Fundamentals of Atmospheric Science

3.4 Solving Energy Problems Involving Phase Changes and Temperature Changes


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 as cpd. 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.

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.

melting icicles
Icicles melting. The energy for the phase change from ice to liquid water comes from the air, which must be warmer than freezing.
Credit: double drops by liz west from Flickr is licensed under CC BY 2.0.

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



1005 1850 4184 2106
Latent Heat (units: J kg–1)

Vaporization @ 0 oC


Vaporization @ 100 oC


Fusion @ 0 oC


Sublimation @ 0 oC

2.501 x 106 2.257 x 106 0.334 x 106 2.834 x 106
Temperature Change
Dry air Water vapor Liquid water Ice
cpd md ΔT = cpd ρdV ΔT cpv mv ΔT= cpv ρvV ΔT cw mliquid ΔT ci mice ΔT
Phase Change
vapor→liquid liquid→vapor vapor→ice ice→vapor liquid→ice ice→liquid
lv mvapor lv mliquid ls mvapor ls mice lf mliquid lf 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= l f m ice =( 0.334× 10 6  J kg 1 )( 10.0 kg )=3.34× 10 6 J MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaamaavacabeWcbeqaaiaaygW7a0qaaabbaaaaaaaaIXwyJTgapeGaey4kIipaaOGaamyuaiaadsgacaWG0bGaeyypa0JaamiBa8aadaWgaaWcbaWdbiaadAgaa8aabeaak8qacaWGTbWdamaaBaaaleaapeGaamyAaiaadogacaWGLbaapaqabaGcpeGaeyypa0ZaaeWaaeaacaaIWaGaaiOlaiaaiodacaaIZaGaaGinaiabgEna0kaaigdacaaIWaWdamaaCaaaleqabaWdbiaaiAdaaaGcpaGaaeiiaiaabQeacaqGGaGaae4AaiaabEgadaahaaWcbeqaaiabgkHiTiaaigdaaaaak8qacaGLOaGaayzkaaGaaGPaVpaabmaabaGaaGymaiaaicdacaGGUaGaaGima8aacaqGGaGaae4AaiaabEgaa8qacaGLOaGaayzkaaGaeyypa0JaaG4maiaac6cacaaIZaGaaGinaiabgEna0kaaigdacaaIWaWdamaaCaaaleqabaWdbiaaiAdaaaGccaaMc8UaaeOsaaaa@688A@

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= l f m ice + c i m ice ΔT =( 0.334× 10 6  J kg 1 )( 10.0 kg )+( 2106  J kg 1  K 1 )( 10.0 kg )( 0.020.0 )K =3.76× 10 6  J MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOabaeqabaWaaubiaeqaleqabaGaaGzaVdqdbaaeeaaaaaaaaigBHn2Aa8qacqGHRiI8aaGccaWGrbGaamizaiaadshacqGH9aqpcaWGSbWdamaaBaaaleaapeGaamOzaaWdaeqaaOWdbiaad2gapaWaaSbaaSqaa8qacaWGPbGaam4yaiaadwgaa8aabeaak8qacqGHRaWkcaWGJbWdamaaBaaaleaapeGaamyAaaWdaeqaaOWdbiaad2gapaWaaSbaaSqaa8qacaWGPbGaam4yaiaadwgaa8aabeaak8qacqqHuoarcaWGubaabaGaeyypa0ZaaeWaaeaacaaIWaGaaiOlaiaaiodacaaIZaGaaGinaiabgEna0kaaigdacaaIWaWdamaaCaaaleqabaWdbiaaiAdaaaGcpaGaaeiiaiaabQeacaqGGaGaae4AaiaabEgadaahaaWcbeqaaiabgkHiTiaaigdaaaaak8qacaGLOaGaayzkaaGaaGPaVpaabmaabaGaaGymaiaaicdacaGGUaGaaGima8aacaqGGaGaae4AaiaabEgaa8qacaGLOaGaayzkaaGaey4kaSYaaeWaaeaacaaIYaGaaGymaiaaicdacaaI2aWdaiaabccacaqGkbGaaeiiaiaabUgacaqGNbWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaaeiiaiaabUeadaahaaWcbeqaaiabgkHiTiaaigdaaaaak8qacaGLOaGaayzkaaWaaeWaaeaacaaIXaGaaGimaiaac6cacaaIWaWdaiaabccacaqGRbGaae4zaaWdbiaawIcacaGLPaaacaaMc8+aaeWaaeaacaaIWaGaaiOlaiaaicdacqGHsislcqGHsislcaaIYaGaaGimaiaac6cacaaIWaaacaGLOaGaayzkaaWdaiaabUeaaeaapeGaeyypa0JaaG4maiaac6cacaaI3aGaaGOnaiabgEna0kaaigdacaaIWaWdamaaCaaaleqabaWdbiaaiAdaaaGcpaGaaeiiaiaabQeaaaaa@91CC@

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 ¯ ΔA AΔt= l f m ice + c i m ice ΔT Δt= l f m ice + c i m ice ΔT ΔQ ¯ ΔA A = 3.76× 10 6  J ( 191  J m 2  s 1 )( 2.09  m 2 ) =9.42× 10 3 s=2.6 h MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@A230@

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= c pd m air Δ T air = l f m ice + c i m ice Δ T ice m air = l f m ice + c i m ice Δ T ice c pd Δ T air = 3.76× 10 6  J ( 1005  J kg 1  K 1 )( 20.0 K ) =187kg MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@A663@

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

Melting Ice
Credit: Dutton Institute. "METEO 300: Melting Ice Example." YouTube. May 14, 2015.
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!