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The Nature of Science and Physics

Kinematics

Introduction to One-Dimensional KinematicsDisplacementVectors, Scalars, and Coordinate SystemsTime, Velocity, and SpeedVideo: One-Dimensional KinematicsAccelerationMotion Equations for Constant Acceleration in One DimensionProblem-Solving Basics for One-Dimensional KinematicsFalling ObjectsGraphical Analysis of One-Dimensional Motion

Two-Dimensional Kinematics

Dynamics: Force and Newton's Laws of Motion

Introduction to Dynamics: Newton's Laws of MotionDevelopment of Force ConceptNewton's First Law of Motion: InertiaNewton's Second Law of Motion: Concept of a SystemNewton's Third Law of Motion: Symmetry in ForcesVideo: Newton's LawsNormal, Tension, and Other Examples of ForcesProblem-Solving StrategiesFurther Applications of Newton's Laws of MotionExtended Topic: The Four Basic Forces—An Introduction

Further Applications of Newton's Laws: Friction, Drag, and Elasticity

Uniform Circular Motion and Gravitation

Introduction to Uniform Circular Motion and GravitationRotation Angle and Angular VelocityCentripetal AccelerationCentripetal ForceFictitious Forces and Non-inertial Frames: The Coriolis ForceNewton's Universal Law of GravitationVideo: GravitationSatellites and Kepler's Laws: An Argument for Simplicity

Work, Energy, and Energy Resources

Introduction to Work, Energy, and Energy ResourcesWork: The Scientific DefinitionKinetic Energy and the Work-Energy TheoremGravitational Potential EnergyVideo: Potential and Kinetic EnergyConservative Forces and Potential EnergyNonconservative ForcesConservation of EnergyPowerWork, Energy, and Power in HumansWorld Energy Use

Linear Momentum and Collisions

Rotational Motion and Angular Momentum

Introduction to Rotational Motion and Angular MomentumAngular AccelerationKinematics of Rotational MotionVideo: Rotational MotionDynamics of Rotational Motion: Rotational InertiaRotational Kinetic Energy: Work and Energy RevisitedAngular Momentum and Its ConservationVideo: Angular MomentumCollisions of Extended Bodies in Two DimensionsGyroscopic Effects: Vector Aspects of Angular Momentum

Statics and Torque

Fluid Statics

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Fluid Dynamics and Its Biological and Medical Applications

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Temperature, Kinetic Theory, and the Gas Laws

Heat and Heat Transfer Methods

Thermodynamics

Introduction to ThermodynamicsThe First Law of ThermodynamicsThe First Law of Thermodynamics and Some Simple ProcessesIntroduction to the Second Law of Thermodynamics: Heat Engines and Their EfficiencyCarnot's Perfect Heat Engine: The Second Law of Thermodynamics RestatedApplications of Thermodynamics: Heat Pumps and RefrigeratorsEntropy and the Second Law of Thermodynamics: Disorder and the Unavailability of EnergyStatistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation

Oscillatory Motion and Waves

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Physics of Hearing

- Describe the use of heat engines in heat pumps and refrigerators.
- Demonstrate how a heat pump works to warm an interior space.
- Explain the differences between heat pumps and refrigerators.
- Calculate a heat pump’s coefficient of performance.

Heat pumps, air conditioners, and refrigerators utilize heat transfer from cold to hot. They are heat engines run backward. We say backward, rather than reverse, because except for Carnot engines, all heat engines, though they can be run backward, cannot truly be reversed. Heat transfer occurs from a cold reservoir

The basic components of a heat pump in its heating mode are shown in Figure 3. A working fluid such as a non-CFC refrigerant is used. In the outdoor coils (the evaporator), heat transfer

The electrically driven compressor (work input

The quality of a heat pump is judged by how much heat transfer

$COP_{\text{hp}}=\frac{Q_{\text{h}}}{W}\\$

.Since the efficiency of a heat engine is

$Eff=\frac{W}{Q_{\text{h}}}\\$

, we see that $COP_{\text{hp}}=\frac{1}{Eff}\\$

, an important and interesting fact. First, since the efficiency of any heat engine is less than 1, it means that $Eff_{\text{C}}=1-\left(\frac{T_{\text{c}}}{T_{\text{h}}}\right)\\$

; thus, the smaller the temperature difference, the smaller the efficiency and the greater the $COP_{\text{hp}}=\frac{1}{Eff}\\$

). In other words, heat pumps do not work as well in very cold climates as they do in more moderate climates.Friction and other irreversible processes reduce heat engine efficiency, but they do

$COP_{\text{hp}}\\$

of a Heat Pump for Home Use$COP_{\text{hp}}=\frac{1}{Eff}\\$

, so that we need to first calculate the Carnot efficiency to solve this problem.$Eff_{\text{C}}=1-\frac{T_{\text{c}}}{T_{\text{h}}}\\$

.
The temperatures in kelvins are $Eff_{\text{C}}=1-\frac{258\text{ K}}{318\text{ K}}=0.1887\\$

.
Thus, from the discussion above,$COP_{\text{hp}}=\frac{1}{Eff}=\frac{1}{0.1887}=5.30\\$

, or $COP_{\text{hp}}=\frac{Q_{\text{h}}}{W}=\frac{1}{0.1887}=5.30\\$

so that Real heat pumps do not perform quite as well as the ideal one in the previous example; their values of

${COP}_{\text{ref}}=\frac{Q_{\text{c}}}{W}\\$

.
Noting again that ${COP}_{\text{hp}}=\frac{Q_{\text{h}}}{W}\\$

and A type of

The

$\displaystyle{EER}=\frac{\frac{Q_{\text{c}}}{t_1}}{\frac{W}{t_2}}\\$

,
where *Examine the situation to determine whether heat, work, or internal energy are involved*. Look for any system where the primary methods of transferring energy are heat and work. Heat engines, heat pumps, refrigerators, and air conditioners are examples of such systems.*Identify the system of interest and draw a labeled diagram of the system showing energy flow.**Identify exactly what needs to be determined in the problem (identify the unknowns)*. A written list is useful. Maximum efficiency means a Carnot engine is involved. Efficiency is not the same as the coefficient of performance.*Make a list of what is given or can be inferred from the problem as stated (identify the knowns).*Be sure to distinguish heat transfer into a system from heat transfer out of the system, as well as work input from work output. In many situations, it is useful to determine the type of process, such as isothermal or adiabatic.*Solve the appropriate equation for the quantity to be determined (the unknown).**Substitute the known quantities along with their units into the appropriate equation and obtain numerical solutions complete with units.**Check the answer to see if it is reasonable: Does it make sense?*For example, efficiency is always less than 1, whereas coefficients of performance are greater than 1.

- An artifact of the second law of thermodynamics is the ability to heat an interior space using a heat pump. Heat pumps compress cold ambient air and, in so doing, heat it to room temperature without violation of conservation principles.
- To calculate the heat pump’s coefficient of performance, use the equation ${\text{COP}}_{\text{hp}}=\frac{{Q}_{\text{h}}}{W}\\$.
- A refrigerator is a heat pump; it takes warm ambient air and expands it to chill it.

- Explain why heat pumps do not work as well in very cold climates as they do in milder ones. Is the same true of refrigerators?
- In some Northern European nations, homes are being built without heating systems of any type. They are very well insulated and are kept warm by the body heat of the residents. However, when the residents are not at home, it is still warm in these houses. What is a possible explanation?
- Why do refrigerators, air conditioners, and heat pumps operate most cost-effectively for cycles with a small difference between
*T*_{h}and*T*_{c}? (Note that the temperatures of the cycle employed are crucial to its*COP*.) - Grocery store managers contend that there is less total energy consumption in the summer if the store is kept at a low temperature. Make arguments to support or refute this claim, taking into account that there are numerous refrigerators and freezers in the store.
- Can you cool a kitchen by leaving the refrigerator door open?

- What is the coefficient of performance of an ideal heat pump that has heat transfer from a cold temperature of −25.0ºC to a hot temperature of 40.0ºC?
- Suppose you have an ideal refrigerator that cools an environment at−20.0ºC and has heat transfer to another environment at 50.0ºC. What is its coefficient of performance?
- What is the best coefficient of performance possible for a hypothetical refrigerator that could make liquid nitrogen at −200ºC and has heat transfer to the environment at 35.0ºC?
- In a very mild winter climate, a heat pump has heat transfer from an environment at 5.00ºC to one at 35.0ºC. What is the best possible coefficient of performance for these temperatures? Explicitly show how you follow the steps in the Problem-Solving Strategies for Thermodynamics.
- (a) What is the best coefficient of performance for a heat pump that has a hot reservoir temperature of 50.0ºC and a cold reservoir temperature of −20.0ºC? (b) How much heat transfer occurs into the warm environment if 3.60 × 10
^{7}J of work (10.0 kW · h) is put into it? (c) If the cost of this work input is 10.0 cents/kW · h, how does its cost compare with the direct heat transfer achieved by burning natural gas at a cost of 85.0 cents per therm. (A therm is a common unit of energy for natural gas and equals 1.055 × 10^{8}J.) - (a) What is the best coefficient of performance for a refrigerator that cools an environment at −30.0ºC and has heat transfer to another environment at 45.0ºC? (b) How much work in joules must be done for a heat transfer of 4186 kJ from the cold environment? (c) What is the cost of doing this if the work costs 10.0 cents per 3.60 × 10
^{6}J (a kilowatt-hour)? (d) How many kJ of heat transfer occurs into the warm environment? (e) Discuss what type of refrigerator might operate between these temperatures. - Suppose you want to operate an ideal refrigerator with a cold temperature of −10.0ºC, and you would like it to have a coefficient of performance of 7.00. What is the hot reservoir temperature for such a refrigerator?
- An ideal heat pump is being considered for use in heating an environment with a temperature of 22.0ºC. What is the cold reservoir temperature if the pump is to have a coefficient of performance of 12.0?
- A 4-ton air conditioner removes 5.06 × 10
^{7}J (48,000 British thermal units) from a cold environment in 1.00 h. (a) What energy input in joules is necessary to do this if the air conditioner has an energy efficiency rating (*EER*) of 12.0? (b) What is the cost of doing this if the work costs 10.0 cents per 3.60 × 10^{6}J (one kilowatt-hour)? (c) Discuss whether this cost seems realistic. Note that the energy efficiency rating (*EER*) of an air conditioner or refrigerator is defined to be the number of British thermal units of heat transfer from a cold environment per hour divided by the watts of power input. - Show that the coefficients of performance of refrigerators and heat pumps are related by
*COP*_{ref}=*COP*_{hp}− 1. Start with the definitions of the*COP*s and the conservation of energy relationship between*Q*_{h},*Q*_{c}, and*W*.

3. 0.311

5. (a) 4.61; (b) 1.66 × 10

7. 27.6ºC

9. (a) 1.44 × 10