16.5: Heat and Changes of State
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
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
Between the temperatures of
Once all the water is in the liquid form, the heat is again absorbed as kinetic energy. Between the temperatures of
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
Example:
How much heat is required to raise the temperature of
Solution:



Q=mCΔT=(25 g)(4.18 J/g⋅∘C)(40.∘C)=4180 J=4.18 kJ


Melting and Freezing Points
Solids, like liquids, have a vapor pressure. Like liquids, the vapor pressure of a solid increases with temperature. At
The vapor pressure of a solid is generally very low because the forces of attraction in solids are strong. For example, at
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
Example:
How much heat must be added to
Solution:



Q=(mass)(ΔHFUSION)=(25 g)(334 J/g)=8350 J=8.4 kJ


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
Example:
How much heat in kJ is necessary to vaporize
Solution:



Q=(mass)(ΔHVAP)=(100. g)(1.38 kJ/g)=138 kJ


The boiling point of ammonia is
To test your understanding of this point, determine which would produce a more severe burn: spilling boiling water at
Example:
How much heat is required to raise the temperature of
Solution:
In this problem, you have to calculate the heat needed to raise the temperature of the liquid water from


 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 5minute 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.
 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*}



 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*}



 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*}



 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*}



 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_X0 (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.
Further Reading / Supplemental Links
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.
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*} 
 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?
 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?
 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?
 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)?
 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*}?
 Why does the boiling point of water increase with increasing surrounding pressure?
 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.
 What is happening to the water in section B?
 What is happening to the water in section A?
 Why are the slopes of the lines in sections A, C, and E different?
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