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For this assignment, you will complete the Unit VI Problem Solving Worksheet. This assignment will allow you to demonstrate what you have learned in this unit. Instructions for completing this assignment are located on the worksheet.

Use the  Unit VI Problem Solving Worksheet(SeeAttached)for this assignment. Save all of your work directly to the worksheet, and submit it in Blackboard for grading.

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.

Course/Unit Learning Outcomes

Learning Activity

6.1

Unit Lesson Chapter 11 Chapter 12 Unit VI Problem Solving

6.2

Unit Lesson Chapter 11 Chapter 12 Unit VI Problem Solving

6.3

Unit Lesson Chapter 11 Chapter 12 Unit VI Problem Solving

Required Unit Resources Chapter 11: Fluids, pp. 289–325 Chapter 12: Temperature and Heat, pp. 326–359

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

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 on p. 293 in the textbook.

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 11.14 on page 298 in the textbook.

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 on page 290 in the textbook. So, the volume is the mass divided by the density.

(100 kg) / (3520 kg / m3) = 0.0284 m3.

Force = Pressure x Area

PHY 1301, Physics I 3

UNIT x STUDY GUIDE

Title

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- 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.

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

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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. 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.

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. 98.6oF is (98.4 – 32) / 1.8 = 36.778oC. The difference is 37 – 36.78 = 0.22oC.

PHY 1301, Physics I 5

UNIT x STUDY GUIDE

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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 on p. 344 in the textbook 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 on p. 343 in the textbook for the phase changes between any two of the three phases of matter.

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 106 and 107 on pp. 324 and 325 in the textbook. 2. Solve questions 104 and 105 on pp. 358 and 359 in the textbook.

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
  • Learning Activities (Nongraded)

,

This assignment will allow you to demonstrate the following objectives:

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.

Instructions: Solve the problems below. Each question is worth 10 points. You must show your work with as much detail as possible. Answer the questions directly in this template. Before doing so, it is highly recommended that you thoroughly review the Unit VI Lesson in the study guide.

1. If the mass of air inside a room is 1 kg, what is the volume of the air? Use Table 11.1 on page 290 in the textbook. Hint: Review Sample Question 1 in the Unit VI Lesson.

2. A massless cube container holds water whose density is 1000 kg/m3. The length of the side of the cube is 7 meters. What is the mass of the water? Hint: The volume of a cube is obtained by R x R x R, where R is the length of the side of the cube. Use the formula mass = density x volume.

3. A 10-kg piece of metal displaces 0.002 m3 of water when submerged. What is the density of the metal? Hint: Use the formula mass = density x volume.

4. The pressure acting on a floating piece of wood is measured by 12345 Pascal, and its surface area is 0.6789 m2. What is the magnitude of the force in Newtons? Hint: Review Example 2 on page 292 in the textbook.

5. A boy’s height is 1.76 meters. The vertical distance from his head to his heart is measured as 0.39 m. Find the 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. What is the blood pressure difference between them? The density of blood is 1060 kg/m3, and the blood is assumed as being a static fluid. Hint: Review Sample Question 2 in the Unit VI Lesson.

6. You bought a 1 kg solid gold statue from a merchant in Italy while you are on vacation. When you get home, you decided to test if this statue is real gold or not. After submerging the gold statue in a large water container, you will measure the volume of displaced water. What is the expected volume if the statue is made of pure gold? For the density of gold, use Table 11.1 on page 290 in the textbook.

7. A traveler at the North Pole measured the temperature as -40oC. Can you convert this temperature to the Fahrenheit scale? What is the temperature on the Kelvin scale? Hint: Use the converting formula: oF = 1.8oC + 32. The relation between the Celsius scale and the Kelvin scale is K = oC + 273.15.

8. A runner who weighs 50 kg produces 500,000 J of heat for a half hour, but this heat is removed by various mechanisms inside of her body to adjust to the conditions. However, if the heat was not removed, what is the increment of temperature? The specific heat capacity of the human body is 3500 J / kg oC from Table 12.2 on page 340 in the textbook. Hint: Review Sample Question 4 in the Unit VI Lesson and Example 9 on page 340 to 341 in the textbook.

9. In order to freeze 2 kg of water at 0oC into ice at 0oC, how much heat is required? The latent heat of fusion for water L = 335,000 J / kg. Hint: Review Sample Question 5 in the Unit VI Lesson.

10. In 1965, Penzias and Wilson discovered the isotropic cosmic background radiation of the microwave and earned the Nobel Prize in 1978. The cosmic microwave background radiation is measured by 2.725 Kelvin through the entire sky. Convert this temperature into the Celsius scale as well as the Fahrenheit scale. Hint: Use the converting formula: oF = 1.8oC + 32. The relation between the Celsius scale and the Kelvin scale is K = oC + 273.15.