use can make a big difference for community water supplies—and your water bill. WaterSense has tips to help you reduce your water use when it's hot outside.

Unit 3.4: Energy And Water Resourcesgeography

Water utilities operate with this higher, summertime use in mind because they must be able to provide for all the water a community needs over an extended period. Some systems may be forced to restrict outdoor watering during the peak to ensure that water is available for more important community needs.


How does your use compare to that of your neighbor?

Torrent free movies download sites. Some utilities provide information on how your household compares to that of your neighbors. This can help you see how your usages stacks up versus other users in your same climate area and can be a helpful way of gauging your 'WaterSense.' Some utilities use bills that compare your use to a random group of your neighbors while some utilities use a 'tiered system' to differentiate users such as in the example below.

Image courtesy of Coachella Valley Water District.EXIT

How are you being charged?

Water utilities need to charge customers to build and maintain infrastructure—the water storage tanks, treatment plants, and underground pipes that deliver water to homes and businesses. The revenue is also used to pay the workers who provide you with water service day or night. There are a wide variety of rate structures that are used to bill customers, some of which are described below.

Rate Types

Flat Fee is a rate structure where all customers are charged the same fee, regardless of the amount of water used. Flat fees are the simplest type of rate structure and are rarely used today. They generally don’t provide revenue sufficient to operate the utility and are not good at promoting water efficiency.

Uniform Rate is a structure that has a constant per unit price for all metered units of water consumed on a year-round basis. It differs from a flat fee in that it requires metered service. Some utilities charge varying user groups different rates such as charging residential households one rate and industrial users a different rate. Constant block rates provide some stability for utilities and encourage conservation because the consumer bill varies with water usage.

Increasing Block Rates is a rate structure in which the unit price of each succeeding block of usage is charged at a higher unit rate than the previous block(s). Increasing block rates are designed to promote conservation and are most often found in urban areas and areas with limited water supplies. The graphic to the right is an example of an increasing block rate structure.

Declining Block Rates are the opposite of increasing block rates where the unit price of each succeeding block of usage is charged at a lower unit rate than the previous block(s). This rate structures are popular in rural areas that service large farming populations or areas with large users such as heavy industry and where water is plentiful.

Seasonal Rates are rates that cover a specific time period. They are established to encourage conservation during peak use periods. Examples of seasonal rates may be increases for the summer season due to increased demand associated with lawn watering and outside activities.

Drought Rates are similar to seasonal rates but instead of applying higher rates during an entire time period, they adjust rates based on the local area's drought level. Higher levels of drought result in higher prices for water in order to encourage conservation.

Water Budget Based Rates is a rate structure where households are given a 'water budget' based on the anticipated needs of that household either by the number of people living in the house and/or property size. Users are charged a certain rate for use within their budget and a higher rate for use that exceeds their budget.

Many utilities use a combination of a fixed fee (base) and a variable fee (volume) for their water rate structure. Fixed charges generally include the price the customer pays as a base charge to help cover costs for maintaining existing infrastructure and repaying loans and bonds used to build that infrastructure. Variable charges are the price the customer pays per volume of water used, which reflect the costs of providing water, such as costs for chemical treatment to provide safe water and energy to move and deliver water.


What are my charges going towards?

Most utilities will provide you with a breakdown of charges in your 'billing detail' or 'summary of charges' section. Note that some utilities measure both water entering the house and waste leaving to the sewer, but many utilities have only one meter on location and will charge both volumes based on water entering the house. This is yet another reason to reduce your own water use. If you're curious about what various surcharges and other charges on your utility bill mean, you can usually find that information either on the back or appendix of the bill or on your local water utility's website. Two examples are provided below.

Uniform Rate Example - in the first example, roughly half of the $147.62 being charged is directly related to water use. Most utilities charge a set flat fee (the 'Water Base Facility Charge' in the example) that helps to pay for the base costs of providing water including the electricity needed to transport and clean the water, the personnel and others costs of daily maintenance of the delivery system, and other fixed operating costs.

This utility uses a uniform rate structure that charges the user $0.00295 per gallon (or roughly 3 cents for every 10 gallons) used during the billing period. The bill also shows a similar facility charge for sewer and a 'rate case expense surcharge' to help pay for the utility's rate setting process. The 'regulatory assessment fee' helps the utility pay for costs associated with maintaining regulatory compliance with clean water statutes. Finally, some utilities charge fees similar to the 'Deferred Capital Expense Surcharge' which puts money into a fund to help pay for long term investments in improvements to infrastructure such as new pipes, treatment facilities or reservoirs.

Increasing Block Rate Example - this second bill is an example of an efficient user with an increasing block rate structure. You can see that the utility has even labeled the various blocks with its corresponding water use efficiency level. The above user falls into the 'Efficient' group and so avoids the much higher per unit costs of the next three tiers. Some utilities will forgive various surcharges for its most efficient users because their below average water use places less burden on the system and reduces demand for new sources of water and pipes to transport this water.

Unit 3.4: Energy And Water Resources Geography Answer

More Information

Utilities will often use the back of the bill as a 'message area.' This area will sometimes have information on rebate programs, water efficient products, or other tips on water conservation.

If you're looking for more information on how your bill functions, you can visit the following sites:

Unit 3.4: Energy And Water Resources Geography Textbook

  • For interactive examples of bills visit Understanding your Water Bill pages from the East Bay Municipal Bay Utility District (CA) and Cleveland (OH) Water.Exit
  • To learn more about what services are being paid for from water bills, visit the Financing Sustainable Water page for concerned citizens. Exit
  • For an example of an interactive, comparative utility bill, visit WaterSmart Software.Exit