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    see attached **files (3)** – ## UnitVIAssignment.docx – ## UnitVI.pdf – UnitVIExampleConceptMap.png **## UnitVIAssignment.docx** Unit VI This concept map assignment measures ULO 6.1, 6.2, and 6.3. Among Archimedes, Pascal, and Bernoulli, who do you think is the most important contributor to developi…

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    **files (3)**

    – ## UnitVIAssignment.docx

    – ## UnitVI.pdf

    – UnitVIExampleConceptMap.png

    **## UnitVIAssignment.docx**

    Unit VI

    This concept map assignment measures ULO 6.1, 6.2, and 6.3.

    Among Archimedes, Pascal, and Bernoulli, who do you think is the most important contributor to developing modern physics in fluid dynamics? Choose one of them, then create a concept map to describe his equations with examples of applications. A concept map or conceptual diagram is a diagram that expresses the concepts or delivers the important ideas of a subject. Review the
    [Unit VI Example Concept Map](https://online.columbiasouthern.edu/bbcswebdav/pid-17830837-dt-asiobject-rid-190111713_1/xid-190111713_1). It does not have to be fancy, but you are encouraged to be as creative as possible to emphasize the content that you want to depict. You may use PowerPoint slides, document files, pdf files, Photoshop, or even a piece of paper with coloring pencils.

    **## UnitVI.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit VI

    Upon completion of this unit, students should be able to:

    6. Explain numerous phenomena using fluid mechanics laws.
    6.1 Utilize the relationship between mass, density, and volume.
    6.2 Apply the concept of Pascal’s principle and Archimedes’ principle.
    6.3 Recognize heat and energy with phase changes of matter.

    Required Unit Resources

    Chapter 11: Fluids

    Chapter 12: Temperature and Heat

    Unit Lesson

    What Are Fluids?

    Unlike solids, liquids and gasses do not have a definite shape unless they are kept in a container. Something
    is referred to as fluid when it can flow without specific configuration. Both liquids and gases are examples of
    fluids. Both water and air move from place to place if they are not confined in a vessel. Both of them show a
    similar pattern of motion, but their density is very different because it depends on the property of matter.
    Usually, gases have lower densities than liquids have because the average distance between molecules is
    greater in gases than in liquids. The density of gases is greatly dependent on pressure and temperature.

    Mass density is defined as the mass of an object divided by the volume of the object. The SI unit of mass is
    kg, that of density is kg / m3, and that of volume is m3.

    UNIT VI STUDY GUIDE
    Basic Fluid Dynamics

    Sample Question 1: If the mass of a diamond box is 100 kg, what is the volume of the box?

    Solution: Use the formula mass = density x volume. The density of diamond is 3520 kg/m3 from
    Table 11.1 under Section 11.1 in Chapter 11 of the eTextbook (Cutnell et al., 2022). So, the
    volume is the mass divided by the density.

    Mass = Density x Volume

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Pressure in Fluids

    When a force (F) acts on area (A) in a fluid, the pressure (P) can be expressed as P = F / A. Here, we only
    consider the magnitude of F, so P is a scalar and its unit is N / m2 = Pa (Pascal). Atmospheric pressure at sea
    level is about 101,300 Pa = 1 atm.

    You may experience greater pressure as you go deeper into a swimming pool or in an ocean. What is the
    relation between pressure and depth? Let’s consider one column of water with height h below the figure in a
    large swimming pool. The area (A) of the top face is the same as that of the bottom face. The pressure (Pt) on
    the top face creates a downward force or PtA. The pressure (Pb) creates an upward force or PbA. Also, the
    weight (mg) due to gravity points downward.

    Water is at rest, and thus its acceleration is zero. That is, the column is in equilibrium. We can apply Newton’s
    second law, and the summation of the vertical forces is zero: PbA – PtA – mg = 0. Use m = ρV = ρAh. Then,
    Pb = Pt + ρgh. You can clearly see that the pressure at a deeper level is greater than the pressure at a shallow
    level if the density is not changing or if it is incompressible. In the case of gas, the density varies according to
    the vertical distances or if it is compressible, so the formula only works when h is very small. For instance, the
    density of our atmosphere varies significantly from the Earth’s surface to higher altitudes. The important thing
    is that the pressure difference between lower and higher levels comes from the height, or the vertical
    distance, not the horizontal distance within the fluid. See Figure 11.5 in Section 11.3 of Chapter 11 in the
    eTextbook (Cutnell et al., 2022).

    Force = Pressure x Area

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Pascal’s and Archimedes’ Principles

    Pascal’s principle states that any externally applied pressure is transmitted undiminished to everywhere in a
    completely enclosed fluid at rest. This is the same analysis of the above equation: Pb = Pt + ρgh. The bottom
    pressure is equal to the sum of top pressure, which is the externally applied pressure, and the static fluid
    pressure due to the weight of the fluid. For instance, this is the case with the mechanism of a hydraulic car lift
    when the static fluid pressure is zero. See Figure 11.15 in Section 11.5 of Chapter 11 in the eTextbook
    (Cutnell et al., 2022).

    You may experience that it is very hard to push a beach ball under the surface of the water. The water, in fact,
    pushes back, and this upward force is called the buoyant force, which arose because of the pressure of fluids
    depending on the depths. In the figure above, the net upward force is called the buoyant force F = PbA – PtA
    = ρghA = mg = weight. Notice that the buoyant force does not depend on the shape of the object. Archimedes
    discovered this property more than 2,000 years ago. Archimedes’ principle states that a fluid exerts a buoyant
    force to an immersed object. The magnitude of the buoyant force equals the weight of the displaced fluid.

    Bernoulli’s Equation and Its Application

    When fluids are in motion, they move with a variety of options. You may have observed that water flows
    calmly in a shallow stream and violently in a steep valley. The air blows very gently sometimes and vigorously
    with great speed at other times. In order to characterize the type of fluids, compressible/incompressible and
    viscous/nonviscous categories are used. When the density of a fluid is almost constant, the fluid is said to be
    incompressible. Luckily, most liquids are incompressible, but all gases are not.

    When fluid like honey does not flow easily, it is a viscous fluid. On the other hand, water flows very easily
    because its viscosity is low. Incompressible and nonviscous fluids are called ideal fluids; this is great to
    describe the motion of fluids with mathematical equations. For steady flow, Bernoulli studied the behavior of
    ideal fluids. Its result is in his equation: P + 1/2 ρv2 + ρgy = constant. In this equation, y is the elevation at any
    point, and v is the fluid speed. If the flow speed is not changing, the above equation goes back to the
    pressure equation when water is at rest. If the flow is horizontal, that is, there is no elevation between two
    points, the pressure is related to the speed. The higher fluid pressure makes the slow-moving flow and vice
    versa. In addition, when the volume flow rate Av = constant, if the cross-sectional area of a tube is large, the
    fluid speed is small and vice versa.

    With these fluid equations, we can describe the motion of liquids in a plumbing system, the speed changes of
    oil in a pipe, and even the dynamics of an airplane wing.

    Three Temperature Scales

    A thermometer is used to measure temperature. We can make a thermometer due to the fact that most
    materials expand when the heat is added. For example, a mercury thermometer, which consists of a mercury-

    Sample Question 2: A young lady is visiting a medical research center to measure her blood
    pressure difference between the blood pressure in the anterior tibial artery at the foot and the blood
    pressure in the aorta at the heart. Let’s assume that the blood in her body is a static fluid, and the
    vertical distance between the feet and the heart is 1.1 meter. What is the blood pressure difference
    between them? The density of blood is 1060 kg/m3.

    Solution: Let Pb be the blood pressure in the anterior tibial artery at the foot, and Pt be the blood
    pressure in the aorta at the heart. According to Pascal’s principle, Pb = Pt + ρgh. Here, ρ(= 1060
    kg/m3) is the density of blood. The acceleration g(= 9.8 m/s2) is due to gravity, and h is given as 1.1
    m. So, the difference Pb – Pt = ρgh = 1060 x 9.8 x 1.1 = 11426.8 Pa.

    Magnitude of Buoyant Force = Weight of Displaced Fluid

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    filled glass bulb connected to a capillary tube, is widely used. When mercury is heated, the expanded amount
    of mercury is directly proportional to the increased temperature.

    Two common temperature scales are the Fahrenheit and Celsius scales. Both scales are based on boiling
    and freezing points of water at atmospheric pressure. In the case of the Celsius scale, the boiling point is
    100oC, and the freezing point is 0oC. For the Fahrenheit scale, the boiling point is 212o F, and the freezing
    point is 32oF. The distance between these two points are divided equally to indicate the temperature scales.
    The separation between the freezing and boiling points on the Celsius scale is 100 degrees while that on the
    Fahrenheit scale is 180 degrees. The converting formula for the two scales is oF = 1.8oC + 32.

    In the scientific world, the Kelvin temperature scale, or absolute temperature, has more meaning. This is the
    SI unit for temperature, and it is defined as 1 K is 1/273.16 of the thermodynamic temperature of the triple
    point of water at which the three phases coexist with equilibrium.

    From the ideal gas law PV = const.*T. If the volume is not changing, the pressure (P) is directly proportional
    to the temperature (T). That is, we can use this relation to indicate temperature while adjusting the pressure of
    the gas container. If we plot between the temperature (T) and the pressure (P), the result indicates that –
    273.15oC, absolute zero points, has an important value because it is the corresponding value when the
    pressure is zero. It is impossible to reach lower than the absolute zero points, 0 K, the lowest temperature.
    The relation between the Celsius scale and the Kelvin scale is K = oC + 273.15.

    A Weird Property of Water

    Most substances contract as the temperature decreases and expand as the temperature increases; however,
    this is not the case for water. As the temperature decreases from room temperature (20oC = 68oF), water
    contracts until the temperature arrives at 4oC (= 39.2oF), and then it begins to expand as the temperature
    decreases to below 4oC. This is a unique characteristic of water.

    It has the maximum density (or minimum volume) at 4oC, not 0oC (= 32oF; freezing temperature) so
    organisms in the water can survive! Let’s consider a lake in winter. As the temperature decreases to 4oC, the
    top surface of the water is denser, so it goes down to the bottom. The next warm layer of water is now at the
    top surface of the water, and when this layer’s temperature drops toward 4oC, it will sink. This process will
    continue until the entire lake’s temperature is 4oC.

    As the air temperature decreases further, the density of the surface of the water in the lake will decrease, but
    the volume increases. No more sinking business occurs. The surface water is freezing when the temperature
    decreases to 0oC. That is, ice is forming on the surface, but under the surface, water is not freezing at all due
    to the special property of water. In fact, the role of the sheet of ice on the top surface of the lake is to be an
    insulator to preserve heat under it, so fish can live even though it is cold outside.

    Heat and Internal Energy

    An object has internal energy that is proportional to the temperature. The internal energy is the total energy of
    the object due to the molecular random motion, forces between molecules or atoms, and so on. Neglecting
    the work done, the internal energy of the hot object decreases and that of the cold object increases as heat
    transfers.

    Sample Question 3: The normal human body temperature is 98.6oF, which was determined in the
    19th century. A more recent study announced that it is 98.2oF. Express the difference in the
    temperature in Celsius.

    Solution: The converting formula for the two scales is oF = 1.8oC + 32.
    That is, oC = (oF -32) / 1.8. So, 98.6oF is (98.6 – 32) / 1.8 = 37oC.
    98.2oF is (98.2 – 32) / 1.8 = 36.778oC. The difference is 37 – 36.78 = 0.22oC.

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    The heat (Q) must be added or removed to change the temperature of a material of mass (m): Q = cm dT,
    where c is the specific heat capacity, and dT is the temperature difference.

    Heat and Phase Change

    As the heat is added to or removed from a material, the internal energy varies, which causes a change in
    temperature or a change in phase. However, it is possible to maintain a constant temperature whether the
    heat is added or not. Consider a cup of water with ice in a warmer room. See Figure 12.27 in Section 12.8 of
    Chapter 12 in the eTextbook (Cutnell et al., 2022) for a graphic representation of the energy involved in the
    phase changes of water.

    The heat will be added to the cup, but the temperature does not increase above 0oC as long as there are ice
    cubes. In fact, the heat is used to melt the ice. When the ice is completely melted, the temperature is now
    increasing. The heat (Q) used to change from one phase to another phase of the matter is Q = mL, where L is
    the latent heat. Its unit is J / kg.

    The latent heat of fusion refers to the change between solid and liquid phases. The latent heat of vaporization
    refers to the change between liquid and solid phases, and the latent heat of sublimation refers to the change
    between solid and gas phases. See Figure 12.26 in Section 12.8 of Chapter 12 in the eTextbook (Cutnell et
    al., 2022) for the phase changes between any two of the three phases of matter.

    Reference

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 77–94 under Physics in Biology, Medicine, and Sports in Chapter 11 of your
    eTextbook.

    2. Solve Questions 82–93 under Physics in Biology, Medicine, and Sports in Chapter 12 of your
    eTextbook.

    Sample Question 4: Suppose 0.5 kg of blood flows from the interior to the surface of John’s body while
    he is exercising. The released energy is 2000 J. The specific heat capacity of blood is 4186 J / kgoC.
    What is the temperature difference between when the blood arrives at the body surface and returns back
    to the interior of the body?

    Solution: Use the formula, Q = cm dT. Here, the specific heat capacity c is given as 4186 J / kgoC. The
    mass m is 0.5 kg. Q is 2000 J. So, the temperature difference dT = Q / cm = 2000 / (4186 x 0.5) =
    0.96oC.

    Sample Question 5: How much energy is needed to change 100 g of 0oC ice to 0oC water? The
    latent heat of fusion for water L = 335,000 J / kg.

    Solution: The heat (Q) used to change from one phase to another phase of the matter is Q = mL,
    where L is the latent heat. Its unit is J / kg. Here, the mass m is given as 100g = 0.1 kg. The latent
    heat L is given as 335,000 J / kg. So, the energy needed is Q = mL = 0.1 x 335000 = 33500 J.

    – Course Learning Outcomes for Unit VI
    – Required Unit Resources
    – Unit Lesson

    – What Are Fluids?
    – Pressure in Fluids
    – Pascal’s and Archimedes’ Principles
    – Bernoulli’s Equation and Its Application
    – Three Temperature Scales
    – A Weird Property of Water
    – Heat and Internal Energy
    – Heat and Phase Change
    – Reference

    – Learning Activities (Nongraded)

    **UnitVIExampleConceptMap.png**

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    **UnitVIExampleConceptMap.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/unitviexampleconceptmap-png)

    **## UnitVIAssignment.docx**

    Unit VI

    This concept map assignment measures ULO 6.1, 6.2, and 6.3.

    Among Archimedes, Pascal, and Bernoulli, who do you think is the most important contributor to developing modern physics in fluid dynamics? Choose one of them, then create a concept map to describe his equations with examples of applications. A concept map or conceptual diagram is a diagram that expresses the concepts or delivers the important ideas of a subject. Review the
    [Unit VI Example Concept Map](https://online.columbiasouthern.edu/bbcswebdav/pid-17830837-dt-asiobject-rid-190111713_1/xid-190111713_1). It does not have to be fancy, but you are encouraged to be as creative as possible to emphasize the content that you want to depict. You may use PowerPoint slides, document files, pdf files, Photoshop, or even a piece of paper with coloring pencils.

    **## UnitVI.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit VI

    Upon completion of this unit, students should be able to:

    6. Explain numerous phenomena using fluid mechanics laws.
    6.1 Utilize the relationship between mass, density, and volume.
    6.2 Apply the concept of Pascal’s principle and Archimedes’ principle.
    6.3 Recognize heat and energy with phase changes of matter.

    Required Unit Resources

    Chapter 11: Fluids

    Chapter 12: Temperature and Heat

    Unit Lesson

    What Are Fluids?

    Unlike solids, liquids and gasses do not have a definite shape unless they are kept in a container. Something
    is referred to as fluid when it can flow without specific configuration. Both liquids and gases are examples of
    fluids. Both water and air move from place to place if they are not confined in a vessel. Both of them show a
    similar pattern of motion, but their density is very different because it depends on the property of matter.
    Usually, gases have lower densities than liquids have because the average distance between molecules is
    greater in gases than in liquids. The density of gases is greatly dependent on pressure and temperature.

    Mass density is defined as the mass of an object divided by the volume of the object. The SI unit of mass is
    kg, that of density is kg / m3, and that of volume is m3.

    UNIT VI STUDY GUIDE
    Basic Fluid Dynamics

    Sample Question 1: If the mass of a diamond box is 100 kg, what is the volume of the box?

    Solution: Use the formula mass = density x volume. The density of diamond is 3520 kg/m3 from
    Table 11.1 under Section 11.1 in Chapter 11 of the eTextbook (Cutnell et al., 2022). So, the
    volume is the mass divided by the density.

    Mass = Density x Volume

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Pressure in Fluids

    When a force (F) acts on area (A) in a fluid, the pressure (P) can be expressed as P = F / A. Here, we only
    consider the magnitude of F, so P is a scalar and its unit is N / m2 = Pa (Pascal). Atmospheric pressure at sea
    level is about 101,300 Pa = 1 atm.

    You may experience greater pressure as you go deeper into a swimming pool or in an ocean. What is the
    relation between pressure and depth? Let’s consider one column of water with height h below the figure in a
    large swimming pool. The area (A) of the top face is the same as that of the bottom face. The pressure (Pt) on
    the top face creates a downward force or PtA. The pressure (Pb) creates an upward force or PbA. Also, the
    weight (mg) due to gravity points downward.

    Water is at rest, and thus its acceleration is zero. That is, the column is in equilibrium. We can apply Newton’s
    second law, and the summation of the vertical forces is zero: PbA – PtA – mg = 0. Use m = ρV = ρAh. Then,
    Pb = Pt + ρgh. You can clearly see that the pressure at a deeper level is greater than the pressure at a shallow
    level if the density is not changing or if it is incompressible. In the case of gas, the density varies according to
    the vertical distances or if it is compressible, so the formula only works when h is very small. For instance, the
    density of our atmosphere varies significantly from the Earth’s surface to higher altitudes. The important thing
    is that the pressure difference between lower and higher levels comes from the height, or the vertical
    distance, not the horizontal distance within the fluid. See Figure 11.5 in Section 11.3 of Chapter 11 in the
    eTextbook (Cutnell et al., 2022).

    Force = Pressure x Area

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Pascal’s and Archimedes’ Principles

    Pascal’s principle states that any externally applied pressure is transmitted undiminished to everywhere in a
    completely enclosed fluid at rest. This is the same analysis of the above equation: Pb = Pt + ρgh. The bottom
    pressure is equal to the sum of top pressure, which is the externally applied pressure, and the static fluid
    pressure due to the weight of the fluid. For instance, this is the case with the mechanism of a hydraulic car lift
    when the static fluid pressure is zero. See Figure 11.15 in Section 11.5 of Chapter 11 in the eTextbook
    (Cutnell et al., 2022).

    You may experience that it is very hard to push a beach ball under the surface of the water. The water, in fact,
    pushes back, and this upward force is called the buoyant force, which arose because of the pressure of fluids
    depending on the depths. In the figure above, the net upward force is called the buoyant force F = PbA – PtA
    = ρghA = mg = weight. Notice that the buoyant force does not depend on the shape of the object. Archimedes
    discovered this property more than 2,000 years ago. Archimedes’ principle states that a fluid exerts a buoyant
    force to an immersed object. The magnitude of the buoyant force equals the weight of the displaced fluid.

    Bernoulli’s Equation and Its Application

    When fluids are in motion, they move with a variety of options. You may have observed that water flows
    calmly in a shallow stream and violently in a steep valley. The air blows very gently sometimes and vigorously
    with great speed at other times. In order to characterize the type of fluids, compressible/incompressible and
    viscous/nonviscous categories are used. When the density of a fluid is almost constant, the fluid is said to be
    incompressible. Luckily, most liquids are incompressible, but all gases are not.

    When fluid like honey does not flow easily, it is a viscous fluid. On the other hand, water flows very easily
    because its viscosity is low. Incompressible and nonviscous fluids are called ideal fluids; this is great to
    describe the motion of fluids with mathematical equations. For steady flow, Bernoulli studied the behavior of
    ideal fluids. Its result is in his equation: P + 1/2 ρv2 + ρgy = constant. In this equation, y is the elevation at any
    point, and v is the fluid speed. If the flow speed is not changing, the above equation goes back to the
    pressure equation when water is at rest. If the flow is horizontal, that is, there is no elevation between two
    points, the pressure is related to the speed. The higher fluid pressure makes the slow-moving flow and vice
    versa. In addition, when the volume flow rate Av = constant, if the cross-sectional area of a tube is large, the
    fluid speed is small and vice versa.

    With these fluid equations, we can describe the motion of liquids in a plumbing system, the speed changes of
    oil in a pipe, and even the dynamics of an airplane wing.

    Three Temperature Scales

    A thermometer is used to measure temperature. We can make a thermometer due to the fact that most
    materials expand when the heat is added. For example, a mercury thermometer, which consists of a mercury-

    Sample Question 2: A young lady is visiting a medical research center to measure her blood
    pressure difference between the blood pressure in the anterior tibial artery at the foot and the blood
    pressure in the aorta at the heart. Let’s assume that the blood in her body is a static fluid, and the
    vertical distance between the feet and the heart is 1.1 meter. What is the blood pressure difference
    between them? The density of blood is 1060 kg/m3.

    Solution: Let Pb be the blood pressure in the anterior tibial artery at the foot, and Pt be the blood
    pressure in the aorta at the heart. According to Pascal’s principle, Pb = Pt + ρgh. Here, ρ(= 1060
    kg/m3) is the density of blood. The acceleration g(= 9.8 m/s2) is due to gravity, and h is given as 1.1
    m. So, the difference Pb – Pt = ρgh = 1060 x 9.8 x 1.1 = 11426.8 Pa.

    Magnitude of Buoyant Force = Weight of Displaced Fluid

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    filled glass bulb connected to a capillary tube, is widely used. When mercury is heated, the expanded amount
    of mercury is directly proportional to the increased temperature.

    Two common temperature scales are the Fahrenheit and Celsius scales. Both scales are based on boiling
    and freezing points of water at atmospheric pressure. In the case of the Celsius scale, the boiling point is
    100oC, and the freezing point is 0oC. For the Fahrenheit scale, the boiling point is 212o F, and the freezing
    point is 32oF. The distance between these two points are divided equally to indicate the temperature scales.
    The separation between the freezing and boiling points on the Celsius scale is 100 degrees while that on the
    Fahrenheit scale is 180 degrees. The converting formula for the two scales is oF = 1.8oC + 32.

    In the scientific world, the Kelvin temperature scale, or absolute temperature, has more meaning. This is the
    SI unit for temperature, and it is defined as 1 K is 1/273.16 of the thermodynamic temperature of the triple
    point of water at which the three phases coexist with equilibrium.

    From the ideal gas law PV = const.*T. If the volume is not changing, the pressure (P) is directly proportional
    to the temperature (T). That is, we can use this relation to indicate temperature while adjusting the pressure of
    the gas container. If we plot between the temperature (T) and the pressure (P), the result indicates that –
    273.15oC, absolute zero points, has an important value because it is the corresponding value when the
    pressure is zero. It is impossible to reach lower than the absolute zero points, 0 K, the lowest temperature.
    The relation between the Celsius scale and the Kelvin scale is K = oC + 273.15.

    A Weird Property of Water

    Most substances contract as the temperature decreases and expand as the temperature increases; however,
    this is not the case for water. As the temperature decreases from room temperature (20oC = 68oF), water
    contracts until the temperature arrives at 4oC (= 39.2oF), and then it begins to expand as the temperature
    decreases to below 4oC. This is a unique characteristic of water.

    It has the maximum density (or minimum volume) at 4oC, not 0oC (= 32oF; freezing temperature) so
    organisms in the water can survive! Let’s consider a lake in winter. As the temperature decreases to 4oC, the
    top surface of the water is denser, so it goes down to the bottom. The next warm layer of water is now at the
    top surface of the water, and when this layer’s temperature drops toward 4oC, it will sink. This process will
    continue until the entire lake’s temperature is 4oC.

    As the air temperature decreases further, the density of the surface of the water in the lake will decrease, but
    the volume increases. No more sinking business occurs. The surface water is freezing when the temperature
    decreases to 0oC. That is, ice is forming on the surface, but under the surface, water is not freezing at all due
    to the special property of water. In fact, the role of the sheet of ice on the top surface of the lake is to be an
    insulator to preserve heat under it, so fish can live even though it is cold outside.

    Heat and Internal Energy

    An object has internal energy that is proportional to the temperature. The internal energy is the total energy of
    the object due to the molecular random motion, forces between molecules or atoms, and so on. Neglecting
    the work done, the internal energy of the hot object decreases and that of the cold object increases as heat
    transfers.

    Sample Question 3: The normal human body temperature is 98.6oF, which was determined in the
    19th century. A more recent study announced that it is 98.2oF. Express the difference in the
    temperature in Celsius.

    Solution: The converting formula for the two scales is oF = 1.8oC + 32.
    That is, oC = (oF -32) / 1.8. So, 98.6oF is (98.6 – 32) / 1.8 = 37oC.
    98.2oF is (98.2 – 32) / 1.8 = 36.778oC. The difference is 37 – 36.78 = 0.22oC.

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    The heat (Q) must be added or removed to change the temperature of a material of mass (m): Q = cm dT,
    where c is the specific heat capacity, and dT is the temperature difference.

    Heat and Phase Change

    As the heat is added to or removed from a material, the internal energy varies, which causes a change in
    temperature or a change in phase. However, it is possible to maintain a constant temperature whether the
    heat is added or not. Consider a cup of water with ice in a warmer room. See Figure 12.27 in Section 12.8 of
    Chapter 12 in the eTextbook (Cutnell et al., 2022) for a graphic representation of the energy involved in the
    phase changes of water.

    The heat will be added to the cup, but the temperature does not increase above 0oC as long as there are ice
    cubes. In fact, the heat is used to melt the ice. When the ice is completely melted, the temperature is now
    increasing. The heat (Q) used to change from one phase to another phase of the matter is Q = mL, where L is
    the latent heat. Its unit is J / kg.

    The latent heat of fusion refers to the change between solid and liquid phases. The latent heat of vaporization
    refers to the change between liquid and solid phases, and the latent heat of sublimation refers to the change
    between solid and gas phases. See Figure 12.26 in Section 12.8 of Chapter 12 in the eTextbook (Cutnell et
    al., 2022) for the phase changes between any two of the three phases of matter.

    Reference

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 77–94 under Physics in Biology, Medicine, and Sports in Chapter 11 of your
    eTextbook.

    2. Solve Questions 82–93 under Physics in Biology, Medicine, and Sports in Chapter 12 of your
    eTextbook.

    Sample Question 4: Suppose 0.5 kg of blood flows from the interior to the surface of John’s body while
    he is exercising. The released energy is 2000 J. The specific heat capacity of blood is 4186 J / kgoC.
    What is the temperature difference between when the blood arrives at the body surface and returns back
    to the interior of the body?

    Solution: Use the formula, Q = cm dT. Here, the specific heat capacity c is given as 4186 J / kgoC. The
    mass m is 0.5 kg. Q is 2000 J. So, the temperature difference dT = Q / cm = 2000 / (4186 x 0.5) =
    0.96oC.

    Sample Question 5: How much energy is needed to change 100 g of 0oC ice to 0oC water? The
    latent heat of fusion for water L = 335,000 J / kg.

    Solution: The heat (Q) used to change from one phase to another phase of the matter is Q = mL,
    where L is the latent heat. Its unit is J / kg. Here, the mass m is given as 100g = 0.1 kg. The latent
    heat L is given as 335,000 J / kg. So, the energy needed is Q = mL = 0.1 x 335000 = 33500 J.

    – Course Learning Outcomes for Unit VI
    – Required Unit Resources
    – Unit Lesson

    – What Are Fluids?
    – Pressure in Fluids
    – Pascal’s and Archimedes’ Principles
    – Bernoulli’s Equation and Its Application
    – Three Temperature Scales
    – A Weird Property of Water
    – Heat and Internal Energy
    – Heat and Phase Change
    – Reference

    – Learning Activities (Nongraded)

    **UnitVIExampleConceptMap.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/unitviexampleconceptmap-png)

    **## UnitVIAssignment.docx**

    Unit VI

    This concept map assignment measures ULO 6.1, 6.2, and 6.3.

    Among Archimedes, Pascal, and Bernoulli, who do you think is the most important contributor to developing modern physics in fluid dynamics? Choose one of them, then create a concept map to describe his equations with examples of applications. A concept map or conceptual diagram is a diagram that expresses the concepts or delivers the important ideas of a subject. Review the
    [Unit VI Example Concept Map](https://online.columbiasouthern.edu/bbcswebdav/pid-17830837-dt-asiobject-rid-190111713_1/xid-190111713_1). It does not have to be fancy, but you are encouraged to be as creative as possible to emphasize the content that you want to depict. You may use PowerPoint slides, document files, pdf files, Photoshop, or even a piece of paper with coloring pencils.

    **## UnitVI.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit VI

    Upon completion of this unit, students should be able to:

    6. Explain numerous phenomena using fluid mechanics laws.
    6.1 Utilize the relationship between mass, density, and volume.
    6.2 Apply the concept of Pascal’s principle and Archimedes’ principle.
    6.3 Recognize heat and energy with phase changes of matter.

    Required Unit Resources

    Chapter 11: Fluids

    Chapter 12: Temperature and Heat

    Unit Lesson

    What Are Fluids?

    Unlike solids, liquids and gasses do not have a definite shape unless they are kept in a container. Something
    is referred to as fluid when it can flow without specific configuration. Both liquids and gases are examples of
    fluids. Both water and air move from place to place if they are not confined in a vessel. Both of them show a
    similar pattern of motion, but their density is very different because it depends on the property of matter.
    Usually, gases have lower densities than liquids have because the average distance between molecules is
    greater in gases than in liquids. The density of gases is greatly dependent on pressure and temperature.

    Mass density is defined as the mass of an object divided by the volume of the object. The SI unit of mass is
    kg, that of density is kg / m3, and that of volume is m3.

    UNIT VI STUDY GUIDE
    Basic Fluid Dynamics

    Sample Question 1: If the mass of a diamond box is 100 kg, what is the volume of the box?

    Solution: Use the formula mass = density x volume. The density of diamond is 3520 kg/m3 from
    Table 11.1 under Section 11.1 in Chapter 11 of the eTextbook (Cutnell et al., 2022). So, the
    volume is the mass divided by the density.

    Mass = Density x Volume

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Pressure in Fluids

    When a force (F) acts on area (A) in a fluid, the pressure (P) can be expressed as P = F / A. Here, we only
    consider the magnitude of F, so P is a scalar and its unit is N / m2 = Pa (Pascal). Atmospheric pressure at sea
    level is about 101,300 Pa = 1 atm.

    You may experience greater pressure as you go deeper into a swimming pool or in an ocean. What is the
    relation between pressure and depth? Let’s consider one column of water with height h below the figure in a
    large swimming pool. The area (A) of the top face is the same as that of the bottom face. The pressure (Pt) on
    the top face creates a downward force or PtA. The pressure (Pb) creates an upward force or PbA. Also, the
    weight (mg) due to gravity points downward.

    Water is at rest, and thus its acceleration is zero. That is, the column is in equilibrium. We can apply Newton’s
    second law, and the summation of the vertical forces is zero: PbA – PtA – mg = 0. Use m = ρV = ρAh. Then,
    Pb = Pt + ρgh. You can clearly see that the pressure at a deeper level is greater than the pressure at a shallow
    level if the density is not changing or if it is incompressible. In the case of gas, the density varies according to
    the vertical distances or if it is compressible, so the formula only works when h is very small. For instance, the
    density of our atmosphere varies significantly from the Earth’s surface to higher altitudes. The important thing
    is that the pressure difference between lower and higher levels comes from the height, or the vertical
    distance, not the horizontal distance within the fluid. See Figure 11.5 in Section 11.3 of Chapter 11 in the
    eTextbook (Cutnell et al., 2022).

    Force = Pressure x Area

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Pascal’s and Archimedes’ Principles

    Pascal’s principle states that any externally applied pressure is transmitted undiminished to everywhere in a
    completely enclosed fluid at rest. This is the same analysis of the above equation: Pb = Pt + ρgh. The bottom
    pressure is equal to the sum of top pressure, which is the externally applied pressure, and the static fluid
    pressure due to the weight of the fluid. For instance, this is the case with the mechanism of a hydraulic car lift
    when the static fluid pressure is zero. See Figure 11.15 in Section 11.5 of Chapter 11 in the eTextbook
    (Cutnell et al., 2022).

    You may experience that it is very hard to push a beach ball under the surface of the water. The water, in fact,
    pushes back, and this upward force is called the buoyant force, which arose because of the pressure of fluids
    depending on the depths. In the figure above, the net upward force is called the buoyant force F = PbA – PtA
    = ρghA = mg = weight. Notice that the buoyant force does not depend on the shape of the object. Archimedes
    discovered this property more than 2,000 years ago. Archimedes’ principle states that a fluid exerts a buoyant
    force to an immersed object. The magnitude of the buoyant force equals the weight of the displaced fluid.

    Bernoulli’s Equation and Its Application

    When fluids are in motion, they move with a variety of options. You may have observed that water flows
    calmly in a shallow stream and violently in a steep valley. The air blows very gently sometimes and vigorously
    with great speed at other times. In order to characterize the type of fluids, compressible/incompressible and
    viscous/nonviscous categories are used. When the density of a fluid is almost constant, the fluid is said to be
    incompressible. Luckily, most liquids are incompressible, but all gases are not.

    When fluid like honey does not flow easily, it is a viscous fluid. On the other hand, water flows very easily
    because its viscosity is low. Incompressible and nonviscous fluids are called ideal fluids; this is great to
    describe the motion of fluids with mathematical equations. For steady flow, Bernoulli studied the behavior of
    ideal fluids. Its result is in his equation: P + 1/2 ρv2 + ρgy = constant. In this equation, y is the elevation at any
    point, and v is the fluid speed. If the flow speed is not changing, the above equation goes back to the
    pressure equation when water is at rest. If the flow is horizontal, that is, there is no elevation between two
    points, the pressure is related to the speed. The higher fluid pressure makes the slow-moving flow and vice
    versa. In addition, when the volume flow rate Av = constant, if the cross-sectional area of a tube is large, the
    fluid speed is small and vice versa.

    With these fluid equations, we can describe the motion of liquids in a plumbing system, the speed changes of
    oil in a pipe, and even the dynamics of an airplane wing.

    Three Temperature Scales

    A thermometer is used to measure temperature. We can make a thermometer due to the fact that most
    materials expand when the heat is added. For example, a mercury thermometer, which consists of a mercury-

    Sample Question 2: A young lady is visiting a medical research center to measure her blood
    pressure difference between the blood pressure in the anterior tibial artery at the foot and the blood
    pressure in the aorta at the heart. Let’s assume that the blood in her body is a static fluid, and the
    vertical distance between the feet and the heart is 1.1 meter. What is the blood pressure difference
    between them? The density of blood is 1060 kg/m3.

    Solution: Let Pb be the blood pressure in the anterior tibial artery at the foot, and Pt be the blood
    pressure in the aorta at the heart. According to Pascal’s principle, Pb = Pt + ρgh. Here, ρ(= 1060
    kg/m3) is the density of blood. The acceleration g(= 9.8 m/s2) is due to gravity, and h is given as 1.1
    m. So, the difference Pb – Pt = ρgh = 1060 x 9.8 x 1.1 = 11426.8 Pa.

    Magnitude of Buoyant Force = Weight of Displaced Fluid

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    filled glass bulb connected to a capillary tube, is widely used. When mercury is heated, the expanded amount
    of mercury is directly proportional to the increased temperature.

    Two common temperature scales are the Fahrenheit and Celsius scales. Both scales are based on boiling
    and freezing points of water at atmospheric pressure. In the case of the Celsius scale, the boiling point is
    100oC, and the freezing point is 0oC. For the Fahrenheit scale, the boiling point is 212o F, and the freezing
    point is 32oF. The distance between these two points are divided equally to indicate the temperature scales.
    The separation between the freezing and boiling points on the Celsius scale is 100 degrees while that on the
    Fahrenheit scale is 180 degrees. The converting formula for the two scales is oF = 1.8oC + 32.

    In the scientific world, the Kelvin temperature scale, or absolute temperature, has more meaning. This is the
    SI unit for temperature, and it is defined as 1 K is 1/273.16 of the thermodynamic temperature of the triple
    point of water at which the three phases coexist with equilibrium.

    From the ideal gas law PV = const.*T. If the volume is not changing, the pressure (P) is directly proportional
    to the temperature (T). That is, we can use this relation to indicate temperature while adjusting the pressure of
    the gas container. If we plot between the temperature (T) and the pressure (P), the result indicates that –
    273.15oC, absolute zero points, has an important value because it is the corresponding value when the
    pressure is zero. It is impossible to reach lower than the absolute zero points, 0 K, the lowest temperature.
    The relation between the Celsius scale and the Kelvin scale is K = oC + 273.15.

    A Weird Property of Water

    Most substances contract as the temperature decreases and expand as the temperature increases; however,
    this is not the case for water. As the temperature decreases from room temperature (20oC = 68oF), water
    contracts until the temperature arrives at 4oC (= 39.2oF), and then it begins to expand as the temperature
    decreases to below 4oC. This is a unique characteristic of water.

    It has the maximum density (or minimum volume) at 4oC, not 0oC (= 32oF; freezing temperature) so
    organisms in the water can survive! Let’s consider a lake in winter. As the temperature decreases to 4oC, the
    top surface of the water is denser, so it goes down to the bottom. The next warm layer of water is now at the
    top surface of the water, and when this layer’s temperature drops toward 4oC, it will sink. This process will
    continue until the entire lake’s temperature is 4oC.

    As the air temperature decreases further, the density of the surface of the water in the lake will decrease, but
    the volume increases. No more sinking business occurs. The surface water is freezing when the temperature
    decreases to 0oC. That is, ice is forming on the surface, but under the surface, water is not freezing at all due
    to the special property of water. In fact, the role of the sheet of ice on the top surface of the lake is to be an
    insulator to preserve heat under it, so fish can live even though it is cold outside.

    Heat and Internal Energy

    An object has internal energy that is proportional to the temperature. The internal energy is the total energy of
    the object due to the molecular random motion, forces between molecules or atoms, and so on. Neglecting
    the work done, the internal energy of the hot object decreases and that of the cold object increases as heat
    transfers.

    Sample Question 3: The normal human body temperature is 98.6oF, which was determined in the
    19th century. A more recent study announced that it is 98.2oF. Express the difference in the
    temperature in Celsius.

    Solution: The converting formula for the two scales is oF = 1.8oC + 32.
    That is, oC = (oF -32) / 1.8. So, 98.6oF is (98.6 – 32) / 1.8 = 37oC.
    98.2oF is (98.2 – 32) / 1.8 = 36.778oC. The difference is 37 – 36.78 = 0.22oC.

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    The heat (Q) must be added or removed to change the temperature of a material of mass (m): Q = cm dT,
    where c is the specific heat capacity, and dT is the temperature difference.

    Heat and Phase Change

    As the heat is added to or removed from a material, the internal energy varies, which causes a change in
    temperature or a change in phase. However, it is possible to maintain a constant temperature whether the
    heat is added or not. Consider a cup of water with ice in a warmer room. See Figure 12.27 in Section 12.8 of
    Chapter 12 in the eTextbook (Cutnell et al., 2022) for a graphic representation of the energy involved in the
    phase changes of water.

    The heat will be added to the cup, but the temperature does not increase above 0oC as long as there are ice
    cubes. In fact, the heat is used to melt the ice. When the ice is completely melted, the temperature is now
    increasing. The heat (Q) used to change from one phase to another phase of the matter is Q = mL, where L is
    the latent heat. Its unit is J / kg.

    The latent heat of fusion refers to the change between solid and liquid phases. The latent heat of vaporization
    refers to the change between liquid and solid phases, and the latent heat of sublimation refers to the change
    between solid and gas phases. See Figure 12.26 in Section 12.8 of Chapter 12 in the eTextbook (Cutnell et
    al., 2022) for the phase changes between any two of the three phases of matter.

    Reference

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 77–94 under Physics in Biology, Medicine, and Sports in Chapter 11 of your
    eTextbook.

    2. Solve Questions 82–93 under Physics in Biology, Medicine, and Sports in Chapter 12 of your
    eTextbook.

    Sample Question 4: Suppose 0.5 kg of blood flows from the interior to the surface of John’s body while
    he is exercising. The released energy is 2000 J. The specific heat capacity of blood is 4186 J / kgoC.
    What is the temperature difference between when the blood arrives at the body surface and returns back
    to the interior of the body?

    Solution: Use the formula, Q = cm dT. Here, the specific heat capacity c is given as 4186 J / kgoC. The
    mass m is 0.5 kg. Q is 2000 J. So, the temperature difference dT = Q / cm = 2000 / (4186 x 0.5) =
    0.96oC.

    Sample Question 5: How much energy is needed to change 100 g of 0oC ice to 0oC water? The
    latent heat of fusion for water L = 335,000 J / kg.

    Solution: The heat (Q) used to change from one phase to another phase of the matter is Q = mL,
    where L is the latent heat. Its unit is J / kg. Here, the mass m is given as 100g = 0.1 kg. The latent
    heat L is given as 335,000 J / kg. So, the energy needed is Q = mL = 0.1 x 335000 = 33500 J.

    – Course Learning Outcomes for Unit VI
    – Required Unit Resources
    – Unit Lesson

    – What Are Fluids?
    – Pressure in Fluids
    – Pascal’s and Archimedes’ Principles
    – Bernoulli’s Equation and Its Application
    – Three Temperature Scales
    – A Weird Property of Water
    – Heat and Internal Energy
    – Heat and Phase Change
    – Reference

    – Learning Activities (Nongraded)

    **UnitVIExampleConceptMap.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/unitviexampleconceptmap-png)

  • [Solved] Physics JL VIII

    ✓ Verified Academic Solution Available

    Executive Solution Overview

    See attached **files (2)** – ## UnitVIIIJournal.docx – ## UnitVIII.pdf **## UnitVIIIJournal.docx** Unit VIII Journal This journal measures your mastery of Unit VIII’s ULO 7.1. If you were asked to choose one of the four thermodynamics laws, which law do you think is the most important? Explain your …

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    – ## UnitVIIIJournal.docx

    – ## UnitVIII.pdf

    **## UnitVIIIJournal.docx**

    Unit VIII Journal

    This journal measures your mastery of Unit VIII’s ULO 7.1.

    If you were asked to choose one of the four thermodynamics laws, which law do you think is the most important? Explain your reasoning based on your daily activity. Rank the importance of each law based on your insight. Justify your reason.

    Reflect on what you have learned in this class. How can these concepts be applied to real-world situations and increase your chances of career or life success?

    Your journal entry must be at least 200 words in length. No references or citations are necessary.

    **## UnitVIII.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit VIII

    Upon completion of this unit, students should be able to:

    1. Describe the system of units for various physical quantities.

    2. Illustrate the scientific method within everyday situations.

    3. Explain Newton's laws of motion at work in common phenomena.

    4. Apply the concept of momentum conservations to daily life.

    5. Identify the total mechanical energy conservation.

    6. Explain numerous phenomena using fluid mechanics laws.

    7. Describe fundamental thermodynamic concepts.
    7.1 Recognize the first law of thermodynamics.
    7.2 Identify the maximum efficiency of a heat engine.
    7.3 Interpret the various thermal processes.

    Required Unit Resources

    Chapter 15: Thermodynamics

    Unit Lesson

    The Zeroth Law of Thermodynamics

    If there is no heat transfer (adiabatic process) between two systems, they are said to be in thermal equilibrium
    so that their temperatures are identical. The adiabatic process is a thermal process with no heat flow. If a third
    system is in thermal equilibrium with the two systems simultaneously, they are all in thermal equilibrium with
    the same temperature. This is known as the zeroth law of thermodynamics. The temperature is the indicator
    of thermal equilibrium, in which any flow of heat is prohibited in the system. Please look at Figure 15.2 under
    Section 15.2 of Chapter 15 in the eTextbook (Cutnell et al., 2022). The mercury level is the same in the
    thermometer in Systems A and B.

    The First Law of Thermodynamics

    This law is just the energy conservation law applied to heat, work, and internal energy. The internal energy of
    a system changes due to heat (Q) and work (W): U = Q – W. The change in internal energy is equivalent to
    the difference between the heat added to the system and the work done by the system (Cutnell et al., 2022).

    Please note, Q is positive when the system gains heat, and Q is negative when the system loses heat.
    Similarly, W is positive when work is done by the system, and W is negative when work is done on the
    system.

    UNIT VIII STUDY GUIDE
    Thermodynamics

    Change in Internal Energy = Heat – Work

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Thermal Processes

    In general, there are four types of thermal processes when considering the ideal gas: isobaric, isochoric,
    isothermal, and adiabatic. The isobaric process is when the pressure of a system is constant. The work done,
    in this case, is the product of the pressure and the change in volume. The change in volume is calculated by
    the final volume minus the initial volume. The isochoric process is when the volume is constant, so the work
    done is none. Consequently, the change in internal energy is the same as the heat. The isothermal process is
    when the temperature is constant. Thus, the final internal energy is equal to the initial internal energy. Lastly,
    the adiabatic process is when the heat does not transfer to anywhere. The adiabatic expansion or
    compression of the ideal gas is related to the ratio of the specific heat capacities at constant pressure and
    constant volume.

    The Second Law of Thermodynamics

    The second law can be stated in various ways. Let’s consider the heat flow statement. It states that heat flows
    spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not
    flow spontaneously in the reverse direction. In order to understand the meaning of this statement, consider
    heat engines. A heat engine has three main components: a hot reservoir, a cold reservoir, and work output.
    See Figure 15.13 under Section 15.8 of Chapter 15 in eTextbook (Cutnell et al., 2022) for a graphic depiction
    of this concept.

    It is impossible to create a perfect engine that uses all of the input heat to produce the work done. Some of
    the input heat must flow to the cold reservoir. It is also impossible to create a perfect refrigerator. Without any
    work, the heat cannot flow from the cold reservoir to the hot reservoir.

    Carnot suggested the condition for the maximum efficiency of a heat engine in a reversible process
    (Erlichson, 1999). The reversible process refers to when the final state can be returned to its original state.
    The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input
    heat QH: e = W / QH. W = QH – QC, where Qc is the output heat. That is, e = 1 – Qc / QH = 1 – Tc / TH, where Tc
    stands for the temperature of the cold reservoir, and TH stands for the temperature of the hot reservoir. The
    unit of temperature must be in Kelvin.

    Sample Question 1: The internal energy U of a system is 7000 J. The work done W by the
    system is given as 8000 J. Determine if the heat is gained or lost to the system.

    Solution: Use the formula U = Q – W. W is positive when work is done by the system, so W =
    +8000 J. The heat Q = U + W = 7000 + 8000= 15,000 J. It is positive, so the system gains 15,000
    J of heat from the surroundings.

    Sample Question 2: We are going to use a tropical ocean as a heat engine. The surface
    temperature is 300 K, and the underwater temperature is 280 K. Determine the maximum possible
    efficiency of this engine.

    Solution: Use the formula e = 1 – Tc / TH. Here, Tc is 280 K, and TH is 300 K. The warm water
    should be a hot reservoir, and the cool water should be a cold reservoir.

    So, e = 1 – 280 / 300 = 1 / 15 ≈ 0.06 = 6%.

    Work Done = Efficiency x Input Heat

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Carnot’s principle or an alternative statement of the second law of thermodynamics states that no irreversible
    engine operating between two reservoirs at constant temperatures can have a greater efficiency than a
    reversible engine operating between the same temperatures (Erlichson, 1999). In addition, all reversible
    engines operating between the same temperatures have the same efficiency.

    Entropy is introduced to describe the order of a system. The second law of thermodynamics can be stated
    using the concept of entropy. The change in entropy (S) is defined as when the heat (Q) flows reversibly at a
    constant (T): S = Q / T. In the reversible process, the total entropy of the universe is constant: S = 0. In an
    irreversible process, the entropy is always increasing: S > 0. For instance, suppose the total energy q that is
    transferred from a hot region in the universe at temperature T1 to a cold region at temperature T2. From the
    definition of the entropy, the hot region’s entropy S1 is the total energy q divided by the hot region’s
    temperature T1. In the same manner, the cold region’s entropy S2 is the total energy q divided by the cold
    region’s temperature T2. In other words, S1 = q / T1 and S2 = q / T2. The change in entropy is obtained by S2
    – S1 = q / T2 – q / T1 = q(1 / T2 – 1 / T1). Please note, that T2 is less than T1. That is, 1 / T2 is greater than 1
    / T1. Thus, the change in entropy is positive. Therefore, the total entropy of the universe is increasing.
    Furthermore, the entropy is strongly related to the probability. When considering a gas in a box, the
    probability that all the gas particles exist in a particular place at the same time is extremely small. That is, it
    has a very low entropy state. However, the probability that the distribution of all the particles is random is
    much more likely. This illustrates a higher entropy state.

    The Third Law of Thermodynamics

    The third law states that it is impossible to arrive at the absolute zero temperature (T = 0 K) in any system
    where there is no motion and no kinetic energy. The zero Kelvin is the lowest possible temperature in the
    universe. This temperature corresponds to -273.15° Celsius or -459.7° Fahrenheit. As the temperature
    approaches 0 Kelvin, the entropy becomes constant. The entropy of a pure perfect crystal is zero at zero
    Kelvin, which is impossible because the entropy cannot become zero based on the second law of
    thermodynamics.

    Thermal History of the Universe

    In the early universe, photons (radiation) were hot enough to produce matter-antimatter pairs. The two
    particles (matter-antimatter pair) can annihilate each other to reproduce a photon. When the temperature was
    greater than 1013 K, there were active interactions among photons, protons, anti-protons, neutrons, and anti-
    neutrons. When the temperature was less than 109 K, the interactions between photons, electrons, and anti-
    electrons were active. Radiation produces matter-antimatter pairs; that is, there were active productions

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    between lights and baryons (ordinary matter like neutrons and protons). That is, when the temperature is
    greater than 1013 K, photons can be created by the interaction between protons and anti-protons or neutrons
    and anti-neutrons. Also, the interaction between two photons can create a pair of particles of protons and anti-
    protons or that of neutrons and anti-neutrons. There are dynamic interactions between lights and baryons
    (ordinary matter like neutrons and protons) in the hot universe.

    As the universe expanded, the temperature decreased, and soon photons did not have enough energy to
    form new matter and antimatter pairs. Most of elementary particles (neutrons and protons) are frozen out at
    the time t ~ 10-6 second when the temperature T dropped below 1013 K. Also, electrons freeze out at t ~ 1 sec
    when T drops below 109 K. The universe keeps expanding and cooling.

    If the number of matter and anti-matter particles had been exactly equal, then complete annihilation of matter
    would have occurred and none of us would be here today. Luckily for us, due to symmetry breaking, there
    was a slight excess of matter over anti-matter, (1 excess matter particle for every billion matter-antimatter
    pairs). Thus, matter particles, neutrons and protons (H nuclei) form by t < 10-6 s and electrons by t = 1 s. We
    owe our existence to this symmetry breaking.

    Energy density in photons (radiation) is larger than energy in matter before about 3 × 104 years. As the
    universe expands, matter energy density overcomes radiation energy density because of different expansion
    rate properties between them. Radiation energy density is proportional to the fourth power of the temperature
    of the universe while matter energy density is proportional to the third power of the temperature.

    As the universe expands, the photon λ stretches, and its energy falls (E ∝ 1 / λ). Matter energy falls because
    the separation of matter particles increases. They drop faster in the case of photons, and soon the energy
    density in matter exceeds the energy in photons. The universe transits to being matter-dominated and
    matter/gas then responds to gravitational forces and forms clumps. These clumps are the seed from which
    larger structures such as stars and proto-galaxies will come.

    Before recombination, the radiation-dominated universe was opaque because the photons collided frequently
    with electrons and sometimes got trapped. Also, electrons and protons existed separately and did not unite as
    neutral hydrogen. As the temperature dropped to about 3000 K, it became a proper condition to form a neutral
    hydrogen atom out of proton and electron. In addition, photons interacted only rarely with neutral atoms, and
    they traveled freely; the universe became transparent. This free streaming of photons can now be observed
    as cosmic background radiation (CBR) of 3 K.

    In this final unit, we studied the definition of thermodynamic systems and their surroundings with four
    thermodynamics laws. We also applied them to the various thermal processes using an ideal gas with specific
    heat capacities and examined the concept of entropy in the universe.

    References

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Erlichson, H. (1999). Sadi Carnot, ‘founder of the second law of thermodynamics’. European Journal of

    Physics, 20(1999), 183–192.
    http://geosci.uchicago.edu/~moyer/GEOS24705/Readings/Carnot_article_1998.pdf

    – Course Learning Outcomes for Unit VIII
    – Required Unit Resources
    – Unit Lesson
    – The Zeroth Law of Thermodynamics
    – The First Law of Thermodynamics
    – Thermal Processes
    – The Second Law of Thermodynamics
    – The Third Law of Thermodynamics
    – Thermal History of the Universe
    – References

    **## UnitVIII.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit VIII

    Upon completion of this unit, students should be able to:

    1. Describe the system of units for various physical quantities.

    2. Illustrate the scientific method within everyday situations.

    3. Explain Newton's laws of motion at work in common phenomena.

    4. Apply the concept of momentum conservations to daily life.

    5. Identify the total mechanical energy conservation.

    6. Explain numerous phenomena using fluid mechanics laws.

    7. Describe fundamental thermodynamic concepts.
    7.1 Recognize the first law of thermodynamics.
    7.2 Identify the maximum efficiency of a heat engine.
    7.3 Interpret the various thermal processes.

    Required Unit Resources

    Chapter 15: Thermodynamics

    Unit Lesson

    The Zeroth Law of Thermodynamics

    If there is no heat transfer (adiabatic process) between two systems, they are said to be in thermal equilibrium
    so that their temperatures are identical. The adiabatic process is a thermal process with no heat flow. If a third
    system is in thermal equilibrium with the two systems simultaneously, they are all in thermal equilibrium with
    the same temperature. This is known as the zeroth law of thermodynamics. The temperature is the indicator
    of thermal equilibrium, in which any flow of heat is prohibited in the system. Please look at Figure 15.2 under
    Section 15.2 of Chapter 15 in the eTextbook (Cutnell et al., 2022). The mercury level is the same in the
    thermometer in Systems A and B.

    The First Law of Thermodynamics

    This law is just the energy conservation law applied to heat, work, and internal energy. The internal energy of
    a system changes due to heat (Q) and work (W): U = Q – W. The change in internal energy is equivalent to
    the difference between the heat added to the system and the work done by the system (Cutnell et al., 2022).

    Please note, Q is positive when the system gains heat, and Q is negative when the system loses heat.
    Similarly, W is positive when work is done by the system, and W is negative when work is done on the
    system.

    UNIT VIII STUDY GUIDE
    Thermodynamics

    Change in Internal Energy = Heat – Work

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Thermal Processes

    In general, there are four types of thermal processes when considering the ideal gas: isobaric, isochoric,
    isothermal, and adiabatic. The isobaric process is when the pressure of a system is constant. The work done,
    in this case, is the product of the pressure and the change in volume. The change in volume is calculated by
    the final volume minus the initial volume. The isochoric process is when the volume is constant, so the work
    done is none. Consequently, the change in internal energy is the same as the heat. The isothermal process is
    when the temperature is constant. Thus, the final internal energy is equal to the initial internal energy. Lastly,
    the adiabatic process is when the heat does not transfer to anywhere. The adiabatic expansion or
    compression of the ideal gas is related to the ratio of the specific heat capacities at constant pressure and
    constant volume.

    The Second Law of Thermodynamics

    The second law can be stated in various ways. Let’s consider the heat flow statement. It states that heat flows
    spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not
    flow spontaneously in the reverse direction. In order to understand the meaning of this statement, consider
    heat engines. A heat engine has three main components: a hot reservoir, a cold reservoir, and work output.
    See Figure 15.13 under Section 15.8 of Chapter 15 in eTextbook (Cutnell et al., 2022) for a graphic depiction
    of this concept.

    It is impossible to create a perfect engine that uses all of the input heat to produce the work done. Some of
    the input heat must flow to the cold reservoir. It is also impossible to create a perfect refrigerator. Without any
    work, the heat cannot flow from the cold reservoir to the hot reservoir.

    Carnot suggested the condition for the maximum efficiency of a heat engine in a reversible process
    (Erlichson, 1999). The reversible process refers to when the final state can be returned to its original state.
    The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input
    heat QH: e = W / QH. W = QH – QC, where Qc is the output heat. That is, e = 1 – Qc / QH = 1 – Tc / TH, where Tc
    stands for the temperature of the cold reservoir, and TH stands for the temperature of the hot reservoir. The
    unit of temperature must be in Kelvin.

    Sample Question 1: The internal energy U of a system is 7000 J. The work done W by the
    system is given as 8000 J. Determine if the heat is gained or lost to the system.

    Solution: Use the formula U = Q – W. W is positive when work is done by the system, so W =
    +8000 J. The heat Q = U + W = 7000 + 8000= 15,000 J. It is positive, so the system gains 15,000
    J of heat from the surroundings.

    Sample Question 2: We are going to use a tropical ocean as a heat engine. The surface
    temperature is 300 K, and the underwater temperature is 280 K. Determine the maximum possible
    efficiency of this engine.

    Solution: Use the formula e = 1 – Tc / TH. Here, Tc is 280 K, and TH is 300 K. The warm water
    should be a hot reservoir, and the cool water should be a cold reservoir.

    So, e = 1 – 280 / 300 = 1 / 15 ≈ 0.06 = 6%.

    Work Done = Efficiency x Input Heat

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Carnot’s principle or an alternative statement of the second law of thermodynamics states that no irreversible
    engine operating between two reservoirs at constant temperatures can have a greater efficiency than a
    reversible engine operating between the same temperatures (Erlichson, 1999). In addition, all reversible
    engines operating between the same temperatures have the same efficiency.

    Entropy is introduced to describe the order of a system. The second law of thermodynamics can be stated
    using the concept of entropy. The change in entropy (S) is defined as when the heat (Q) flows reversibly at a
    constant (T): S = Q / T. In the reversible process, the total entropy of the universe is constant: S = 0. In an
    irreversible process, the entropy is always increasing: S > 0. For instance, suppose the total energy q that is
    transferred from a hot region in the universe at temperature T1 to a cold region at temperature T2. From the
    definition of the entropy, the hot region’s entropy S1 is the total energy q divided by the hot region’s
    temperature T1. In the same manner, the cold region’s entropy S2 is the total energy q divided by the cold
    region’s temperature T2. In other words, S1 = q / T1 and S2 = q / T2. The change in entropy is obtained by S2
    – S1 = q / T2 – q / T1 = q(1 / T2 – 1 / T1). Please note, that T2 is less than T1. That is, 1 / T2 is greater than 1
    / T1. Thus, the change in entropy is positive. Therefore, the total entropy of the universe is increasing.
    Furthermore, the entropy is strongly related to the probability. When considering a gas in a box, the
    probability that all the gas particles exist in a particular place at the same time is extremely small. That is, it
    has a very low entropy state. However, the probability that the distribution of all the particles is random is
    much more likely. This illustrates a higher entropy state.

    The Third Law of Thermodynamics

    The third law states that it is impossible to arrive at the absolute zero temperature (T = 0 K) in any system
    where there is no motion and no kinetic energy. The zero Kelvin is the lowest possible temperature in the
    universe. This temperature corresponds to -273.15° Celsius or -459.7° Fahrenheit. As the temperature
    approaches 0 Kelvin, the entropy becomes constant. The entropy of a pure perfect crystal is zero at zero
    Kelvin, which is impossible because the entropy cannot become zero based on the second law of
    thermodynamics.

    Thermal History of the Universe

    In the early universe, photons (radiation) were hot enough to produce matter-antimatter pairs. The two
    particles (matter-antimatter pair) can annihilate each other to reproduce a photon. When the temperature was
    greater than 1013 K, there were active interactions among photons, protons, anti-protons, neutrons, and anti-
    neutrons. When the temperature was less than 109 K, the interactions between photons, electrons, and anti-
    electrons were active. Radiation produces matter-antimatter pairs; that is, there were active productions

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    between lights and baryons (ordinary matter like neutrons and protons). That is, when the temperature is
    greater than 1013 K, photons can be created by the interaction between protons and anti-protons or neutrons
    and anti-neutrons. Also, the interaction between two photons can create a pair of particles of protons and anti-
    protons or that of neutrons and anti-neutrons. There are dynamic interactions between lights and baryons
    (ordinary matter like neutrons and protons) in the hot universe.

    As the universe expanded, the temperature decreased, and soon photons did not have enough energy to
    form new matter and antimatter pairs. Most of elementary particles (neutrons and protons) are frozen out at
    the time t ~ 10-6 second when the temperature T dropped below 1013 K. Also, electrons freeze out at t ~ 1 sec
    when T drops below 109 K. The universe keeps expanding and cooling.

    If the number of matter and anti-matter particles had been exactly equal, then complete annihilation of matter
    would have occurred and none of us would be here today. Luckily for us, due to symmetry breaking, there
    was a slight excess of matter over anti-matter, (1 excess matter particle for every billion matter-antimatter
    pairs). Thus, matter particles, neutrons and protons (H nuclei) form by t < 10-6 s and electrons by t = 1 s. We
    owe our existence to this symmetry breaking.

    Energy density in photons (radiation) is larger than energy in matter before about 3 × 104 years. As the
    universe expands, matter energy density overcomes radiation energy density because of different expansion
    rate properties between them. Radiation energy density is proportional to the fourth power of the temperature
    of the universe while matter energy density is proportional to the third power of the temperature.

    As the universe expands, the photon λ stretches, and its energy falls (E ∝ 1 / λ). Matter energy falls because
    the separation of matter particles increases. They drop faster in the case of photons, and soon the energy
    density in matter exceeds the energy in photons. The universe transits to being matter-dominated and
    matter/gas then responds to gravitational forces and forms clumps. These clumps are the seed from which
    larger structures such as stars and proto-galaxies will come.

    Before recombination, the radiation-dominated universe was opaque because the photons collided frequently
    with electrons and sometimes got trapped. Also, electrons and protons existed separately and did not unite as
    neutral hydrogen. As the temperature dropped to about 3000 K, it became a proper condition to form a neutral
    hydrogen atom out of proton and electron. In addition, photons interacted only rarely with neutral atoms, and
    they traveled freely; the universe became transparent. This free streaming of photons can now be observed
    as cosmic background radiation (CBR) of 3 K.

    In this final unit, we studied the definition of thermodynamic systems and their surroundings with four
    thermodynamics laws. We also applied them to the various thermal processes using an ideal gas with specific
    heat capacities and examined the concept of entropy in the universe.

    References

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Erlichson, H. (1999). Sadi Carnot, ‘founder of the second law of thermodynamics’. European Journal of

    Physics, 20(1999), 183–192.
    http://geosci.uchicago.edu/~moyer/GEOS24705/Readings/Carnot_article_1998.pdf

    – Course Learning Outcomes for Unit VIII
    – Required Unit Resources
    – Unit Lesson
    – The Zeroth Law of Thermodynamics
    – The First Law of Thermodynamics
    – Thermal Processes
    – The Second Law of Thermodynamics
    – The Third Law of Thermodynamics
    – Thermal History of the Universe
    – References

    **## UnitVIIIJournal.docx**

    Unit VIII Journal

    This journal measures your mastery of Unit VIII’s ULO 7.1.

    If you were asked to choose one of the four thermodynamics laws, which law do you think is the most important? Explain your reasoning based on your daily activity. Rank the importance of each law based on your insight. Justify your reason.

    Reflect on what you have learned in this class. How can these concepts be applied to real-world situations and increase your chances of career or life success?

    Your journal entry must be at least 200 words in length. No references or citations are necessary.

    **## UnitVIII.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit VIII

    Upon completion of this unit, students should be able to:

    1. Describe the system of units for various physical quantities.

    2. Illustrate the scientific method within everyday situations.

    3. Explain Newton's laws of motion at work in common phenomena.

    4. Apply the concept of momentum conservations to daily life.

    5. Identify the total mechanical energy conservation.

    6. Explain numerous phenomena using fluid mechanics laws.

    7. Describe fundamental thermodynamic concepts.
    7.1 Recognize the first law of thermodynamics.
    7.2 Identify the maximum efficiency of a heat engine.
    7.3 Interpret the various thermal processes.

    Required Unit Resources

    Chapter 15: Thermodynamics

    Unit Lesson

    The Zeroth Law of Thermodynamics

    If there is no heat transfer (adiabatic process) between two systems, they are said to be in thermal equilibrium
    so that their temperatures are identical. The adiabatic process is a thermal process with no heat flow. If a third
    system is in thermal equilibrium with the two systems simultaneously, they are all in thermal equilibrium with
    the same temperature. This is known as the zeroth law of thermodynamics. The temperature is the indicator
    of thermal equilibrium, in which any flow of heat is prohibited in the system. Please look at Figure 15.2 under
    Section 15.2 of Chapter 15 in the eTextbook (Cutnell et al., 2022). The mercury level is the same in the
    thermometer in Systems A and B.

    The First Law of Thermodynamics

    This law is just the energy conservation law applied to heat, work, and internal energy. The internal energy of
    a system changes due to heat (Q) and work (W): U = Q – W. The change in internal energy is equivalent to
    the difference between the heat added to the system and the work done by the system (Cutnell et al., 2022).

    Please note, Q is positive when the system gains heat, and Q is negative when the system loses heat.
    Similarly, W is positive when work is done by the system, and W is negative when work is done on the
    system.

    UNIT VIII STUDY GUIDE
    Thermodynamics

    Change in Internal Energy = Heat – Work

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Thermal Processes

    In general, there are four types of thermal processes when considering the ideal gas: isobaric, isochoric,
    isothermal, and adiabatic. The isobaric process is when the pressure of a system is constant. The work done,
    in this case, is the product of the pressure and the change in volume. The change in volume is calculated by
    the final volume minus the initial volume. The isochoric process is when the volume is constant, so the work
    done is none. Consequently, the change in internal energy is the same as the heat. The isothermal process is
    when the temperature is constant. Thus, the final internal energy is equal to the initial internal energy. Lastly,
    the adiabatic process is when the heat does not transfer to anywhere. The adiabatic expansion or
    compression of the ideal gas is related to the ratio of the specific heat capacities at constant pressure and
    constant volume.

    The Second Law of Thermodynamics

    The second law can be stated in various ways. Let’s consider the heat flow statement. It states that heat flows
    spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not
    flow spontaneously in the reverse direction. In order to understand the meaning of this statement, consider
    heat engines. A heat engine has three main components: a hot reservoir, a cold reservoir, and work output.
    See Figure 15.13 under Section 15.8 of Chapter 15 in eTextbook (Cutnell et al., 2022) for a graphic depiction
    of this concept.

    It is impossible to create a perfect engine that uses all of the input heat to produce the work done. Some of
    the input heat must flow to the cold reservoir. It is also impossible to create a perfect refrigerator. Without any
    work, the heat cannot flow from the cold reservoir to the hot reservoir.

    Carnot suggested the condition for the maximum efficiency of a heat engine in a reversible process
    (Erlichson, 1999). The reversible process refers to when the final state can be returned to its original state.
    The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input
    heat QH: e = W / QH. W = QH – QC, where Qc is the output heat. That is, e = 1 – Qc / QH = 1 – Tc / TH, where Tc
    stands for the temperature of the cold reservoir, and TH stands for the temperature of the hot reservoir. The
    unit of temperature must be in Kelvin.

    Sample Question 1: The internal energy U of a system is 7000 J. The work done W by the
    system is given as 8000 J. Determine if the heat is gained or lost to the system.

    Solution: Use the formula U = Q – W. W is positive when work is done by the system, so W =
    +8000 J. The heat Q = U + W = 7000 + 8000= 15,000 J. It is positive, so the system gains 15,000
    J of heat from the surroundings.

    Sample Question 2: We are going to use a tropical ocean as a heat engine. The surface
    temperature is 300 K, and the underwater temperature is 280 K. Determine the maximum possible
    efficiency of this engine.

    Solution: Use the formula e = 1 – Tc / TH. Here, Tc is 280 K, and TH is 300 K. The warm water
    should be a hot reservoir, and the cool water should be a cold reservoir.

    So, e = 1 – 280 / 300 = 1 / 15 ≈ 0.06 = 6%.

    Work Done = Efficiency x Input Heat

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Carnot’s principle or an alternative statement of the second law of thermodynamics states that no irreversible
    engine operating between two reservoirs at constant temperatures can have a greater efficiency than a
    reversible engine operating between the same temperatures (Erlichson, 1999). In addition, all reversible
    engines operating between the same temperatures have the same efficiency.

    Entropy is introduced to describe the order of a system. The second law of thermodynamics can be stated
    using the concept of entropy. The change in entropy (S) is defined as when the heat (Q) flows reversibly at a
    constant (T): S = Q / T. In the reversible process, the total entropy of the universe is constant: S = 0. In an
    irreversible process, the entropy is always increasing: S > 0. For instance, suppose the total energy q that is
    transferred from a hot region in the universe at temperature T1 to a cold region at temperature T2. From the
    definition of the entropy, the hot region’s entropy S1 is the total energy q divided by the hot region’s
    temperature T1. In the same manner, the cold region’s entropy S2 is the total energy q divided by the cold
    region’s temperature T2. In other words, S1 = q / T1 and S2 = q / T2. The change in entropy is obtained by S2
    – S1 = q / T2 – q / T1 = q(1 / T2 – 1 / T1). Please note, that T2 is less than T1. That is, 1 / T2 is greater than 1
    / T1. Thus, the change in entropy is positive. Therefore, the total entropy of the universe is increasing.
    Furthermore, the entropy is strongly related to the probability. When considering a gas in a box, the
    probability that all the gas particles exist in a particular place at the same time is extremely small. That is, it
    has a very low entropy state. However, the probability that the distribution of all the particles is random is
    much more likely. This illustrates a higher entropy state.

    The Third Law of Thermodynamics

    The third law states that it is impossible to arrive at the absolute zero temperature (T = 0 K) in any system
    where there is no motion and no kinetic energy. The zero Kelvin is the lowest possible temperature in the
    universe. This temperature corresponds to -273.15° Celsius or -459.7° Fahrenheit. As the temperature
    approaches 0 Kelvin, the entropy becomes constant. The entropy of a pure perfect crystal is zero at zero
    Kelvin, which is impossible because the entropy cannot become zero based on the second law of
    thermodynamics.

    Thermal History of the Universe

    In the early universe, photons (radiation) were hot enough to produce matter-antimatter pairs. The two
    particles (matter-antimatter pair) can annihilate each other to reproduce a photon. When the temperature was
    greater than 1013 K, there were active interactions among photons, protons, anti-protons, neutrons, and anti-
    neutrons. When the temperature was less than 109 K, the interactions between photons, electrons, and anti-
    electrons were active. Radiation produces matter-antimatter pairs; that is, there were active productions

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    between lights and baryons (ordinary matter like neutrons and protons). That is, when the temperature is
    greater than 1013 K, photons can be created by the interaction between protons and anti-protons or neutrons
    and anti-neutrons. Also, the interaction between two photons can create a pair of particles of protons and anti-
    protons or that of neutrons and anti-neutrons. There are dynamic interactions between lights and baryons
    (ordinary matter like neutrons and protons) in the hot universe.

    As the universe expanded, the temperature decreased, and soon photons did not have enough energy to
    form new matter and antimatter pairs. Most of elementary particles (neutrons and protons) are frozen out at
    the time t ~ 10-6 second when the temperature T dropped below 1013 K. Also, electrons freeze out at t ~ 1 sec
    when T drops below 109 K. The universe keeps expanding and cooling.

    If the number of matter and anti-matter particles had been exactly equal, then complete annihilation of matter
    would have occurred and none of us would be here today. Luckily for us, due to symmetry breaking, there
    was a slight excess of matter over anti-matter, (1 excess matter particle for every billion matter-antimatter
    pairs). Thus, matter particles, neutrons and protons (H nuclei) form by t < 10-6 s and electrons by t = 1 s. We
    owe our existence to this symmetry breaking.

    Energy density in photons (radiation) is larger than energy in matter before about 3 × 104 years. As the
    universe expands, matter energy density overcomes radiation energy density because of different expansion
    rate properties between them. Radiation energy density is proportional to the fourth power of the temperature
    of the universe while matter energy density is proportional to the third power of the temperature.

    As the universe expands, the photon λ stretches, and its energy falls (E ∝ 1 / λ). Matter energy falls because
    the separation of matter particles increases. They drop faster in the case of photons, and soon the energy
    density in matter exceeds the energy in photons. The universe transits to being matter-dominated and
    matter/gas then responds to gravitational forces and forms clumps. These clumps are the seed from which
    larger structures such as stars and proto-galaxies will come.

    Before recombination, the radiation-dominated universe was opaque because the photons collided frequently
    with electrons and sometimes got trapped. Also, electrons and protons existed separately and did not unite as
    neutral hydrogen. As the temperature dropped to about 3000 K, it became a proper condition to form a neutral
    hydrogen atom out of proton and electron. In addition, photons interacted only rarely with neutral atoms, and
    they traveled freely; the universe became transparent. This free streaming of photons can now be observed
    as cosmic background radiation (CBR) of 3 K.

    In this final unit, we studied the definition of thermodynamic systems and their surroundings with four
    thermodynamics laws. We also applied them to the various thermal processes using an ideal gas with specific
    heat capacities and examined the concept of entropy in the universe.

    References

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Erlichson, H. (1999). Sadi Carnot, ‘founder of the second law of thermodynamics’. European Journal of

    Physics, 20(1999), 183–192.
    http://geosci.uchicago.edu/~moyer/GEOS24705/Readings/Carnot_article_1998.pdf

    – Course Learning Outcomes for Unit VIII
    – Required Unit Resources
    – Unit Lesson
    – The Zeroth Law of Thermodynamics
    – The First Law of Thermodynamics
    – Thermal Processes
    – The Second Law of Thermodynamics
    – The Third Law of Thermodynamics
    – Thermal History of the Universe
    – References

    **## UnitVIIIJournal.docx**

    Unit VIII Journal

    This journal measures your mastery of Unit VIII’s ULO 7.1.

    If you were asked to choose one of the four thermodynamics laws, which law do you think is the most important? Explain your reasoning based on your daily activity. Rank the importance of each law based on your insight. Justify your reason.

    Reflect on what you have learned in this class. How can these concepts be applied to real-world situations and increase your chances of career or life success?

    Your journal entry must be at least 200 words in length. No references or citations are necessary.

    **## UnitVIII.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit VIII

    Upon completion of this unit, students should be able to:

    1. Describe the system of units for various physical quantities.

    2. Illustrate the scientific method within everyday situations.

    3. Explain Newton's laws of motion at work in common phenomena.

    4. Apply the concept of momentum conservations to daily life.

    5. Identify the total mechanical energy conservation.

    6. Explain numerous phenomena using fluid mechanics laws.

    7. Describe fundamental thermodynamic concepts.
    7.1 Recognize the first law of thermodynamics.
    7.2 Identify the maximum efficiency of a heat engine.
    7.3 Interpret the various thermal processes.

    Required Unit Resources

    Chapter 15: Thermodynamics

    Unit Lesson

    The Zeroth Law of Thermodynamics

    If there is no heat transfer (adiabatic process) between two systems, they are said to be in thermal equilibrium
    so that their temperatures are identical. The adiabatic process is a thermal process with no heat flow. If a third
    system is in thermal equilibrium with the two systems simultaneously, they are all in thermal equilibrium with
    the same temperature. This is known as the zeroth law of thermodynamics. The temperature is the indicator
    of thermal equilibrium, in which any flow of heat is prohibited in the system. Please look at Figure 15.2 under
    Section 15.2 of Chapter 15 in the eTextbook (Cutnell et al., 2022). The mercury level is the same in the
    thermometer in Systems A and B.

    The First Law of Thermodynamics

    This law is just the energy conservation law applied to heat, work, and internal energy. The internal energy of
    a system changes due to heat (Q) and work (W): U = Q – W. The change in internal energy is equivalent to
    the difference between the heat added to the system and the work done by the system (Cutnell et al., 2022).

    Please note, Q is positive when the system gains heat, and Q is negative when the system loses heat.
    Similarly, W is positive when work is done by the system, and W is negative when work is done on the
    system.

    UNIT VIII STUDY GUIDE
    Thermodynamics

    Change in Internal Energy = Heat – Work

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Thermal Processes

    In general, there are four types of thermal processes when considering the ideal gas: isobaric, isochoric,
    isothermal, and adiabatic. The isobaric process is when the pressure of a system is constant. The work done,
    in this case, is the product of the pressure and the change in volume. The change in volume is calculated by
    the final volume minus the initial volume. The isochoric process is when the volume is constant, so the work
    done is none. Consequently, the change in internal energy is the same as the heat. The isothermal process is
    when the temperature is constant. Thus, the final internal energy is equal to the initial internal energy. Lastly,
    the adiabatic process is when the heat does not transfer to anywhere. The adiabatic expansion or
    compression of the ideal gas is related to the ratio of the specific heat capacities at constant pressure and
    constant volume.

    The Second Law of Thermodynamics

    The second law can be stated in various ways. Let’s consider the heat flow statement. It states that heat flows
    spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not
    flow spontaneously in the reverse direction. In order to understand the meaning of this statement, consider
    heat engines. A heat engine has three main components: a hot reservoir, a cold reservoir, and work output.
    See Figure 15.13 under Section 15.8 of Chapter 15 in eTextbook (Cutnell et al., 2022) for a graphic depiction
    of this concept.

    It is impossible to create a perfect engine that uses all of the input heat to produce the work done. Some of
    the input heat must flow to the cold reservoir. It is also impossible to create a perfect refrigerator. Without any
    work, the heat cannot flow from the cold reservoir to the hot reservoir.

    Carnot suggested the condition for the maximum efficiency of a heat engine in a reversible process
    (Erlichson, 1999). The reversible process refers to when the final state can be returned to its original state.
    The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input
    heat QH: e = W / QH. W = QH – QC, where Qc is the output heat. That is, e = 1 – Qc / QH = 1 – Tc / TH, where Tc
    stands for the temperature of the cold reservoir, and TH stands for the temperature of the hot reservoir. The
    unit of temperature must be in Kelvin.

    Sample Question 1: The internal energy U of a system is 7000 J. The work done W by the
    system is given as 8000 J. Determine if the heat is gained or lost to the system.

    Solution: Use the formula U = Q – W. W is positive when work is done by the system, so W =
    +8000 J. The heat Q = U + W = 7000 + 8000= 15,000 J. It is positive, so the system gains 15,000
    J of heat from the surroundings.

    Sample Question 2: We are going to use a tropical ocean as a heat engine. The surface
    temperature is 300 K, and the underwater temperature is 280 K. Determine the maximum possible
    efficiency of this engine.

    Solution: Use the formula e = 1 – Tc / TH. Here, Tc is 280 K, and TH is 300 K. The warm water
    should be a hot reservoir, and the cool water should be a cold reservoir.

    So, e = 1 – 280 / 300 = 1 / 15 ≈ 0.06 = 6%.

    Work Done = Efficiency x Input Heat

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Carnot’s principle or an alternative statement of the second law of thermodynamics states that no irreversible
    engine operating between two reservoirs at constant temperatures can have a greater efficiency than a
    reversible engine operating between the same temperatures (Erlichson, 1999). In addition, all reversible
    engines operating between the same temperatures have the same efficiency.

    Entropy is introduced to describe the order of a system. The second law of thermodynamics can be stated
    using the concept of entropy. The change in entropy (S) is defined as when the heat (Q) flows reversibly at a
    constant (T): S = Q / T. In the reversible process, the total entropy of the universe is constant: S = 0. In an
    irreversible process, the entropy is always increasing: S > 0. For instance, suppose the total energy q that is
    transferred from a hot region in the universe at temperature T1 to a cold region at temperature T2. From the
    definition of the entropy, the hot region’s entropy S1 is the total energy q divided by the hot region’s
    temperature T1. In the same manner, the cold region’s entropy S2 is the total energy q divided by the cold
    region’s temperature T2. In other words, S1 = q / T1 and S2 = q / T2. The change in entropy is obtained by S2
    – S1 = q / T2 – q / T1 = q(1 / T2 – 1 / T1). Please note, that T2 is less than T1. That is, 1 / T2 is greater than 1
    / T1. Thus, the change in entropy is positive. Therefore, the total entropy of the universe is increasing.
    Furthermore, the entropy is strongly related to the probability. When considering a gas in a box, the
    probability that all the gas particles exist in a particular place at the same time is extremely small. That is, it
    has a very low entropy state. However, the probability that the distribution of all the particles is random is
    much more likely. This illustrates a higher entropy state.

    The Third Law of Thermodynamics

    The third law states that it is impossible to arrive at the absolute zero temperature (T = 0 K) in any system
    where there is no motion and no kinetic energy. The zero Kelvin is the lowest possible temperature in the
    universe. This temperature corresponds to -273.15° Celsius or -459.7° Fahrenheit. As the temperature
    approaches 0 Kelvin, the entropy becomes constant. The entropy of a pure perfect crystal is zero at zero
    Kelvin, which is impossible because the entropy cannot become zero based on the second law of
    thermodynamics.

    Thermal History of the Universe

    In the early universe, photons (radiation) were hot enough to produce matter-antimatter pairs. The two
    particles (matter-antimatter pair) can annihilate each other to reproduce a photon. When the temperature was
    greater than 1013 K, there were active interactions among photons, protons, anti-protons, neutrons, and anti-
    neutrons. When the temperature was less than 109 K, the interactions between photons, electrons, and anti-
    electrons were active. Radiation produces matter-antimatter pairs; that is, there were active productions

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    between lights and baryons (ordinary matter like neutrons and protons). That is, when the temperature is
    greater than 1013 K, photons can be created by the interaction between protons and anti-protons or neutrons
    and anti-neutrons. Also, the interaction between two photons can create a pair of particles of protons and anti-
    protons or that of neutrons and anti-neutrons. There are dynamic interactions between lights and baryons
    (ordinary matter like neutrons and protons) in the hot universe.

    As the universe expanded, the temperature decreased, and soon photons did not have enough energy to
    form new matter and antimatter pairs. Most of elementary particles (neutrons and protons) are frozen out at
    the time t ~ 10-6 second when the temperature T dropped below 1013 K. Also, electrons freeze out at t ~ 1 sec
    when T drops below 109 K. The universe keeps expanding and cooling.

    If the number of matter and anti-matter particles had been exactly equal, then complete annihilation of matter
    would have occurred and none of us would be here today. Luckily for us, due to symmetry breaking, there
    was a slight excess of matter over anti-matter, (1 excess matter particle for every billion matter-antimatter
    pairs). Thus, matter particles, neutrons and protons (H nuclei) form by t < 10-6 s and electrons by t = 1 s. We
    owe our existence to this symmetry breaking.

    Energy density in photons (radiation) is larger than energy in matter before about 3 × 104 years. As the
    universe expands, matter energy density overcomes radiation energy density because of different expansion
    rate properties between them. Radiation energy density is proportional to the fourth power of the temperature
    of the universe while matter energy density is proportional to the third power of the temperature.

    As the universe expands, the photon λ stretches, and its energy falls (E ∝ 1 / λ). Matter energy falls because
    the separation of matter particles increases. They drop faster in the case of photons, and soon the energy
    density in matter exceeds the energy in photons. The universe transits to being matter-dominated and
    matter/gas then responds to gravitational forces and forms clumps. These clumps are the seed from which
    larger structures such as stars and proto-galaxies will come.

    Before recombination, the radiation-dominated universe was opaque because the photons collided frequently
    with electrons and sometimes got trapped. Also, electrons and protons existed separately and did not unite as
    neutral hydrogen. As the temperature dropped to about 3000 K, it became a proper condition to form a neutral
    hydrogen atom out of proton and electron. In addition, photons interacted only rarely with neutral atoms, and
    they traveled freely; the universe became transparent. This free streaming of photons can now be observed
    as cosmic background radiation (CBR) of 3 K.

    In this final unit, we studied the definition of thermodynamic systems and their surroundings with four
    thermodynamics laws. We also applied them to the various thermal processes using an ideal gas with specific
    heat capacities and examined the concept of entropy in the universe.

    References

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Erlichson, H. (1999). Sadi Carnot, ‘founder of the second law of thermodynamics’. European Journal of

    Physics, 20(1999), 183–192.
    http://geosci.uchicago.edu/~moyer/GEOS24705/Readings/Carnot_article_1998.pdf

    – Course Learning Outcomes for Unit VIII
    – Required Unit Resources
    – Unit Lesson
    – The Zeroth Law of Thermodynamics
    – The First Law of Thermodynamics
    – Thermal Processes
    – The Second Law of Thermodynamics
    – The Third Law of Thermodynamics
    – Thermal History of the Universe
    – References

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