36113095 Extra Materials Carnot Engine

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Physics 5D Fall 2008

Assignment 8

Due at 4:00pm on Thursday, December 11, 2008

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Carnot Cycle: Just How Ideal Is It?

Description: A mix of conceptual and quantitative questions about ideal heat engines.

Learning Goal: To understand the quantitative relationships related to ideal (Carnot) engines and the limitations of suchdevices imposed by the second law of thermodynamics.

In 1824, Sadi Carnot, a French engineer, introduced a theoretical engine that has been since then called a Carnot engine , themost efficient engine possible. The following statement is known as Carnot's theorem :

No engine operating between a hot and a cold reservoir can be more efficient than the Carnot engine thatoperates between the same two reservoirs.

The Carnot engine operates cyclically, just like any real engine.The Carnot cycle includes four reversible steps: two isothermalprocesses and two adiabatic ones.

In this problem, you will be asked several questions about Carnot engines. We will use the following symbols:

: the absolute value (magnitude) of the heat absorbed from the hot reservoir during one cycle or during some time

specified in the problem;: the absolute value (magnitude) of the heat delivered to the cold reservoir during one cycle or during some time

specified in the problem;: the amount of work done by the engine during one cycle or during some time specified in the problem;

: the absolute temperature of the hot reservoir; and

: the absolute temperature of the cold reservoir.

Part A

In general terms, the efficiency of a system can be thought of as the output per unit input. Which of the expressions is a goodmathematical representation of efficiency of any heat engine?

ANSWER:

Part B

During the Carnot cycle, the overall entropy ________.

Hint B.1 Some useful equations

Recall that in a Carnot cycle,

,

where we have taken all heat exchanged to be positive. Also, the definition of the entropy change when heat is

transferred at a temperature is

.
http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1171902#


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: increasesdecreasesremains constant

Part C

Which of the following gives the efficiency of the Carnot engine?

Hint C.1 Some useful equations


,

where we have taken all heat exchanged to be positive. Substitute for and in the earlier expression for the efficiency.

Also recall that

.

ANSWER:

Part D

Consider a Carnot engine operating between the melting point of lead ( ) and the melting point of ice ( ). What is the

efficiency of such an engine?

ANSWER: 00.4550.5451infinity

Part EWe have stressed that the Carnot engine does not exist in real life: It is a purely theoretical device, useful for understanding thelimitations of heat engines. Real engines never operate on the Carnot cycle; their efficiency is hence lower than that of theCarnot engine. However, no attempts to build a Carnot engine are being made. Why is that?

ANSWER: A Carnot engine would generate too much thermal pollution.Building the Carnot engine is possible but is too expensive.The Carnot engine has zero power.The Carnot engine has too low an efficiency.

The Carnot cycle contains only reversible processes. To be reversible, a process must allow the system to equilibratewith its surroundings at every step, which makes it infinitely slow; therefore, the Carnot engine, although the most efficient, is the least powerful one ! Its power is indeed zero, since work is being done at an infinitely slow rate.

Part F

A real heat engine operates between temperatures and . During a certain time, an amount of heat is released to thecold reservoir. During that time, what is the maximum amount of work that the engine might have performed?

Hint F.1 Real vs. ideal

How much work could an ideal (Carnot) engine perform? This is the maximum possible, since the Carnot engine is the mostefficient one.

Hint F.2 Some useful equations


,


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Also recall that

.

Express your answer in terms of , , and .

ANSWER:

=

From Hot to Cool: The Second Law of Thermodynamics

Description: Discussion of the second law of thermodynamics: several statements, examples of irreversible processes.Introduction to entropy; some basic calculations involving entropy change at constant temperature.

Learning Goal: To understand the meaning and applications of the second law of thermodynamics, to understand the meaningof entropy, and perform some basic calculations involving entropy changes.

The first law of thermodynamics (which states that energy is conserved) does not specify the direction in which thermodynamicprocesses in nature can spontaneously occur. For example, imagine an object initially at rest suddenly taking off along a roughhorizontal surface and speeding up (gaining kinetic energy) while cooling down (losing thermal energy). Although such aprocess would not violate conservation of energy, it is, of course, impossible and could never take place spontaneously .

The second law of thermodynamics dictates which processes in nature may occur spontaneously and which ones may not. Thesecond law can be stated in many ways, one of which uses the concept of entropy .

Entropy

Entropy can be thought of as a measure of a system's disorder : A lower degree of disorder implies lower entropy, and viceversa. For example, a highly ordered ice crystal has a relatively low entropy, whereas the same amount of water in a much lessordered state, such as water vapor, has a much higher entropy. Entropy is usually denoted by , and has units of energy

divided by temperature ( ). For an isothermal process (the temperature of the system remains constant as it exchanges heat

with its surroundings), the change in a system's entropy is given by

,

where is the amount of heat involved in the process and is the absolute temperature of the system. The heat is positive

if thermal energy is absorbed by the system from its surroundings, and is negative if thermal energy is transferred from thesystem to its surroundings.

Using the idea of entropy, the second law can be stated as follows:

The entropy of an isolated system may not decrease. It either increases as the system approaches equilibrium, or staysconstant if the system is already in equilibrium.

Any process that would tend to decrease the entropy of an isolated system could never occur spontaneously in nature. For asystem that is not isolated, however, the entropy can increase, stay the same, or decrease.

Part A

What happens to the entropy of a bucket of water as it is cooled down (but not frozen)?

ANSWER: It increases.It decreases.It stays the same.

Presumably, the bucket is not isolated: Heat must be transferred to another object, which is most likely at a lowertemperature than that of the bucket.

Part B

What happens to the entropy of a cube of ice as it is melted?


Part C

What happens to the entropy of a piece of wood as it is burned?


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When a solid object is turned into a gas, the degree of disorder increases, so the entropy increases.

Let us try some calculations now.

Part D

An object at 20 absorbs 25.0 of heat. What is the change in entropy of the object?

Express your answer numerically in joules per kelvin.

ANSWER:=

Part E

An object at 500 dissipates 25.0 of heat into the surroundings. What is the change in entropy of the object? Assume

that the temperature of the object does not change appreciably in the process.


ANSWER:=

Part F

An object at 400 absorbs 25.0 of heat from the surroundings. What is the change in entropy of the object? Assume

that the temperature of the object does not change appreciably in the process.


ANSWER:=

Part G

Two objects form a closed system. One object, which is at 400 , absorbs 25.0 of heat from the other object,which is at

500 . What is the net change in entropy of the system? Assume that the temperatures of the objects do not changeappreciably in the process.


ANSWER:=

Note that the net entropy change is positive as the heat is transferred from the hotter object to the colder one. If heat weretransferred in the other direction, the change in entropy would have been negative ; that is, the entropy of the systemwould have decreased. This observation, not surprisingly, is in full accord with the second law of thermodynamics.

Heat Pumps and Refrigerators

Description: Simple questions to illustrate the concept and basic principles of heat pumps and refrigerators.

Learning Goal: To understand that a heat engine run backward is a heat pump that can be used as a refrigerator.

By now you should be familiar with heat engines--devices, theoretical or actual, designed to convert heat into work. You shouldunderstand the following:

1. Heat engines must be cyclical; that is, they must return to their original state some time after having absorbed some heatand done some work).

2. Heat engines cannot convert heat into work without generating some waste heat in the process.

The second characteristic is a rigorous result even for a perfect engine and follows from thermodynamics. A perfect heat engineis reversible, another result of the laws of thermodynamics.

If a heat engine is run backward (i.e., with every input and output reversed), it becomes a heat pump (as pictured schematically ).


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Work must be put into a heat pump, and it then pumps heat

from a colder temperature to a hotter temperature , that is,

against the usual direction of heat flow (which explains why it iscalled a "heat pump").

The heat coming out the hot side of a heat pump or the heat

going in to the cold side of a refrigerator is more than the work

put in; in fact it can be many times larger. For this reason, the ratioof the heat to the work in heat pumps and refrigerators is called thecoefficient of performance , . In a refrigerator, this is the ratio of

heat removed from the cold side to work put in:

.

In a heat pump the coefficient of performance is the ratio of heat exiting the hot side to the work put in:

.

Take , and to be the magnitudes of the heat emitted and absorbed respectively.

Part A

What is the relationship of to the work done by the system?

Hint A.1 Note the differences in wording

Recall that is the work done by the system; is the work done on the system.

Express in terms of and other quantities given in the introduction.

ANSWER:=

Part B

Find , the heat pumped out by the ideal heat pump.

Hint B.1 Conservation of energy and the first law

Apply conservation of energy. If you think in terms of the first law of thermodynamics, remember the sign conventions forheat and work and note that the internal energy does not change in an engine after one cycle.

Express in terms of and .

ANSWER:=

Part C

A heat pump is used to heat a house in winter; the inside radiators are at and the outside heat exchanger is at . If it is a

perfect (i.e., Carnot cycle) heat pump, what is , its coefficient of performance?

Hint C.1 Heat pump efficiency in terms of and

What is the efficiency of a heat pump in terms of the heats in and out? Use the expression for the efficiency of the

heat pump and the expression that you found involving the work done on the system, , and the outside heats, and

.

Give your answer in terms of and .

ANSWER:

=

Hint C.2 Relation between and in a Carnot cycle



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.

Give your answer in terms of and .

ANSWER:

=

Part DThe heat pump is designed to move heat. This is only possible if certain relationships between the heats and temperatures atthe hot and cold sides hold true. Indicate the statement that must apply for the heat pump to work.

ANSWER: and .

and .

and .

and .

Part E

Assume that you heat your home with a heat pump whose heat exchanger is at , and which maintains the baseboard

radiators at . If it would cost $1000 to heat the house for one winter with ideal electric heaters (which have a

coefficient of performance of 1), how much would it cost if the actual coefficient of performance of the heat pump were 75%

of that allowed by thermodynamics?

Hint E.1 Money, heat, and the efficiency

The amount of money one has to pay for the heat is directly proportional to the work done to generate the heat. Thus, themore efficient the heat generation the less work needs to be done and the lower the heating bill.

You are given that the cost of is $1000. You also have an equation for in terms of the temperatures:

% .

Set this equal to and solve for the monetary value of , the amount of external energy input the pump requires.

You can measure energies in units of currency for this calculation.

Hint E.2 Units of and

Keep in mind that when calculating an efficiency of a thermodynamic device you need to use temperature in kelvins. That is,.

Express the cost in dollars.

ANSWER:Cost =

dollars

This savings is accompanied by more initial capital costs, both for the heat pump and for the generous area of baseboardheaters needed to transfer enough heat to the house without raising , which would reduce the coefficient of

performance. An additional problem is icing of the outside heat exchanger, which is very difficult to avoid if the outsideair is humid and not much above zero degrees Celsius. Therefore heat pumps are most useful in temperate climates orwhere the heat can be obtained from a groundwater that is abundant or flowing (e.g., an underground stream).

Second Law of ThermodynamicsDescription: Conceptual questions probing general understanding of the second law of thermodynamics. Offers threestatements of the second laws and stresses (though does not explain) their equivalence.

Learning Goal: To understand the implications of the second law of thermodynamics.

The second law of thermodynamics explains the direction in which the thermodynamic processes tend to go. That is, it limits thetypes of final states of the system that naturally evolve from a given initial state. The second law has many practicalapplications. For example it explains the limits of efficiency for heat engines and refrigerators. To develop a betterunderstanding of this law, try these conceptual questions.

Part A

The thermodynamic processes that occur in nature ____________.


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ANSWER: convert thermal energy into mechanical energylead to a more ordered statecannot be reverseddo not conserve energy

Only infinitely slow, or quasi-equilibrium , processes can be reversible; such processes exist only in the imaginations of scientists.

One of the ways to state the second law of thermodynamics is as follows:

Any process occurring in a closed system either increases the entropy (disorder) of the system or leaves itconstant. For irreveresible processes, the entropy increases; for the reversible ones, the entropy remainsconstant.

Part B

According to the second law of thermodynamics, it is impossible for ____________.

ANSWER: heat energy to flow from a colder body to a hotter bodyan ideal heat engine to have the efficiency of 99%an ideal heat engine to have power greater than 0.1

a physical process to yield more energy than what is put in

The ideal engine follows a reversible cycle--therefore, an infinitely slow one. If the work is being done at the infinitelyslow rate, the power of such an engine is zero.

An alternative way to state the second law of thermodynamics is as follows:

It is impossible to construct a cyclical heat engine whose sole effect is the continuous transfer of heatenergy from a colder object to a hotter one.

This statement is known as Clausius statement of the second law . Note the word "sole." Of course, it is possible toconstruct a machine in which a heat flow from a colder to a hotter object is accompanied by another process, such aswork input.

Part C

If the coefficient of performance of a refrigerator is 1, which the following statements is true?

Hint C.1 Definition of the coefficient of performance of a refrigerator

The coefficient of performance of a refrigerator is defined as the heat extracted from the colder region per unit of work done.

ANSWER: The temperature outside equals the temperature inside of the refrigerator.The rate at which heat is removed from the inside equals the rate at which heat is delivered outside.The power consumed by the refrigerator equals the rate at which heat is removed from the inside.The power consumed by the refrigerator equals the rate at which heat is delivered to the outside.

Part D

To increase the efficiency of an ideal heat engine, one must increase which of the following?

Hint D.1 Formula for the efficiency of an ideal engine

The efficiency of an ideal (Carnot) engine that absorbs heat from a reservoir at a temperature , and releases heat

to another reservoir at a temperature , while doing work , is

.

ANSWER: the amount of heat consumed per secondthe temperature of the cold reservoirthe temperature of the hot reservoirthe size of the cold reservoirthe size of the hot reservoir

Part E

How would you increase the coefficient of performance of an ideal refrigerator?


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Hint E.1 Graphical approach to the problem

The figure shows a pV diagram for a refrigerator undergoing anideal (Carnot) cycle. To increase the efficiency, you wouldhave to decrease the work input required (i.e., the area of theshaded region) for a given . How would you achieve this?

ANSWER: Increase the mechanical work input.Decrease the outside temperature.Decrease the inside temperature.Increase the outside temperature.

The efficiency of an ideal/Carnot refrigerator that absorbs heat from a reservoir at a temperature , and releases

heat to another reservoir at a temperature , with a work input , is

.

From this equation, you can see that decreasing and/or increasing lead to an increase in the efficiency .

Part F

Why must every heat engine have a cold reservoir?

ANSWER: Because it is impossible for even a perfect engine to convert heat entirely into mechanical work.Because the cold reservoir keeps the engine from overheating.Because the cold reservoir keeps the engine from overcooling.Because the cold reservoir increases the power of the engine.

Another way to state the second law of thermodynamics is as follows:

It is impossible to construct a cyclical heat engine whose sole effect is absorption of energy from the hotreservoir and the performance of the equal amount of work.

This statement is known as the Kelvin-Planck statement of the second law . Note the word "sole."

You have now seen three different statements of the second law. Understanding the equivalence of these three statementsis important. However, it is not a trivial matter: Your textbook and discussions should help you to get a better grasp of this equivalency.

A Carnot Refrigerator

Description: A mix of conceptual and quantitative questions about ideal refrigerators.

The ideal Carnot engine operates cyclically, just like any real engine.The Carnot cycle includes four reversible steps: twoisothermal processes and two adiabatic ones. If the Carnot cycle is reversed, a Carnot refrigerator is created. This theoretical

device has the highest coefficient of performance among all refrigerators operating between given inside and outsidetemperatures.

Throughout this problem, we will use the following symbols:

: the (positive) amount of heat delivered to the "outside" during one cycle or during some time specified in the

problem;: the (positive) amount of heat absorbed from the "inside" during one cycle or during some time specified in the

problem;: the amount of external work input during one cycle or during some time specified in the problem;

: the absolute temperature of the outside; and

: the absolute temperature of the inside.


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In general terms, the performance of a system can be thought of as the output per unit input. The coefficient of performance

of any refrigerator can be represented mathematically as

.

Part A

What is the coefficient of performance of the Carnot refrigerator?

Hint A.1 Some useful equationsRecall that in a Carnot cycle,

,


Also recall that

.

ANSWER:

Part B

Imagine an ideal (Carnot) refrigerator that keeps helium in its liquid state, at a temperature of about 4.00 . The refrigerator,

with the helium container inside, is in a laboratory where the temperature is about 294 . Because of the imperfection of the

insulation, 2.00 of heat is absorbed by the refrigerator each hour. How much electrical energy must be used by the

refrigerator to maintain a temperature of 4.00 inside for one hour?

Hint B.1 How to approach the problem

In Part A, you found an expression for in a Carnot refrigerator in terms of the temperatures. Since you have thetemperatures, you can calculate for this problem. You also know that is defined by . Therefore, if you

find the value of and calculate , you can find the work required. The work is supplied by the electrical energy, so

.

Hint B.2 Find the value of

Use the formula you found in Part A to determine for this refrigerator.

Express your answer numerically.

ANSWER:=

Hint B.3 What is the value of ?

The refrigerator, i.e., the helium in it, absorbs 2.00 of heat every hour. For the helium to be maintained at the same

temperature, the same amount of heat must be removed from it also. Therefore .

Express your answer in joules to three significant figures.

ANSWER:=

When the temperature difference is great, and is very low, the coefficient of performance is also very low, and the


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refrigerator requires a great deal of energy input to do the job. This partially explains why it is so expensive to produceand maintain liquid helium. In this example, 145 of energy is required to extract only 2 from the refrigerator.

However, when the temperature difference is not so great, the refrigerators become much more efficient, as illustrated bythe next problem.

Part C

Imagine an ideal (Carnot) refrigerator that keeps soda bottles chilled to a temperature of about 280 . The refrigerator is

located in a hot room with a temperature of about 300 . Because of the imperfect insulation, 5.00 of heat is absorbed by

the refrigerator each hour. How much electrical energy must be used by the refrigerator to maintain the temperature of 280

inside for one hour?

Hint C.1 Find the value of

Use the formula you found in Part A to determine for this refrigerator.

Express your answer numerically.

ANSWER:=

Express your answer in joules to three significant figures.

ANSWER:=

When the temperature difference is small, and is not too low, the refrigerator requires very little energy input: In our

example, it uses less than 0.4 to exhaust 5 from the refrigerator.

Carnot Heat Engine Pressure versus Volume Graph Conceptual Question

Description: A conceptual question in which a pressure vs. volume representation of a Carnot heat engine is given. Questionsdeal with change in entropy and heat. (conceptual)

Imagine the Carnot heat engine represented by the vs. diagram given in the figure.

Part A

What is the sign of the change in entropy of the gas after one complete cycle?

Hint A.1 Entropy

The change in entropy for a reversible process that transfers heat energy at temperature is given by

.

Hint A.2 Carnot cycle

The complete cycle given in the diagram is a Carnot cycle, in that each leg of the cycle is reversible. This reversibility iswhat distinguishes the Carnot cycle from other thermodynamic cycles. In a Carnot cycle, the amount of heat energy gainedby the gas is the same as the amount lost by the gas.

ANSWER: positivezero


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negative

In an ideal Carnot engine, where each process is reversible, the change in entropy of the gas and the change in entropy of the surroundings are each exactly zero after each complete cycle. However, let us investigate what happens in a real process.

During the isothermal expansion, an amount of heat energy flows into the gas at a temperature . This results in an

increase in the gas's entropy of

.

However, heat energy will naturally flow into the gas from the surroundings only if there is a (slight) temperature differencebetween the gas and the surroundings.

Part B

In a real isothermal expansion, the temperature of the surroundings must be________the temperature of the gas.

Hint B.1 Heat flow and temperature difference

Heat energy will naturally flow only from a higher to a lower temperature.

Complete the sentence above.

ANSWER: greater thanless than

Part C

Because of this temperature difference, the magnitude of the entropy lost by the surroundings is ________ the magnitude of entropy gained by the gas during a real isothermal expansion.

Hint C.1 Magnitude of the change in entropy

Since the change in entropy for this process that transfers heat energy at temperature is defined by

,

the substance at lower temperature will have a larger change in entropy.

Hint C.2 Comparing the entropy changes

The change in entropy of the gas that transfers heat energy at temperature is given by

.

This is positive because the heat energy flows into the gas.

The change in entropy for the surroundings, which are at temperature , is given by

.

This is negative because the heat energy flows out of the surroundings. Also note that the temperature of the surroundingsmust be slightly higher than the temperature of the gas in order for heat to flow from the surroundings into the gas.

Complete the sentence above.

ANSWER: greater thanequal to

less than

Part D

Because of this difference in entropy change, the net entropy change of the entire system is ________ during a real isothermalexpansion.

Hint D.1 Total change in entropy

The change in entropy of the entire system is the arithmetic sum of the change in entropy of the gas, , and

the change in entropy of the surroundings, ,


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When the partition is removed, the gas expands freely and no work is done. Thus the velocities of the gas molecules areunaffected. However, now each molecule has a bigger volume in which it can move and the number of possible positions of each molecule has increased. By considering the change in volume of the gas after the expansion, you can determine thenumber of possible microscopic states. This information allows you to determine the change in entropy of the gas as afunction of the number of moles in the gas, and finally to find the number of moles in the gas.

Hint A.2 Find the number of possible microscopic states of a gas after a free expansion

Consider a thermally insulated box such as the one described in the introduction. Assume that the smaller compartmentcontains molecules of gas, whereas the other is evacuated. When the partition is removed the gas expands and occupies

the entire box. The number of microscopic states of the system when the gas initially occupies a volume is . Find the

number of microscopic states when the gas occupies the entire box (volume ).

Express your answer in terms of and .

ANSWER:=

Hint A.3 Find the microscopic expression for the change in entropy of a gas

Consider a gas that undergoes a thermodynamic process such that the gas is taken from an initial macroscopic state, wherethe possible microscopic states are , to a final macroscopic state characterized by possible microscopic states. What isthe change in entropy of the gas after this process?

Hint A.3.1 Microscopic expression for entropy

The entropy of a macroscopic state with possible microscopic states is

,

where is the Boltzmann constant.

Express your answer in terms of some or all of the variables , , and the Boltzmann constant .

ANSWER:=

Now, combine this result with the expression for the number of possible microscopic states of a gas after a freeexpansion and solve for the number of molecules. Note that the problem asks for the number of moles of the gas, soyou need to express the number of molecules in terms of the number of moles. Remember that the Boltzmann constantcan be expressed in terms of the gas constant and Avogadro's number.

Hint A.4 Relating to the gas constant

Recall the relation , where is the gas constant, is the Boltzmann constant, and is Avogadro's number.

Express your answer to three significant figures.

ANSWER:

=

1.89

From Hot to Cool: A Change in Entropy

Description: Basic calculation involving entropy change at constant temperature.In a well-insulated calorimeter, 1.0 of water at 20 is mixed with 1.0 of ice at 0 .

Part A

What is the net change in entropy of the system from the time of mixing until the moment the ice completely melts?

The heat of fusion of ice is .

Note that since the amount of ice is relatively small, the temperature of the water remains nearly constant throughout theprocess. Note also that the ice starts out at the melting point, and you are asked about the change in entropy by the time it justmelts. In other words, you can assume that the temperature of the "ice water" remains constant as well.


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Hint A.1 How to approach the problem

Find the changes in entropy for the water and for the ice. Since energy is being transfered from the water into the ice, thechanges in entropy will have opposite signs. Add the changes to find the net change in entropy for the system.

Hint A.2 Description of entropy

For an isothermal process (the temperature of the system remains constant as it exchanges heat with its surroundings), thechange in a system's entropy is given by

,

where is the amount of heat involved in the process and is the absolute temperature of the system. The heat ispositive if thermal energy is absorbed by the system from its surroundings, and is negative if thermal energy is transferredfrom the system to its surroundings.

Hint A.3 Heat needed to melt the ice

The amount of heat necessary to melt a piece of ice of mass is given by , where is the heat of fusion of

ice.

Express your answer numerically in joules per kelvin. Use two significant figures in your answer.

ANSWER:

=

As you would expect, in this spontaneous process the net change in entropy is positive: The entropy increases. This isevident not just from the calculation but also from the fact that a crystal becomes liquid and hence the degree of disorderincreases.

Irreversible versus Reversible Processes

Description: Short conceptual problem on identifying reversible processes. This problem is based on Young/GellerConceptual Analysis 16.1.

Part A

Which of the following conditions should be met to make a process perfectly reversible?

Hint A.1 Reversible processes

A reversible process is a transition from one state of a thermodynamic system to another, during which the system is alwaysvery close to a state of mechanical and thermal equilibrium. This implies that the process should be slow and smooth toallow temperature and pressure to change only by infinitesimal amounts.

Check all that apply.

ANSWER: Any mechanical interactions taking place in the process should be frictionless.Any thermal interactions taking place in the process should occur across infinitesimal temperature orpressure gradients.The system should not be close to equilibrium.

Part B

Based on the results found in the previous part, which of the following processes are not reversible?

Hint B.1 How to approach the problem

As you found in the previous part, a process that takes place through infinitesimal changes in the conditions of a system isreversible. Do any of the processes listed involve nonnegligible changes in temperature and pressure?

Check all that apply.

ANSWER: Melting of ice in an insulated ice-water mixture at 0 .

Lowering a frictionless piston in a cylinder by placing a bag of sand on top of the piston.Lifting the piston described in the previous statement by removing one grain of sand at a time.Freezing water originally at 5 .

Six New Heat Engines Conceptual Question

Description: A conceptual question in which flow diagrams for six heat engines are given. Choose which engines violate the


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first law of thermodynamics, which violate the second law of thermodynamics, and which of the remaining engines has thehighest thermal efficiency. (conceptual)

As part of your job at the patent office, you are asked to evaluate the six designs shown in the figure for innovative new heatengines.

Part A

Which of the designs violate(s) the first law of thermodynamics?

Hint A.1 The first law of thermodynamics applied to a heat engine

By conservation of energy, the heat energy input to an engine must equal the sum of the work output and heat energy output.

Give the letter(s) of the design(s) in alphabetical order, without commas or spaces (e.g., ACD ).

ANSWER:CF

Part B

Which of the remaining designs violate(s) the second law of thermodynamics?

Hint B.1 The second law of thermodynamics applied to a heat engine

By the second law of thermodynamics, a heat engine operating between two reservoirs and has a maximum

thermal efficiency given by

.

An engine with thermal efficiency greater than that given by the above equation violates the second law of thermodynamics.

Give the letter(s) of the design(s) in alphabetical order, without commas or spaces (e.g., ABD ).

ANSWER: BD

Part C

Which of the remaining designs has the highest thermal efficiency?

ANSWER: device Adevice E

Refrigerator Prototypes Ranking Task

Description: A conceptual problem in which power and heat removal are given for six refrigerator prototypes. Rank therefrigerators on the basis of performance coefficient and ability to heat a sealed room. (ranking task)

Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power needed for it to operate

and the maximum heat energy that can be removed per second from its interior are given.

Part A

Rank these refrigerators on the basis of their performance coefficient.


A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. By


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conservation of energy, the energy deposited in the hot reservoir is the sum of the work done on the refrigerator and theenergy removed from the cold reservoir.

A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoirfor a small amount of work input.

Hint A.2 Definition of the performance coefficient

The performance coefficient of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir

to the work input to the refrigerator:

.

Recall that power is defined as work per unit time.

Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:

View

Part BThe six refrigerators are placed in six identical sealed rooms. Rank the refrigerators on the basis of the rate at which they raisethe temperature of the room.

Hint B.1 Temperature and Conservation of energy

A refrigerator uses electrical energy to remove heat from its interior and expel it into the environment. By conservation of energy, the energy expelled into the room must be the sum of the energy extracted from the interior of the refrigerator andthe energy input to run the refrigerator. Notice that the rate at which the temperature of the room rises is directly proportionalto the rate at which energy is expelled into the room.


ANSWER:

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Six Carnot Engines with Varying Reservoirs Ranking Task

Description: A conceptual question in which energy flow diagrams for six Carnot engines with different hot and coldreservoir temperatures are given. Rank the engines on the basis of the change in entropy during the complete cycle and during

just the isothermal expansion phase. (ranking task)

Six Carnot engines operating between different hot and cold reservoirs are described below. The heat energy transferred to thegas during the isothermal expansion phase of each cycle is indicated.

Part A

Rank these engines on the basis of the change in entropy of the gas during the isothermal expansion phase of the cycle.

Hint A.1 Change in entropy

The change in entropy for a reversible process that transfers heat energy at temperature is

.


ANSWER: View
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Part B

Rank these engines on the basis of the change in entropy of the gas during one complete cycle.

Hint B.1 Change in entropy for a complete cycle

Entropy is a state function , like internal energy. This means that the value of the entropy depends only on the state of asystem, not on the process by which it arrived at that state. Notice that the initial and final states of the gas are identical for acomplete cycle. (The initial and final states of the rest of the universe are not.)

Hint B.2 Does the second law of thermodynamics apply?

The second law of thermodynamics applies to all phenomena, as long as we include the entropy changes for the entireinteracting system. If we isolate just a portion of a larger interacting system, the entropy of that subsystem need not alwaysincrease.


ANSWER:

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An Air Conditioner: Refrigerator or Heat Pump?

Description: Find the heat absorbed by an air conditioner given the mass of refrigerant per cycle and the percetange of vaporand liquid at the entrance and exit of the evaporator. Also, find the work done by the compressor given the change in internalenergy of the refrigerant and calculate the heat released by the air conditioner when operated as a heat pump.

The typical operation cycle of a common refrigerator is shown schematically in the figure. Both the condenser coils to the left and the evaporator coils to theright contain a fluid (the working substance) called refrigerant,which is typically in vapor-liquid phase equilibrium. Thecompressor takes in low-pressure, low-temperature vapor andcompresses it adiabatically to high-pressure, high-temperaturevapor, which then reaches the condenser. Here the refrigerant is at ahigher temperature than that of the air surrounding the condensercoils and it releases heat by undergoing a phase change. Therefrigerant leaves the condenser coils as a high-pressure, high-temperature liquid and expands adiabatically at a controlled rate inthe expansion valve. As the fluid expands, it cools down. Thus,when it enters the evaporator coils, the refrigerant is at a lowertemperature than its surroundings and it absorbs heat. The airsurrounding the evaporator cools down and most of the refrigerantin the evaporator coils vaporizes. It then reaches the compressor asa low-pressure, low-temperature vapor and a new cycle begins.

Part A

Air conditioners operate on the same principle as refrigerators. Consider an air conditioner that has 7.00 of refrigerant

flowing through its circuit each cycle. The refrigerant enters the evaporator coils in phase equilibrium, with 54.0 of its mass

as liquid and the rest as vapor. It flows through the evaporator at a constant pressure and when it reaches the compressor 95

of its mass is vapor. In each cycle, how much heat is absorbed by the refrigerant while it is in the evaporator? The heat of

vaporization of the refrigerant is 1.50 ! 10 5 .


To transform a given mass of liquid refrigerant to vapor requires the addition of a certain quantity of heat that depends on theheat of vaporization of the refrigerant and the mass of refrigerant transformed to vapor. Since the refrigerant is in phaseequilibrium when it enters the evaporator, it is already at boiling temperature and all the absorbed heat is used in the liquid-vapor phase change. Note that the pressure is kept constant in the evaporator. Also, a small fraction of refrigerant is stillliquid when it leaves the evaporator, so the refrigerant must still be in phase equilibrium at that point and no change intemperature has occurred in the evaporator.

Hint A.2 Find the percentage of refrigerant transformed to vapor

Consider a refrigerator where, in each cycle, 54.0 of refrigerant enters the evaporator as liquid in liquid-vapor phase

equilibrium and 95 leaves as vapor. Assume that the pressure of the refrigerant is kept constant while it is in the

evaporator. What is the percentage of the total mass of refrigerant transformed to vapor in the evaporator per cycle?

ANSWER:=
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Now calculate the heat needed to transform this amount of refrigerant to vapor. Recall that the heat of vaporization isdefined as the amount of heat required to vaporize a unit mass of a given substance.

Express your answer numerically in joules.

ANSWER:

=

Part B

In each cycle, the change in internal energy of the refrigerant when it leaves the compresser is 1.20 ! 10 5 . What is the work

done by the motor of the compressor?

Hint B.1 Adiabatic compression

Recall that the compressor performs an adiabatic compression; that is, no heat exchange occurs while the refrigerant is in thecompressor. Then use the first law of thermodynamics to determine .

Express your answer in joules.

ANSWER:=

Part C

If the direction of the refrigerant flow is inverted in an air conditioner, the air conditioning unit turns into a heat pump and itcan be used for heating rather than cooling. In this case, the coils where the refrigerant would condense in the air conditionerbecome the evaporator coils when the unit is operated as a heat pump, and, vice versa, the evaporator coils of the airconditioner become the condenser coils in the heat pump. Suppose you operate the air conditioner described in Parts A and Bas a heat pump to heat your bedroom. In each cycle, what is the amount of heat released into the room? You may assume

that the energy changes and work done during the expansion process are negligible compared to those for other processesduring the cycle.

Hint C.1 How to approach the problem

When the air conditioner is operated in the cooling mode, in each cycle it absorbs an amount of heat from the air inside

the room, as you calculated in Part A, and gives off a certain amount of heat to the outside. When it is operated in the

heating mode, instead, the direction of the flow of the refrigerant inside the coil circuit is simply inverted, but the operationcycle remains the same. In the heating mode, then, the same quantity is now absorbed from the air outside the room and

the amount of heat is released to the air inside the room. Note that the compressor does the same amount of work

regardless of the mode of operation, and use the first law to determine .

Hint C.2 Find the right expresssion for the first law of thermodynamics

Suppose you have an ideal system that absorbs heat from a cold reservoir and releases heat to a hot reservoir. The

net input of work required by this system for operation is . Which of the following expressions is correct?

Hint C.2.1 Cyclic processes

Remember that the net internal energy change in a cyclic process is zero, since the system has the same temperature whenit returns to the starting point.

ANSWER:


ANSWER:

=

Does Entropy Really Always Increase?

Description: A hot metal bar is thrown into a lake. Find the entropy change of the lake given the temperatures of the lake and


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the bar and the specific heat capacity of the metal. Also determine the total change of entropy given the entropy change of thebar and discuss the application of the second law of thermodynamics to systems that are not isolated.

An aluminum bar of mass 2.00 at 300 is thrown into a lake. The temperature of the water in the lake is 15.0 ; the

specific heat capacity of aluminum is 900 .

Part A

The bar eventually reaches thermal equilibrium with the lake. What is the entropy change of the lake? Assume that the

lake is so large that its temperature remains virtually constant.


You can calculate the entropy change of the lake using the formula for the entropy change in an isothermal process. Notethat the amount of heat transferred to the water in the lake is equal to the amount of heat lost by the bar.

Hint A.2 Find the heat absorbed by the lake

Find , the amount of heat absorbed by the lake.

Hint A.2.1 Find the temperature of thermal equilibrium

What is the temperature of thermal equilibrium of the aluminum bar and the lake?

Express your answer numerically in kelvins.

ANSWER:=

Hint A.2.2 Find the energy change for the aluminum barWhat is the change of heat energy for the metal bar?

Hint A.2.2.1 Heat and the temperature change

The heat absorbed or given up by an object is given by

,

where is the mass of the object, is the initial temperature, is the final temperature, and is the specific

heat of the substance.


ANSWER:=


ANSWER:=

Hint A.3 Entropy change in an isothermal process

When there is a heat transfer of to a substance at constant temperature , the entropy change of the substance is

given by

,

where is absolute temperature.

Express your answer numerically in joules per kelvin

ANSWER:=

Part B

Has the entropy of the aluminum bar decreased or increased?

Hint B.1 How to approach the question

Since the temperature of the aluminum bar changes during the cooling process, the exact entropy change of the bar cannot beeasily calculated. However, you can use the interpretation of entropy as a measure of randomness to determine whether itschange is positive or negative.


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ANSWER: Since the entropy change of a system is always positive, we can deduce that the entropy of the aluminumbar has increased.Since the final lower temperature of the bar means lower average speed of molecular motion, we candeduce that the entropy of the bar has decreased.We don't have enough information to determine whether the entropy of the aluminum bar has decreasedor increased.

Part C

Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone.Rather, it should be applied to the bar in combination with its surroundings (the lake).Assume that the entropy change of the bar is -73.5 , what is the change in total entropy ?

Hint C.1 Total change of entropy

The change in total entropy is defined as the sum of the entropy changes of the system and the surroundings.

Express your answer numerically in joules per Kelvin

ANSWER:

=

Even though the aluminum bar lowers its entropy, the total entropy change of the bar and its surroundings (the water inthe lake) is positive, and the total entropy increases.

Part D

The second law of thermodynamics states that spontaneous processes tend to be accompanied by entropy increase. Consider,however, the following spontaneous processes:

the growth of plants from simple seeds to well-organized systemsthe growth of a fertilized egg from a single cell to a complex adult organismthe formation of snowflakes from molecules of liquid water with random motion to a highly ordered crystalthe growth of organized knowledge over time

In all these cases, systems evolve to a state of less disorder and lower entropy, apparently violating the second law of thermodynamics. Could we, then, consider them as processes occurring in systems that are not isolated?

ANSWER: TrueFalse

All the processes listed above require energy input to occur just as a refrigerator requires electrical energy to run.Systems can become more ordered and lower their entropy as time passes. However, this can happen only as the entropyof the environment increases, just as we found out in the case of the hot aluminum bar cooling down in the lake.

Melting Ice with a Carnot Engine

Description: If a Carnot engine melts a given amount of ice, calculate the work done by the engine.

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a largetub of ice and water. In 5 minutes of operation of the engine, the heat rejected by the engine melts a mass of ice equal to4.15 ! 10 " 2 .

Throughout this problem use for the heat of fusion for water.

Part A

During this time, how much work is performed by the engine?

Hint A.1 How to approach the problemDetermine the temperature of each reservoir in kelvins and the heat rejected to the cold reservoir. Use these values tocalculate the heat absorbed from the hot reservoir, and then calculate the work done by the engine.

Hint A.2 Temperature conversion

To convert a temperature from degrees Celsius into kelvins, use .

Hint A.3 Calculate the heat rejected

Calculate , the heat rejected to the cold reservoir by the Carnot engine.

Hint A.3.1 How to approach the problem


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Since the amount of ice that melts is known, the heat needed to melt the ice must be equal to the magnitude of the heatrejected by the Carnot engine. Use the equation for the heat of a phase change to determine this, then think about how thisrelates to the heat rejected by the engine.

Hint A.3.2 Equation for melting ice

The equation for ice melting is , where is the heat absorbed by the ice, 4.15 ! 10 " 2 is the mass of ice being

melted, and is the heat of fusion for water.

ANSWER:=

Hint A.4 Calculate the heat absorbed

Calculate , the heat absorbed from the hot reservoir by the Carnot engine.

Hint A.4.1 How to approach the problem

Use the relationship between the temperatures of the reservoirs and the heats absorbed from or rejected to those reservoirsto calculate the heat absorbed from the hot reservoir.

Hint A.4.2 Equation for heat transfer in a Carnot engine

The equation relating the heats and temperatures in a Carnot engine is given by

.

Note the sign used in the equation, and recall that all temperatures should be expressed in kelvins.

ANSWER:

=

Hint A.5 Using the first law of thermodynamics

The work done by the engine is the sum of the heat absorbed by the engine from the hot reservoir and the heat rejected by theengine into the cold reservoir. Be very careful about your signs.

ANSWER:

=

As you can see from this problem, it is very important to keep in mind the signs of the heats exchanged in an engine.When the Carnot engine absorbs heat from a reservoir, the heat will be a positive quantity since the heat is being added tothe engine, before it does any work. Similarly, when the Carnot engine rejects heat to a reservoir, the heat will be anegative quantity since the heat is lost from the engine. The work done by the engine, by the first law of thermodynamics,is therefore the sum of all heat changes in the engine.

Summary 1 of 15 items complete (6.67% avg. score)4 of 70 points

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36113095 Extra Materials Carnot Engine