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5.3: Energy and Phase Transitions

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    52108
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    Learning Objectives

    • Define heat of fusion and heat of vaporization
    • Relate heat of fusion and heat of vaporization to heat of freezing and heat of condensation
    • Label various part of a heating curve
    • Using a heating curve, calculate the heat input/output associated with phase changes and temperature changes using the proper equation(s)
    • Using a heating curve, calculate the final temperature for a given quantity of heat input/out.

    Energy and Phase Changes

    When adding or removing heat from a system one of three things can occur:

    1. Temperature Change within a phase (physical change)
    2. Phase Change between two phases (physical change)
    3. Adding heat can cause a chemical reaction (chemical change)

    In this section we shall look at changes between two phases. In reality there are 4 states of matter, solid, liquid, gas and plasma, although in this class we will focus on the three common states, and their transitions.

    Video \(\PageIndex{1}/): 0'49" YouTube uploaded by kosasihiskandarsjah describing the four states of matter (https://youtu.be/tEY0V8BD8K4)

    Figure \(\PageIndex{1}\) shows the phase transitions between the various phases and not the enthalpy of the system is a measure of the heat energy at constant pressure and will be introduced in section 5.4.3 of this Chapter.

    clipboard_ee00599283a333bddc6f1bc9fc65d7e3a
    Figure \(\PageIndex{1}\): Energy diagrams showing phase transitions. The diagram on the left shows all 4 states, while the one on the right shows the three common states. This class will not be dealing with transitions to the plasma state, but you should be aware of it. There are several things to note on the right diagram.
    1. Transitions going up (increased energy) are endothermic and require energy ,while those going down (decreased energy) are exothermic and release energy.
    2. The energies absorbed in the endothermic steps are identical in magnitude to the energies released in the exothermic steps for the same transition.
      ΔHFusion = -ΔHFreezing
      ΔHVaporization = -ΔHCondensation
      ΔHSublimation = -ΔHDeposition.
    3. Energy is conserved
      ΔHSublimation = ΔHFusion + ΔHVaporization
      ΔHDeposition = ΔHCondensation + ΔHFreezing

    Note: Endothermic and Exothermic Reactions

    We will formally introduce endothermic and exothermic reactions in the section 5.4 (the next section) when we discuss the First Law of Thermodynamics and define the direction of heat transfer (as being positive when something gains energy, and negative when something loses energy). So an exothermic process is one where energy is released to the environment and an endothermic one is where energy is absorbed from the environment. We are introducing the concept early because all the process going in direction of increased energy (S->L ->G) are endothermic (you add energy to boil water), and those of decreasing energy (G -> L -> S) are exothermic

    Exercise \(\PageIndex{1}\)

    Consider the process of water freezing.

    1. Is it an endothermic or exothermic process?
    2. Does it cool or warm the surroundings?
    Answer a

    Exothermic. It is easier for us to think about the opposite process of melting. It takes energy to melt ice, so heat is flowing into the system(the ice cube) and out of the surroundings (the air, let's say). So the surroundings (air) during the process of melting the ice. The opposite of melting ice is freezing water. So, you expect this process to be exothermic, releasing heat to the surroundings, and therefore, warming the surrounds.

    Answer b

    Warm

    Sublimation

    Students often have a hard time understanding sublimation. Carbon dioxide (dry ice) that you can buy at many grocery stores sublimes, that is, it goes directly from a solid to a gas without ever melting (which is why the call it "dry ice"). Video \(\PageIndex{2}\) demonstrates the sublimation and deposition of iodine.

    Video \(\PageIndex{2}\): 1'39" YouTube uploaded by koen2all showing sublimation and deposition of iodine (https://youtu.be/jX9pskbKSw0).

    Quantifying Phase Changes

    We have seen how heat is either released or absorbed during phase changes. Now, we will study the quantity of energy associated with those phase changes. In order to carry out those calculations, we need to establish a standard bases for comparison, the molar heat of the phase transition, or the heat required for one mole of substance to change phase. Particularly, molar heat of fusion, refers to the amount of heat (in kJ) it take for one mole of substance to transition from the solid state to the liquid state (melt). The molar heat of vaporization, is the amount of heat (in kJ) it takes for one mole of a substance to transition from the liquid state to the vapor phase. Since we are not always working with exactly one mole of each substance, we will use this equation to Figure out the heat for the particular number of moles that we have according to the following equation:

    q = nΔHtransition

    where n is the number of moles, and H is the heat or enthalpy required for one mole of a substance to go through the phase transition at the temperature where it occurs (which depends on the pressure). Unless otherwise noted, you can assume these are the boiling points and freezing point at 1 atm pressure. As you can see from Table 1, these values are usually given and tabulated. Since, we typically deal with the mass of substance, there are some tables that also give you the heat of transition of a substance on a per gram bases (usually reported as J/g) Here is a link the following tables heat of fusion and heat of vaporization tables:

    B2: Heats of Vaporization (Reference Table)

    B3: Heats of Fusion (Reference Table)

    clipboard_ea36d6f63b3d6a1900f33777bf742aace

    Table \(\PageIndex{1}\): Enthalpies of Fusion (melting) and Vaporization for some common substances, and the transition temperature where they occur (at normal pressure of 1 atm).

    Exercise \(\PageIndex{2}\)

    For a given substance, which is higher, ΔHFusion = ΔHVaporization and why?

    Answer

    From table \(\PageIndex{1}\) it is clear that the enthalpy of vaporization is larger than fusion. This is because when you vaporize something you overcome all of the intermolecular forces and the material becomes a gas, filling up the entire container. It takes much more energy to vaporize a substance then to melt it.

    Molar Heat of Freezing and Condensation

    You may have wondered why table \(\PageIndex{1}\) only provides the molar heat of fusion and vaporization, and not the opposite processes of freezing and condensation. For example, if we wanted to quantify the amount of heat it takes to freeze a certain mass of N2, what would be the molar heat of freezing? This is actually the negative of molar heat of melting or fusion, which is the heat energy required to melt one mole of a substance. So from the above diagram, it is - 0.71 kJ/mol. Note, this is a consequence of the First Law of Thermodynamics, the law of the conservation of energy, which we will discuss in detail in the next section. What it is stating is that the energy you must add to melt a mole of a substance is the same as the energy released when a mole freezes, because energy is conserved, but the energy flows in the opposite direction. Therefore, the above table not only gives you the enthalpies of fusion and vaporization, but also of freezing and condensation.

    ΔHFusion = - ΔHfreezing
    ΔHvaporization = -ΔHcondensation

    Molar Heat of Sublimation and Condensation

    Table \(\PageIndex{1}\) also provides the heats of sublimation and condensation. Can you Figure them out?

    ΔHSublimation = ΔHFusion + ΔHVaporization
    ΔHDeposition = - ΔHVaporization -ΔHFusion

    Note: Quantifying Heat Transfer within a Phase

    The heat being pumped into the system does not always go toward changing the phase of a substance. It may go into changing the temperature of a substance. For example, if you want to melt an ice cube that is at -10°C, you have to heat the ice to 0°C, before you can actually start the melting process. In order to quantity the amount of heat required to change the temperature (in the same phase, ice) from -10°C to 0°C, you can use the specific heat capacity equation that we covered in 5.2.

    specific heat capacity, q = mCΔT (section 5.2) . Remember, the specific heat capacity (c), is different depending on which phase that the substance is in.

    Heating Curves

    A heating curve is a graph of the temperature of a substance as a function of the amount of heat added (moving to the right) or removed (moving to the left), as depicted in Figure 5.3.2. Note the flat regions represent phase changes, and the slope of the line in the rising regions is related to the capacitance of the sample being measured (it is actually the reciprocal, or 1/mc).

    The concept of the heating curve is very important as it helps us Figure out what equation to use to quantify the heat associated with phase changes and temperature changes (watch the two videos below). At the beginning of this section, we stated that two of the three things that could happen when you add heat to a substance is that either it will undergo a phase change, or its temperature will rise. These are the two effects being described by the heating/cooling curve. The mathematical equations are:

    q=mc ΔT (energy of a temperature change within a phase)
    q=n ΔHtransition (energy of a phase transition)

    It needs to be realized that if you add heat, you move to the right, and if you remove heat, you move to the left.

    image.png
    Figure \(\PageIndex{2}\): Cooling Curve for Water. Note, Increasing heat moves to the left, removing heat moves to the right.

    What is wrong with the slopes of the lines in Figure \(\PageIndex{2}\)?

    For water c(liq) = 4.184(J/g-degC), c(ice) = 2.06(J/g-degC) and c(steam)=2.01(J/g-degC).

    The slopes would be different, and the slopes of the ice and steam would roughly be twice as steep as the water. That is, it takes more energy to raise the temperature of liquid water by 1 deg than it does for either steam or ice.

    What is wrong with the flat region of Figure \(\PageIndex{2}\)?

    They are roughly equal in length. In exercise \(\PageIndex{2}\) it was pointed out that ΔHVaporization >> ΔHFusion and so the flat region for vaporizing should be much longer than the one for melting. As this is such an effect, a zig-zag is often drawn into the flat region to indicate that it is off the scale of the graph (see videos below).

    Combined Temperate and Phase Change Problems

    The following three examples deal with adding different amounts of energy to 100.0g of ice starting at -10.0°C.

    Example \(\PageIndex{1}\) Calculation of Final Temperature

    What is the final temperature when 1.6 kJ of heat is added to 100.0 grams of ice at -10.0°C?

    Solution

    This problem requires you figuring out what phase the substance ends up in. Does 1.6kJ have enough enough energy to bring it to the freezing point? To melt it? To warm the liquid to its boiling point? To boil it? To raise the energy of a super heated steam?

    Video \(\PageIndex{3}\): 5'35" YouTube showing the final temperature after adding 1.6kJ of heat to 100.0g ice at -10.0°C, (https://youtu.be/pjhZVhFjHGc).

    Example \(\PageIndex{2}\) Calculation of Final Temperature

    What is the final temperature when 25.00 kJ of heat is added to 100.0 grams of ice at -10.0°C?

    Solution

    This problem requires you figuring out what phase the substance ends up in. Does 25.00kJ have enough enough energy to bring it to the freezing point? To melt it? To warm the liquid to its boiling point? To boil it? To raise the energy of a super heated steam?

    Video \(\PageIndex{3}\): 3'08" YouTube showing the final temperature after adding 25.00kJ of heat to 100.0g ice at -10.0°C, (https://youtu.be/3cjSloZOsHQ).

    Exercise \(\PageIndex{3}\)

    After all the ice has melted in the above problem, is there more ice, or liquid water?

    Answer

    Since it takes 33.24kJ to melt all the ice, and you have 22.94 left over after bring the supercooled ice to its freezing point, you have effectively applied the 22.94 kJ to the ice at zero degrees, and so the fraction melted is what you applied over what was required to melt it all. So 22.94/33.24 = 0.69, so it is 69% liquid water and 31% ice (69 grams of the 100 grams melted, and 31 are still frozen).

    Example \(\PageIndex{3}\) Calculation of Final Temperature

    What is the final temperature when 310.0 kJ of heat is added to 100.0 grams of ice at -10.0°C?

    Solution

    This problem requires you figuring out what phase the substance ends up in. Does 310.0kJ have enough enough energy to bring it to the freezing point? To melt it? To warm the liquid to its boiling point? To boil it? To raise the energy of a super heated steam?

    After adding 310 kJ you have enough energy to raise the super cooled water from -10.0oC to its freezing point (2.06kJ), melt the ice (33.42kJ), heat the liquid water to its boiling point (41.84kJ) and boil all the water (225.37kJ) using a total of 303.7 kJ, with 7.3kJ left to raise the temperature of the super-heated steam to a final temperature of 136oC. Notice how the majority of the energy was used in the fourth step, the boiling of the the water (225.37kJ out of 310kJ, or 72.7% of the entire energy). This makes sense, as in going from a liquid to a gas you are overcoming all of the forces holding the water molecules together.

    Video \(\PageIndex{3}\): 3'08" YouTube showing the final temperature after adding 310.0 kJ of heat to 100.0g ice at -10.0°C, (https://youtu.be/3cjSloZOsHQ).

    Contributors and Attributions

    • Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

    • Ronia Kattoum (UALR)

    5.3: Energy and Phase Transitions is shared under a not declared license and was authored, remixed, and/or curated by Robert Belford.

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