Download TASK Matrix Module 4 Phase Changes

THERMODYNAMICS
Module 4. PHASE CHANGES
Learning
Outcomes
1. Discussed
calorimetry,
method of
mixtures and
the types of
phase
changes.
2. Applied the
principles of
calorimetry,
and method of
mixtures in
solving
problems.
Topic
Resource Materials
Tasks
Lesson 1:
Calorimetry
and Method
of Mixtures
Main Reading:
Lyublinskaya, I. et al., (2017) College
Physics for AP Courses
https://openstax.org/details/books/collegephysics-ap-courses
Rubric in
Designing an
Experiment
Lesson 2:
Types of
Phase
Changes
3. Designed
experiment
involving
calorimetry
and method of
mixtures.
Serway, R.A & Vuille, C. (2015) College
Physics, 10th Edition. Cengage Learning
Asia, Pte. Ltd.
Cutnell, J.D. & Johnson, K.W. (2013)
Introduction to Physics. 9th Edition, John
Wiley & Sons, Inc.
Supplemental Video:
YouTube link:
Calorimetry Problems, Thermochemistry
Practice, Specific Heat Capacity, Enthalpy
Fusion, Chemistry (2016)
https://www.youtube.com/watch?v=HlvllF
6Ml9c
Latent Heat of Fusion and Vaporization,
Specific Heat Capacity & Calorimetry –
Physics (2016)
https://www.youtube.com/watch?v=dxtz2P
OUTJE
Phase Changes, Heats of Fusion and
Vaporization, and Phase Diagrams (2015)
https://www.youtube.com/watch?v=oc0ype
DELb0
How Can This Chemical Be a Solid,
Liquid, and Gas at the Same Time?
https://www.youtube.com/watch?v=MP6M
VLWuNZQ
Thermodynamics - Explaining the Triple
Point
https://www.youtube.com/watch?v=HEzk
HqWIiKM
See the following attachments for reference.
1. Rubrics for Scoring for Electronic Scrapbook Making and Problem Set
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Module 4: Phase Changes
Problem Set
Time
Frame
Week
5-6
THERMODYNAMICS
2. Supplementary Learning Materials: Activity Sheet, Worksheet and Problem Set
Submission Options:
For Online
1. Submit online your accomplished task via our learning group, or my messenger at your own pace but
not later than the specified deadline or a week after the date you received this task.
2. Compile all your submitted papers (synthesis and reflection paper) and other requirements in your eportfolio and submit on or before the semester ends.
Prepared by:
NAPOLEON P. ALBOR, JR., EdD
Course Facilitator
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Module 4: Phase Changes
THERMODYNAMICS
Module 4: PHASE CHANGES
LEARNING OBJECTIVES
At the end of the module the students must have:
1. discussed calorimetry, method of mixtures and the types of phase changes
2. applied the principles of methods of mixture and calorimetry in solving problems
3. designed experiment involving calorimetry and method of mixtures
INTRODUCTION
Solids are formed due to the cohesive bonds between molecules. Thus, energy is required to melt its
particle to be broken apart such that, in the liquid, the molecules can move around at comparable kinetic
energies; thus, there is no rise in temperature. Similarly, energy is also needed to vaporize a liquid, because
molecules in a liquid interact with each other via attractive forces. There is no temperature change until a
phase change is complete. For instance, energy is released during freezing and condensation, usually in the
form of thermal energy. Work is done by cohesive forces when molecules are brought together. The
corresponding energy must be given off to allow them to stay together.
This module focuses on how heat is measured as matter absorbed and released it and its effects on
matter and processes involving heat and internal energy between a system and its surroundings.
Method of Mixtures and Calorimetry
The principle of the method of mixtures states that the heat lost by a hot body is equal to the heat
gained by the cold body when they are mixed together and attain the same temperature. This principle is
based on the law of conservation of energy.
The law implies that if the energy of an object goes up (its temperature goes up), that energy is not a
newly created energy. The energy comes from the source which is the warm body. The amount of heat lost
by the warm body is the same as the amount gained by the cold body. In some references, the term used is
heat exchange. It can be stated as:
Heat energy lost by the hot body = Heat energy gained by the cold body
Heat lost (by the hot body) + heat gained (by the cold body) = 0
Qlost + Q gained = 0
This is called the principle of the method of mixture. It is based on Newton’s Law of Cooling, which
states that when a liquid is heated of higher temperature and placed to cool. Then the rate of heat lost by a
temperature of the liquid is directly proportional to the difference in temperature of the surrounding.
The other name to it is the principle of calorimetry. This principle is based on the law of
conservation of energy.
The calorimeter is a device used to measure the heat flow of a chemical or physical reaction.
Calorimetry is the process of measuring this heat. It consists of a metal container to hold water above the
combustion chamber and a thermometer to measure the temperature change. The principle of calorimetry is
heat lost is equal to the heat gained.
Calorimetry is a technique used to measure amounts of heat transferred to or from a substance
involved in a chemical or physical process. The measurement of heat transfer using this approach requires
the definition of a system (the substance or substances undergoing the chemical or physical change) and
its surroundings (the other components of the measurement apparatus that serve to either provide heat to
the system or absorb heat from the system).
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Module 4: Phase Changes
THERMODYNAMICS
Figure 4.1. In a calorimetric determination, either (a) an exothermic process occurs and heat, q, is
negative, indicating that thermal energy is transferred from the system to its surroundings, or (b)
an endothermic process occurs and heat, q, is positive, indicating that thermal energy is
transferred from the surroundings to the system.
Source: https://opentextbc.ca/chemistry/chapter/5-2-calorimetry/
The process can be exothermic or endothermic. In an exothermic reaction, the heat produced is
absorbed by the solution, thus temperature increases. On the other hand, in endothermic reaction, the heat is
absorbed from the thermal energy of the solution, thus, temperature decreases (see figure 4.1) (Use the given
links
and
watch
the
video
on
Phase
changes:
Exothermic
and
Endothermic?
https://www.youtube.com/watch?v=0cUK4jcAEaU) (https://www.youtube.com/watch?v=i3mYWB2fNp4)
In calculations, heat Q can be positive or negative. The useful sign convention for heat is as follows:
1. When heat is absorbed by the body, its temperature increase (T 2 > T1). So ΔT and Q are positive.
2. When heat is released by the body, its temperature decreases (T 1 > T2). So ΔT and Q are
negative.
In the discussion of heat and thermodynamics, a system is considered. System is any object or set of
objects you choose to consider, everything else in the universe as it is the ―surroundings‖ or the
―environment‖. The three types of system are:
1. Closed system – one for which no mass enters or leave (but energy maybe exchanged with the
environment).
2. Open system – mass may enter or leave (as energy). Many (idealized) systems we study in
Physics are closed system.
3. Isolated system – is a closed system wherein no energy in any forms (as well as no mass passes
across its boundaries. A perfectly ideal system is an isolated system.
Sample Problem:
Calculating the Final Temperature in Calorimetry
Suppose you pour 0.250 kg of 20.0°C water (about a cup) into a 0.500 kg aluminum pan off the
stove with a temperature of 150°C. Assume no heat transfer takes place to anything else: The pan is placed
on an insulated pad, and heat transfer to the air is neglected in the short time needed to reach equilibrium.
Thus, this is a calorimetry problem, even though no isolating container is specified. Also assume that a
negligible amount of water boils off. What is the temperature when the water and pan reach thermal
equilibrium?
Strategy:
Originally, the pan and water are not in thermal equilibrium: The pan is at a higher temperature than
the water. Heat transfer restores thermal equilibrium once the water and pan are in contact; it stops once
thermal equilibrium between the pan and the water is achieved. The heat lost by the pan is equal to the heat
gained by the water—that is the basic principle of calorimetry.
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Module 4: Phase Changes
THERMODYNAMICS
Solution:
1. Use the equation for heat transfer Q = mcΔT to express the heat lost by the aluminum pan in terms of the
mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final
temperature:
Qhot = mA1 cA1 (Tf − 150°C)
2. Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the
initial temperature of the water, and the final temperature:
Qcold = mw cw (Tf − 20.0°C)
3. Note that Qhot < 0 and Qcold > 0 and that as stated above, they must sum to zero:
Qcold + Qhot = 0
Qcold = −Qhot
mwcw (Tf − 20.0 °C) = −mA1 cA1 (Tf − 150°C)
4. This a linear equation for the unknown final temperature, T f . To solve for Tf ,
(
)
(
)
And insert the numerical values
(
)(
)(
(
)(
)
(
)(
)(
)
(
)(
)
)
= 59.1
Discussion:
Why is the final temperature so much closer to 20.0 °C than to 150°C? The reason is that water has
a greater specific heat than most common substances and thus undergoes a smaller temperature change for
a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its
temperature appreciably. This explains why the temperature of a lake stays relatively constant during the
day even when the temperature change of the air is large. However, the water temperature does not change
over longer times.
PHASE CHANGE AND LATENT HEAT
A phase change occurs when matter changes from one state (solid, liquid, gas,
plasma) to another (see illustration) when energy supply is sufficient to the system
(or a sufficient amount is lost). It also occurs when the pressure on the
system changed. The temperature and pressures under which these
changes happen differ depending on the chemical and physical properties
of the system. The energy associated with these transitions is called Latent
Heat. Below are the types of phase changes:
TYPES OF PHASE CHANGES
1. Freezing (Liquid → Solid)
Freezing means the solidification phase change of a liquid or
the liquid content of a substance, usually due to cooling.
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Module 4: Phase Changes
Source: https://energyeducation.ca/encyclopedia/Phasechange
THERMODYNAMICS
The temperature of a freezing liquid remains constant, even when more heat is removed.
Examples:
 Freezing of water to form ice
 Formation of snow
 Congealing of bacon grease
 Solidification of Candle wax
 Lava hardening into solid rock
2. Melting (Solid → Liquid)
Melting, or fusion, is a physical process that results in the phase transition of a substance from a solid
to a liquid. This occurs when the internal energy of the solid increases due to the application of heat or
pressure, which increases the substance's temperature to the melting point. The substance changes back from
the solid to liquid.





Examples:
Melting of ice into liquid water
Melting of steel (require very high temperature)
Melting of mercury and gallium (both are liquid at room temperature)
Melting of butter
Melting of candle
3. Condensation (Gas → Liquid)
This change is caused by a change of pressure and temperature of the substance.
Examples:
 Morning Dew- This happen when the moisture present in the air condenses on the grasses and leaves
cooling overnight
 Clouds- Another example of condensation without surface clouds
 Foggy Mirror- You mirror fogs up after a hot shower. It happens because the shower moisture
condense on the cool mirror
4. Vaporization (Liquid → Gas)
This process is due to an increase in temperature and/or pressure. Vaporization is also the conversion
of a substance from the liquid or solid phase into the gaseous (vapor) phase. If conditions allow the
formation of vapor bubbles within a liquid, the vaporization process is called boiling. These include
evaporation and boiling. Evaporation is a surface phenomenon while boiling involves bulk phenomena.






Examples:
Drying clothes under The Sun
Ironing Clothes
Cooling down of hot tea and other tea liquid
Melting ice cube
Drying of wet hair
Boiling Water in a hot pot
5. Sublimation (Solid → Gas)
During this process, a transition from the solid phase to the gas phase without passing through an
intermediate liquid phase. It is caused by the absorption of heat which provides enough energy for some
molecules to overcome the attractive forces of their neighbor and escape into the vapor phase.
Examples:
 Dry Ice
 Specialized Printers
 Moth Balls
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Module 4: Phase Changes
THERMODYNAMICS
 Freeze Drying
 Air Fresheners
6. Deposition (Gas →Solid)
The substance changes directly from a gas to a solid without going through the liquid phase. The
reverse of deposition is sublimation; hence sometimes deposition is called desublimation.
Example:
 One example of deposition is the process by which, in sub-freezing air, water vapor changes directly
to ice without first becoming a liquid. This is how frost and hoar frost form on the ground or other
surfaces
7. Ionization (Gas →Plasma)
Ionization is the process by which an atom or a molecule acquires a negative or positive charge by
gaining or losing electrons, often in conjunction with other chemical changes. The resulting electrically
charged atom or molecule is called an ion.


Examples:
Formation of auroras
Removal of electrons from ions
8. Recombination (Plasma → Gas)
Turning off the power to a neon light allows the ionized particles to
return to the gas phase called recombination, the combining of charges or
transfer of electrons in a gas that result in the neutralization of ions.
Example:
Turning off the power to a neon light
 When heat is absorbed or released by a material, it can either
1. change the temperature of a the material without changing its phase
2. change the phase of the material without changing its temperature.
The energy involved in a phase change depends on two major
factors: the number and strength of bonds or force pairs. The number of
bonds is proportional to the number of molecules and thus to the mass of
the sample. The strength of forces depends on the type of molecules.
The heat Q required to change the phase of a sample of mass m is given by
Q = mLf (melting/freezing)
Q = mLv (vaporization/condensation)
where the latent heat of fusion (Lf ) and latent heat of vaporization (Lv) are material constants that are
determined experimentally (see table 4.1).
 When heat is absorbed by materials (Q is positive) when its phase is converted in the direction of
solid – liquid – gas. However, when heat is released (Q is negative) when the phase of the material is
changed in the direction of gas – liquid – solid.
 Latent heat can be classified into the following:
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Module 4: Phase Changes
THERMODYNAMICS
1. Latent heat of fusion (Lf) – the amount of heat needed to change the phase of a unit mass of a
substance from solid to liquid or from liquid to solid.
2. Latent heat of vaporization (Lv) – the amount of heat needed to change the phase of a unit mass of
a substance from liquid to gas or from gas to liquid.
Table 4.1 Heat of Fusion and Vaporization of some substances.(4)
4. Values quoted at the normal melting and boiling temperatures at standard atmospheric pressure (1 atm)
5. At 37.0ºC (body temperature), the heat of vaporization Lv for water is 2430 kJ/kg or 580 kcal/kg
6. At 37.0ºC (body temperature), the heat of vaporization Lv for water is 2430 kJ/kg or 580 kcal/kg
Latent heat is measured in units of J/kg. Lf and Lv are collectively called latent heat of coefficients.
They are latent, or hidden, because in phase changes, energy enters or leaves a system without causing a
temperature change in the system; so, in effect, the energy is hidden.
SAMPLE PROBLEM
Calculating Heat of Vaporization
1. How much heat is released by 20.0 grams of steam as it condenses at 100C?
Given: msteam = 20.0 g
Lv = 2260 J/g
Solution: Q = mLv
= (20.0 g) (2260 J/g)
= 45, 200 J
Calculating Final Temperature from Phase Change:
2. Three ice cubes are used to chill a soda at 20°C with mass msoda = 0.25 kg. The ice is at 0°C and each ice
cube has a mass of 6.0 g. Assume that the soda is kept in a foam container so that heat loss can be ignored
and that the soda has the same specific heat as water. Find the final temperature when all ice has melted.
Strategy:
The ice cubes are at the melting temperature of 0°C. Heat is transferred from the soda to the ice for
melting. Melting yields water at 0°C, so more heat is transferred from the soda to this water until the water
plus soda system reaches thermal equilibrium.
The heat transferred to the ice is
8
Module 4: Phase Changes
THERMODYNAMICS
Qice = mice Lf + mice cW (Tf − 0 ).
The heat given off by the soda is
Qsoda = msoda cW (Tf − 20°C).
Since no heat is lost, Qice = −Qsoda, as in the above example, so that
mice Lf + mice cW (Tf − 0°C) = −msoda cW (Tf − 20°C).
Solve for the unknown quantity T f :
(
)
(
)
Solution:
First, we need to identify the known quantities.
mice = 3 x 6.0 g = 18g. Convert this to Kg.
msoda = 0.25 Kg,
Then we calculate the final temperature. Substitute the numerical values to the above equation.
(
)(
)(
(
)
(
)(
)(
)
)
ASSESSMENT TASK: (see attached rubric)
I. Design an experiment on calorimety/methods of mixture/phase change. You may use the template below:
Experiment Title:___________________________
A. Rationale/Background
B. Science Processes/Skills
C. Activities
I. Objectives II. Materials/Methods III. Procedure IV. Guide/Observation
V. Conclusion
Questions
Your
List
down
the Describe the
Formulate
Include here the
objectives
materials needed for procedure to
guide/observation
questions leading
must
be the activity.
be followed in
questions to facilitate
to drawing of
based from
conducting the discussion of the
generalization.
secondary
experiment.
possible results.
science
competencies
9
Module 4: Phase Changes
THERMODYNAMICS
II. Problem Set.
Direction: Solve the following problems and show your solution.
1. How much heat must be absorbed by 25 g ice at 00C to change it into water without a change in
temperature?
2. Find the heat of fusion of ice, using the following data obtained from an experiment:
Given:
Weight of brass calorimeter
= 50 g
Specific heat of brass calorimeter
= 0.09 cal/g.0C
Mass of water
= 200 g at 300C
Temperature of ice
= 20.45 g at 00C
Temperature of mixture
= 200C
3. How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package of frozen vegetables
originally at 0ºC if their heat of fusion is the same as that of water?
4. In a study of healthy young men, doing 20 pushups in 1 minute burned an amount of energy per kg that
for a 70 kg man corresponds to 8.06 calories (kcal). How much would a 70 kg man’s temperature rise if
he did not lose any heat during that time?
5. A pressure cooker contains water and steam in equilibrium at a pressure greater than atmospheric
pressure. How does this greater pressure increase cooking speed?
REFERENCES
Cutnell, J.D. & Johnson, K.W. (2013) Introduction to Physics. 9th Edition, John Wiley & Sons, Inc.
Serway, R.A & Vuille, C. (2015) College Physics, 10th Edition. Cengage Learning Asia, Pte. Ltd.
Ling, S.J. et al, (2018). University Physics Volume 2. https://openstax.org/details/books/universityphysicsvolume-2
Lyublinskaya, I. et al., (2017) College Physics for AP Courses. https://openstax.org/details/books/collegephysics-ap-courses
Phase change. Energy Education. https://energyeducation.ca/encyclopedia/Phase_change
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Module 4: Phase Changes