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# 16.5: Heat and Changes of State

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• calculate the energy changes during phase changes.
• explain the slopes of the various parts in heating and cooling curves.
• explain why it is necessary for a solid to absorb heat during melting even though no temperature change is occurring.
• calculate, given appropriate thermodynamic data, the heat required to raise temperatures of a given substance with no phase change.
• calculate, given appropriate thermodynamic data, the heat required to melt specific samples of solids with no temperature change.
• calculate, given appropriate thermodynamic data, the heat required to produce both a phase change and temperature change for a given sample of solid.

## Vocabulary

• freezing
• freezing point
• fusion
• heat of fusion
• melting
• melting point
• specific heat

## Introduction

In order to vaporize a liquid, heat must be added to raise the kinetic energy (temperature) to the boiling point. Upon reaching the phase change temperature, additional heat is still needed to provide the potential energy to separate the molecules into gaseous form.

The melting point of a substance, like its boiling point, is directly related to the strength of the forces of attraction between molecules. Low melting points are typical of substances whose forces of attraction are very weak, such as hydrogen gas whose melting point is 259C\begin{align*}-259^\circ\mathrm{C}\end{align*}. High melting points are associated with substances whose forces of attraction are very strong, such as elemental carbon whose melting point is greater than 3500C\begin{align*}3500^\circ\mathrm{C}\end{align*}.

## Heating and Melting Curves

The addition of heat before, during, and after a phase transition can be analyzed with the help of a heating curve. In the heating curve below, a sample of water at 20C\begin{align*}-20^\circ\mathrm{C}\end{align*} and 1.00 atm\begin{align*}1.00 \ \mathrm{atm}\end{align*} pressure is heated at a constant rate.

Between the temperatures of 20C\begin{align*}-20^\circ\mathrm{C}\end{align*} and 0C\begin{align*}0^\circ\mathrm{C}\end{align*}, all the heat added is absorbed as kinetic energy, causing the temperature of the solid to increase. Upon reaching the melting point, even though heat is still being added at the same rate, the temperature does not increase. All the heat added during this time goes into providing the potential energy needed to melt the solid; this energy represents the heat of melting. During this flat line period, an observer would see the water changing from solid to liquid. Both the water in solid form and the water in liquid form would be at exactly 0C\begin{align*}0^\circ\mathrm{C}\end{align*}. Adding more heat causes the water to melt faster, but the temperature does not increase until all of the solid has been converted to liquid. If the liquid is cooled, it follows this same curve in reverse. As it is cooled, the same flat line appears while the heat of fusion is removed before the temperature may go down again.

Once all the water is in the liquid form, the heat is again absorbed as kinetic energy. Between the temperatures of 0C\begin{align*}0^\circ\mathrm{C}\end{align*} and 100C\begin{align*}100^\circ\mathrm{C}\end{align*}, the heat added causes the temperature increases. Upon reaching the boiling point, the temperature again does not increase even though heat is added. All the heat added to the sample during the time that the slope of the line is zero goes into potential energy. This energy represents the heat of vaporization and is used to separate the liquid molecules into the gaseous form. During this flat line period, an observer would see that the water was changing into gas, but both the liquid part and the gaseous part would be at exactly 100C\begin{align*}100^\circ\mathrm{C}\end{align*}. Adding more heat causes the water to boil faster, but its temperature will never exceed 100C\begin{align*}100^\circ\mathrm{C}\end{align*}. Once all the water is in the gaseous form, the heat can again go into raising the kinetic energy and temperature of the gas. When the gas is cooled, it follows this same curve in reverse. As it is cooled, the same flat line appears while the heat of condensation is removed before the temperature may drop.

This video contains a lecture covering the energy involved in phase changes (7c): http://www.youtube.com/watch?v=wx0UAFMZkMw (6:28).

## Specific Heat

Thermodynamic data (melting point, boiling point, heat of melting, heat of vaporization) for almost all the elements and for thousands of compounds are available in various reference books and on the internet. One useful piece of thermodynamic data is the specific heat. Symbolized by the letter C\begin{align*}C\end{align*}, the specific heat for a substance is the amount of heat required to raise 1.00\begin{align*}1.00\end{align*} gram of the substance by 1.00C\begin{align*}1.00^\circ\mathrm{C}\end{align*}. For liquid water, the specific heat is 4.18 J/gC\begin{align*}4.18 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*}.

Example:

How much heat is required to raise the temperature of 25 grams\begin{align*}25 \ \mathrm{grams}\end{align*} of water from 15C\begin{align*}15^\circ\mathrm{C}\end{align*} to 55C\begin{align*}55^\circ\mathrm{C}\end{align*}?

Solution:

Q=mCΔT=(25 g)(4.18 J/gC)(40.C)=4180 J=4.18 kJ\begin{align*}Q = mC \Delta T = (25 \ \mathrm{g})(4.18 \ \mathrm{J/g} \cdot ^\circ\mathrm{C})(40.^\circ\mathrm{C}) = 4180 \ \mathrm{J} = 4.18 \ \mathrm{kJ}\end{align*}

## Melting and Freezing Points

Solids, like liquids, have a vapor pressure. Like liquids, the vapor pressure of a solid increases with temperature. At 100C\begin{align*}100^\circ\mathrm{C}\end{align*}, the vapor pressure of liquid water is 760\begin{align*}760\end{align*} mm of Hg. As the temperature decreases to 0C\begin{align*}0^\circ\mathrm{C}\end{align*}, the vapor pressure decreases (non-linearly) to 4.6\begin{align*}4.6\end{align*} mm of Hg.

The vapor pressure of a solid is generally very low because the forces of attraction in solids are strong. For example, at 83C\begin{align*}-83^\circ\mathrm{C}\end{align*}, the vapor pressure of ice is 0.00025\begin{align*}0.00025\end{align*} mm of Hg. As ice is heated, its vapor pressure increases. At 0C\begin{align*}0^\circ\mathrm{C}\end{align*}, the vapor pressure of ice is 4.6\begin{align*}4.6\end{align*} mm of Hg, which also happens to be the vapor pressure of water at 0C\begin{align*}0^\circ\mathrm{C}\end{align*}. In fact, for all substances, their solids and liquids have the same vapor pressure at the melting point. The melting point of a solid is defined as the temperature at which the vapor pressure of the solid and liquid are the same. Therefore, the melting point of the solid and the freezing point of the liquid are exactly the same temperature.

## Heat of Fusion

Melting, the phase change from solid to liquid, has many similarities to vaporization. The solid must reach its melting point before the molecules can enter the liquid phase. The molecules in liquid phase, however, are farther apart than the molecules in the solid phase. Since the molecules attract each other, increasing the distance between them requires work. The work done in separating the molecules is stored in the molecules as potential energy in the liquid phase. This is the same process that occurs when the heat of vaporization must be added to liquid molecules to get them into the gaseous phase. In the case of melting, this potential energy is called the heat of fusion, or the heat of melting.

The word “fusion” is used several times in science with different meanings. You need to note the context in order to determine which meaning is intended. In this case, fusion refers to the change from the liquid to solid phase. When a solid melts, the heat of fusion must be added, and when a liquid freezes (fuses) back to solid, the heat of fusion is given off. The heat of fusion for water is 334 joules/gram\begin{align*}334 \ \mathrm{joules/gram}\end{align*}.

Example:

How much heat must be added to 25 g\begin{align*}25 \ \mathrm{g}\end{align*} of ice at 0C\begin{align*}0^\circ\mathrm{C}\end{align*} to convert it to liquid water at 0C\begin{align*}0^\circ\mathrm{C}\end{align*}?

Solution:

Q=(mass)(ΔHFUSION)=(25 g)(334 J/g)=8350 J=8.4 kJ\begin{align*}Q = (\mathrm{mass})( \Delta H_{FUSION}) = (25 \ \mathrm{g})(334 \ \mathrm{J/g}) = 8350 \ \mathrm{J} = 8.4 \ \mathrm{kJ}\end{align*}

## Heat of Vaporization

The difference between the liquid phase and the gas phase of a substance is essentially the distance between the molecules. Since the molecules attract each other and are separated by a greater distance in the gaseous phase than in the liquid phase, the molecules in the gaseous phase possess more potential energy than in the liquid phase. When a substance changes from the liquid phase to the gaseous phase, work must be done on the molecules to pull them away from each other. The work done separating the molecules is stored in the molecular structure as potential energy. If the molecules are allowed to condense back into the liquid phase, the potential energy is released – exactly the same amount that was needed to separate the molecules. This potential energy stored in molecules in the gaseous phase is called the heat of vaporization. The heat of vaporization (ΔHVAP)\begin{align*}( \Delta H_{VAP})\end{align*} for water is 540\begin{align*}540\end{align*} calories/gram, which is 2.26 kJ/g\begin{align*}2.26 \ \mathrm{kJ/g}\end{align*} at the normal boiling point. Because of the strength of the polar attractions holding water molecules together in the liquid form, water has a fairly high heat of vaporization. Other examples of polar molecules are ammonia, NH3\begin{align*}\mathrm{NH}_3\end{align*}, and ethanol, C2H5OH\begin{align*}\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\end{align*}, which have heats of vaporization of 1.38 kJ/g\begin{align*}1.38 \ \mathrm{kJ/g}\end{align*} and 0.84 kJ/g\begin{align*}0.84 \ \mathrm{kJ/g}\end{align*}, respectively.

Example:

How much heat in kJ is necessary to vaporize 100. grams\begin{align*}100. \ \mathrm{grams}\end{align*} of ammonia at its boiling point?

Solution:

Q=(mass)(ΔHVAP)=(100. g)(1.38 kJ/g)=138 kJ\begin{align*}Q = (\mathrm{mass})( \Delta H_{VAP}) = (100. \ \mathrm{g})(1.38 \ \mathrm{kJ/g}) = 138 \ \mathrm{kJ}\end{align*}

The boiling point of ammonia is 33C\begin{align*}-33^\circ\mathrm{C}\end{align*}. It is very important to understand that the ammonia is at the boiling point before and after the heat of vaporization is added. All the energy involved in the heat of vaporization is absorbed by the substance as potential energy; none of it goes into kinetic energy, so the temperature cannot change.

To test your understanding of this point, determine which would produce a more severe burn: spilling boiling water at 100C\begin{align*}100^\circ\mathrm{C}\end{align*} on your skin or being burned by gaseous water at 100C\begin{align*}100^\circ\mathrm{C}\end{align*}. At first it may seem that they would do the same damage since they are both at the same temperature, but in fact, the gaseous water would do more damage. The gaseous water would release a tremendous amount of heat (heat of vaporization) to your skin as it condenses to water before burning your skin as 100C\begin{align*}100^\circ\mathrm{C}\end{align*} water.

Example:

How much heat is required to raise the temperature of 25 grams\begin{align*}25 \ \mathrm{grams}\end{align*} of liquid water from 25C\begin{align*}25^\circ\mathrm{C}\end{align*} to gaseous water at 100.C\begin{align*}100.^\circ\mathrm{C}\end{align*}?

Solution:

In this problem, you have to calculate the heat needed to raise the temperature of the liquid water from 25C\begin{align*}25^\circ\mathrm{C}\end{align*} to \begin{align*}100.^\circ\mathrm{C}\end{align*} and then the heat of vaporization for the 25 g of water.

Heating: \begin{align*}Q = mC \Delta T = (25 \ \mathrm{g})(4.18 \ \mathrm{J/g} \cdot ^\circ\mathrm{C})(75^\circ\mathrm{C}) = 7838 \ \mathrm{J} = 7.8 \ \mathrm{kJ}\end{align*}
Vaporizing: \begin{align*}Q = mH_{VAP} =(25 \ \mathrm{g})(2.26 \ \mathrm{kJ/g}) = 56.5 \ \mathrm{kJ}\end{align*}

\begin{align*}Q_T = 7.8 \ \mathrm{kJ} + 56.5 \ \mathrm{kJ} = 64.3 \ \mathrm{kJ}\end{align*}

Example:

A \begin{align*}2,000. \ \mathrm{grams}\end{align*} mass of water in a calorimeter has its temperature raised by \begin{align*}3.00^\circ\mathrm{C}\end{align*} while an exothermic chemical reaction is occurring. How much heat, in joules, is transferred to the water by the heat of reaction?

Solution:

The heat is calculated by determining the heat absorbed by the water. This amount of heat is the product of three factors: 1) the mass of the water, 2) the specific heat of water, and 3) the change in temperature of the water.

\begin{align*}Q = (\mathrm{mass \ of \ water})(C_{water})( \Delta T) = (2000. \ \mathrm{g})(4.18 \ \mathrm{J/g} \cdot ^\circ\mathrm{C})(3.00^\circ\mathrm{C}) = 25,080 \ \mathrm{J} = 25.1 \ \mathrm{kJ}\end{align*}

Example:

A \begin{align*}1,000. \ \mathrm{gram}\end{align*} mass of water whose temperature was \begin{align*}50.^\circ\mathrm{C}\end{align*} lost \begin{align*}33,400 \ \mathrm{J}\end{align*} of heat over a 5-minute period. What was the temperature of the water after the heat loss?

Solution:

\begin{align*} Q = (\mathrm{mass})(C_w)( \Delta T) = \frac {\mathrm{heat}} {(\mathrm{mass})(C_w)} = \frac {-33,400 \ \mathrm{J}} {(1000. \ \mathrm{g})(4.18 \ \mathrm{J/g} \cdot ^\circ\mathrm{C})} = -8.00^\circ\mathrm{C} \end{align*}

If the original temperature was \begin{align*}50.^\circ\mathrm{C}\end{align*} and the temperature decreased by \begin{align*}8^\circ\mathrm{C}\end{align*}, then the final temperature would be \begin{align*}42^\circ\mathrm{C}\end{align*}.

Example:

Use the thermodynamic data given to calculate the total amount of energy necessary to raise \begin{align*}25 \ \mathrm{grams}\end{align*} of ice at \begin{align*}-20.^\circ\mathrm{C}\end{align*} to gaseous water at \begin{align*}120.^\circ\mathrm{C}\end{align*}.

Data:

• Melting point of ice = \begin{align*}0^\circ\mathrm{C}\end{align*}
• Boiling point of water = \begin{align*}100^\circ\mathrm{C}\end{align*}
• \begin{align*} \Delta H_{VAP}\end{align*} for water = \begin{align*}2260 \ \mathrm{J/g}\end{align*}
• \begin{align*} \Delta H_{FUSION}\end{align*} for water = \begin{align*}334 \ \mathrm{J/g}\end{align*}
• \begin{align*}C_{ice} = 2.11 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*}
• \begin{align*}C_{water} = 4.18 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*}
• \begin{align*}C_{water \ vapor} = 1.84 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*}

Solution

There will be five steps in the solution process.

1. We must calculate the heat required to raise the temperature of the ice from \begin{align*}-20.^\circ\mathrm{C}\end{align*} to \begin{align*}0^\circ\mathrm{C}\end{align*}.
Raising the temperature of ice, \begin{align*}Q = mC \Delta T = (25 \ \mathrm{g})(2.11 \ \mathrm{J/g} \cdot ^\circ\mathrm{C})(20^\circ\mathrm{C}) = 1055 \ \mathrm{J}\end{align*}
1. We must calculate the heat required to provide the heat of melting in order to change ice at \begin{align*}0^\circ\mathrm{C}\end{align*} to water at \begin{align*}0^\circ\mathrm{C}\end{align*}.
Melting ice to liquid, \begin{align*}Q = (\mathrm{mass})(\Delta H_{FUSION}) = (25 \ \mathrm{g})(334 \ \mathrm{J/g}) = 8350 \ \mathrm{J}\end{align*}
1. We must calculate the heat required to raise the temperature of the liquid water from \begin{align*}0^\circ\mathrm{C}\end{align*} to \begin{align*}100^\circ\mathrm{C}\end{align*}.
Raising the temperature of liquid, \begin{align*}Q = mC \Delta T = (25 \ \mathrm{g})(4.18 \ \mathrm{J/g} \cdot ^\circ\mathrm{C})(100^\circ\mathrm{C}) = 10450 \ \mathrm{J}\end{align*}
1. We must calculate the heat required to provide the heat of vaporization to change liquid water at \begin{align*}100^\circ\mathrm{C}\end{align*} to gaseous water at \begin{align*}100^\circ\mathrm{C}\end{align*}.
Vaporizing liquid to gas, \begin{align*}Q = (\mathrm{mass})(\Delta H_{VAP}) = (25 \ \mathrm{g})(2260 \ \mathrm{J/g}) = 56500 \ \mathrm{J}\end{align*}
1. Finally, we must calculate the heat required to raise the temperature of the gaseous water from \begin{align*}100^\circ\mathrm{C}\end{align*} to \begin{align*}120^\circ\mathrm{C}\end{align*}.
Raising the temperature of gas, \begin{align*}Q = mC \Delta T = (25 \ \mathrm{g})(1.84 \ \mathrm{J/g} \cdot ^\circ\mathrm{C})(20^\circ\mathrm{C}) = 920 \ \mathrm{J}\end{align*}
The sum of all five steps is \begin{align*}77,275 \ \mathrm{J} = 77 \ \mathrm{kJ}\end{align*}

The cooling process would be the exact reverse of the heating process. If water in the gaseous phase is cooled, each \begin{align*}1.84\end{align*} joules of heat removed would lower the temperature of \begin{align*}1.00 \ \mathrm{g}\end{align*} of gas by \begin{align*}1.00^\circ\mathrm{C}\end{align*}. When the gaseous water reaches the boiling point (also the condensation point), each gram of gaseous water that condenses to liquid will release \begin{align*}2260 \ \mathrm{joules}\end{align*} of heat. Once all the water is in the liquid form, the removal of each \begin{align*}4.18 \ \mathrm{joules}\end{align*} of heat by cooling will cause the temperature of \begin{align*}1.00 \ \mathrm{g}\end{align*} of water to cool by \begin{align*}1.00^\circ\mathrm{C}\end{align*}. At the freezing point (also the melting point) \begin{align*}334 \ \mathrm{joules}\end{align*} of heat must be removed to convert each gram of liquid water to ice. When the entire sample of water is in the form of ice, \begin{align*}2.26 \ \mathrm{joules}\end{align*} of heat must be removed to cool each gram by \begin{align*}1.00^\circ\mathrm{C}\end{align*}.

This video serves a blackboard lecture on the mathematics of heat involved in temperature change and phase change (7d): http://www.youtube.com/watch?v=zz4KbvF_X-0 (14:49).

## Lesson Summary

• In heating and cooling curves, temperature change is associated with a change in kinetic energy, while no temperature change is associated with a change in potential energy.
• The specific heat of a substance is the amount of heat required to raise the temperature of one gram of the substance by \begin{align*}1.0^\circ\mathrm{C}\end{align*}.
• The heat of vaporization of a liquid is the quantity of heat required to vaporize a unit mass of that liquid at constant temperature.
• The energy released when a gas condenses to a liquid is called the heat of condensation.
• Solids melt when the vapor pressure of the solid equals the vapor pressure of the liquid.
• Heat must be absorbed by a solid to become a liquid even though the temperature remains the same. The quantity of heat absorbed per unit mass is called the heat of fusion.
• Stronger forces of attraction between particles in solids produce higher melting points and higher heats of fusion.

This video explores how matter changes state depending on the temperature.

## Review Questions

Use the thermodynamic data given in the following table (Table below) to answer problems 1 - 5.

Thermodynamic Data of Various Substances
Water Cesium, \begin{align*}\mathrm{Cs}\end{align*} Silver, \begin{align*}\mathrm{Ag}\end{align*}
Melting Point \begin{align*}0^\circ\mathrm{C}\end{align*} \begin{align*}29^\circ\mathrm{C}\end{align*} \begin{align*}962^\circ\mathrm{C}\end{align*}
Boiling Point \begin{align*}100.^\circ\mathrm{C}\end{align*} \begin{align*}690.^\circ\mathrm{C}\end{align*} \begin{align*}2162^\circ\mathrm{C}\end{align*}
\begin{align*} \bf{ \triangle H_{\mathrm{fusion}}}\end{align*} \begin{align*}334 \ \mathrm{J/g}\end{align*} \begin{align*}16.3 \ \mathrm{J/g}\end{align*} \begin{align*}105 \ \mathrm{J/g}\end{align*}
\begin{align*} \bf{ \triangle H_{\mathrm{vaporization}}}\end{align*} \begin{align*}2260 \ \mathrm{J/g}\end{align*} \begin{align*}669 \ \mathrm{J/g}\end{align*} \begin{align*}2362 \ \mathrm{J/g}\end{align*}
Specific Heat, \begin{align*}C\end{align*}, for Gas \begin{align*}2.01 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*} \begin{align*}0.167 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*} \begin{align*}0.159 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*}
Specific Heat, \begin{align*}C\end{align*}, for Liquid \begin{align*} 4.18 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*} \begin{align*}0.209 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*} \begin{align*}0.294 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*}
Specific Heat, \begin{align*}C\end{align*}, for Solid \begin{align*}2.09 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*} \begin{align*}0.251 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*} \begin{align*}0.235 \ \mathrm{J/g} \cdot ^\circ\mathrm{C}\end{align*}
1. How many joules are required to melt \begin{align*}100. \ \mathrm{grams}\end{align*} of silver at its normal melting point with no temperature change?
2. How many joules are required to boil \begin{align*}150. \ \mathrm{grams}\end{align*} of cesium at its normal boiling point with no temperature change?
3. How many joules are required to heat \begin{align*}200. \ \mathrm{g}\end{align*} of liquid water from \begin{align*}25^\circ\mathrm{C}\end{align*} to steam at \begin{align*}125^\circ\mathrm{C}\end{align*} under normal pressure?
4. How many joules are required raise the temperature of \begin{align*}1.00 \ \mathrm{gram}\end{align*} of water from \begin{align*}-269^\circ\mathrm{C}\end{align*} (the current temperature of space) to \begin{align*}1.60 \times 10^{15}\ {^{\circ}}\mathrm{C}\end{align*} (the estimated temperature of space immediately after the Big Bang)?
5. How many joules are required to raise the temperature of \begin{align*}1000. \ \mathrm{g}\end{align*} of cesium from \begin{align*}-200.^\circ\mathrm{C}\end{align*} to \begin{align*}+200.^\circ\mathrm{C}\end{align*}?
6. Why does the boiling point of water increase with increasing surrounding pressure?
7. Why must heat be absorbed to melt a solid even though both the solid and the liquid are at the same temperature?

Use the image below for problems 8 - 10.

1. What is happening to the water in section B?
2. What is happening to the water in section A?
3. Why are the slopes of the lines in sections A, C, and E different?

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