Homework Assignment #12

Heat Engines Introduced

Description: Simple questions to illustrate the concept and basic principles of a heat engine.

Learning Goal: To understand what a heat engine is and its theoretical limitations.

Ever since Hero demonstrated a crude steam turbine in ancient Greece, humans have dreamed of

converting heat into work. If a fire can boil a pot and make the lid jump up and down, why can't

heat be made to do useful work?

A heat engine is a device designed to convert heat into work. The heat engines we will study will

be cyclic: The working substance eventually returns to its original state sometime after having

absorbed a quantity of heat and done some work. A cyclic heat engine cannot convert heat into

work without generating some waste heat in the process. Although by no means intuitively

obvious, this is an important fact of nature, since it dramatically affects the technology of energy

generation. If it were possible to convert heat into work without any waste heat, then one would

be able to build refrigerators that are more than 100% efficient!

Consequently, the "impossible heat engine" pictured schematically here

cannot exist, even in theory. Engineers tried

hard for many years to make such a device, but Sadi Carnot proved in 1824 that it was

impossible.

The next figure shows an "ideal" heat engine,

one that obeys the laws of thermodynamics. It takes in heat at a temperature and does work

. In the process of doing this it generates waste heat at a cooler temperature .

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

both quantities are positive.

Part A

A heat engine is designed to do work. This is possible only if certain relationships between the

heats and temperatures at the input and output hold true. Which of the following sets of

statements must apply for the heat engine to do work?

ANSWER:

and

and

and

and

Part B

Find the work done by the "ideal" heat engine.

Express in terms of and .

ANSWER:

=

Part C

The thermal efficiency of a heat engine is defined as follows: .

Express the efficiency in terms of and .

ANSWER:

=

± From Hot to Cool: The Second Law of Thermodynamics

Description: ± Includes Math Remediation. 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 meaning of 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 thermodynamic processes in nature can spontaneously occur. For example,

imagine an object initially at rest suddenly taking off along a rough horizontal surface and

speeding up (gaining kinetic energy) while cooling down (losing thermal energy). Although such

a process 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. The second 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 vice versa. For example, a highly ordered ice crystal has a relatively

low entropy, whereas the same amount of water in a much less ordered 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 the system 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 stays constant 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 a system 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 lower temperature than that of the bucket.

Part B

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

ANSWER:

It increases.

It decreases.

It stays the same.

Part C

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

ANSWER:

It increases.

It decreases.

It stays the same.

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:

=

8.53×10−2

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.

Express your answer numerically in joules per kelvin.

ANSWER:

=

-50

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.

Express your answer numerically in joules per kelvin.

ANSWER:

=

62.5

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 change appreciably in the process.

Express your answer numerically in joules per kelvin.

ANSWER:

=

12.5

Note that the net entropy change is positive as the heat is transferred from the hotter object to the

colder one. If heat were transferred in the other direction, the change in entropy would have been

negative; that is, the entropy of the system would have decreased. This observation, not

surprisingly, is in full accord with the second law of thermodynamics.

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 the bar and the specific heat capacity of the metal. Also determine

the total change of entropy given the entropy change of the bar 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.

Hint A.1 How to approach the problem

You can calculate the entropy change of the lake using the formula for the entropy change in an

isothermal process. Note that 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:

=

288

Hint A.2.2 Find the energy change for the aluminum bar

What 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.

Express your answer numerically in joules.

ANSWER:

=

−5.13×105

Express your answer numerically in joules.

ANSWER:

=

5.13×105

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:

=

1780

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 be easily calculated. However, you can use the interpretation

of entropy as a measure of randomness to determine whether its change is positive or negative.

ANSWER:

Since the entropy change of a system is always positive, we can deduce that

the entropy of the aluminum bar has increased.

Since the final lower temperature of the bar means lower average speed of

molecular motion, we can deduce that the entropy of the bar has decreased.

We don't have enough information to determine whether the entropy of the

aluminum bar has decreased or 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:

=

1710

Even though the aluminum bar lowers its entropy, the total entropy change of the bar and its

surroundings (the water in the 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 systems

the growth of a fertilized egg from a single cell to a complex adult organism

the formation of snowflakes from molecules of liquid water with random motion to a

highly ordered crystal

the 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:

True

False

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 entropy of the environment increases, just as we

found out in the case of the hot aluminum bar cooling down in the lake.

Heat into Work

Description: Series of true/fase questions about isothermal expansion which test knowledge of

the first and second laws of thermodynamics.

An ideal gas is confined within a thermally isolated cylinder. It consists of atoms initially at a

pressure of . A movable piston seals the right end of the cylinder, as shown in the figure.

A given amount of heat is slowly added to

the gas, while the piston allows the gas to expand in such a way that the gas's temperature

remains constant at .

Part A

As heat is added, the pressure in this gas __________.

Hint A.1 Expansion or contraction?

If the gas is heated, positive work must be done, which means that the gas must expand.

Hint A.2 Isothermal process

Recall the ideal gas law: . Remember that the expansion is regulated to keep the

temperature constant.

ANSWER:

increases

decreases

remains constant

cannot be determined

Part B

Is the internal energy of the gas the same before and after is added?

Hint B.1 Find the formula for internal energy

The internal energy of an ideal gas is equal to the total kinetic energy of its constituent atoms.

Determine an expression for the internal energy of the gas per degree of freedom.

Express your answer in terms of Boltzmann's constant and the temperature of the gas.

ANSWER:

=

ANSWER:

yes

no

Part C

Does the second law of thermodynamics forbid converting all of the absorbed heat into work

done by the piston?

Hint C.1 Second law of thermodynamics

The following is the Kelvin-Planck statement of the second law of thermodynamics:

It is impossible for an engine working in a cycle to produce no other effect than that of

extracting heat from a reservoir and performing an equivalent amount of work.

Note especially the words "in a cycle." Does the gas in this problem undergo a complete cycle?

ANSWER:

yes

no

Part D

The (Kelvin-Planck statement of the) second law of thermodynamics reads as follows:

It is impossible for an engine working in a cycle to produce no other effect than that of extracting

heat from a reservoir and performing an equivalent amount of work.

The phrase "in a cycle" does not apply in this situation, so the second law does not forbid heat

being converted entirely into work. For this particular problem, is all of the heat energy absorbed

by the gas in fact turned into work done on the piston?

Hint D.1 Relate , , and

Using the first law of thermodynamics, find in terms of the change in the gas's internal energy

and the net work done by the gas .

ANSWER:

=

ANSWER:

yes

no

Part E

Does the magnitude of the force that the gas exerts on the piston depend on the piston's area?

Hint E.1 Find a formula for the force exerted on the piston

Type an expression for , the force exerted on the piston.

Express your answer in terms of the gas's pressure and the area of the cylinder .

ANSWER:

=

ANSWER:

yes

no

Part F

Is the total work done by the gas independent of the area of the piston?

Hint F.1 Find a formula for the work done on the piston

Find an expression for an infinitesimal amount of work done by the gas on the piston.

Express your answer in terms of the gas's pressure and an infinitesimal change in volume .

ANSWER:

=

Does this expression explicitly depend on the piston's area? Alternatively, consider the

statement of the first law of thermodynamics: . If and the temperature of the gas

are fixed, does depend on the area of the piston?

ANSWER:

yes

no

Given the same initial pressure of the gas, the greater the area of the piston, the larger the force

on it. However, the work done on the piston when heat energy is added is independent of

piston area (because and ). From this we can infer that the force must act

through a shorter distance (i.e., the piston does not move as much) when the piston area is

greater.

Simplified Otto Cycle

Description: Extremely simple, qualitative problem. Basically reinforces the definition of

adiabaticity, the fact that internal energy depends only on temperature, and the sign convention

for work done by/on a gas.

The idealized cycle shown is known as the Otto cycle.

Suppose an engine is executing this Otto cycle,

using a gas (not necessarily ideal) as its working substance. From state A to state B, the gas is

allowed to expand adiabatically. (An adiabatic process is one in which no heat is added to, or

given off by, the working gas.) The gas is then cooled at constant volume until it reaches state C,

at which point it is adiabatically compressed to state D. Finally, it is heated at constant volume

until it returns to state A.

The pressure and volume of the gas in state A are and respectively. The pressure and

volume of the gas in state C are and respectively.

Part A

How much heat is added to the gas as it proceeds from state A to state B?

Hint A.1 Type of thermodynamic process

The process from state A to state B is adiabatic. What does adiabatic mean?

Express your answer in terms of any needed variables from the problem introduction.

ANSWER:

=

0

Part B

How much heat is given up by the gas as it proceeds from state C to state D?

Hint B.1 Type of thermodynamic process

The process from state C to state D is adiabatic. What does adiabatic mean?

Express your answer in terms of any needed variables from the problem introduction.

ANSWER:

=

0

Part C

What is , the change in the gas's internal energy after a complete cycle?

Hint C.1 What does internal energy depend on?

The internal energy is a "state function"; it depends only on the state of the gas (in this case, just

the temperature) and is therefore not path-dependent. After a complete cycle, the gas returns to

its original state.

Express your answer in terms of any needed variables from the problem introduction.

ANSWER:

=

0

Part D

Suppose you want positive work to be done by the gas. In which direction should the Otto cycle

be run?

ANSWER:

clockwise (A to B to C to D to A)

counterclockwise (A to D to C to B to A)

20.6. IDENTIFY: Apply 1

11e

r C

H

| |1

| |

Qe

Q

SET UP: In part (b), H 10,000 J Q The heat discarded is C| |.Q

EXECUTE: (a) 0 40

11 0 594 59 4%

9 50e

(b) C H| | | |(1 ) (10,000 J)(1 0 594) 4060 J.Q Q e

EVALUATE: The work output of the engine is H C| | | | 10,000 J 4060 J 5940 J.W Q Q

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 large tub of ice and water. In 5 minutes of operation of the engine,

the heat rejected by the engine melts a mass of ice equal to 2.00×10−2

.

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 problem

Determine the temperature of each reservoir in kelvins and the heat rejected to the cold

reservoir. Use these values to calculate 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

Since the amount of ice that melts is known, the heat needed to melt the ice must be equal to

the magnitude of the heat rejected by the Carnot engine. Use the equation for the heat of a

phase change to determine this, then think about how this relates 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,

2.00×10−2

is the mass of ice being melted, and is the heat of fusion for water.

ANSWER:

=

-6680

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 reservoirs to 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:

=

9130

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 the engine into the cold reservoir. Be very careful about your

signs.

ANSWER:

=

2450

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 to the engine, before it does any work. Similarly,

when the Carnot engine rejects heat to a reservoir, the heat will be a negative 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.

Test Your Understanding 20.3: Internal-Combustion Engines

Description: [[This problem tests student understanding of internal-combustion engines.

Multiple-choice problem based on Test Your Understanding 20.3.]] (a) If you double the

compression ratio of an Otto-cycle engine from 6.0 to 12, what happens to the engine's...

Part A

If you double the compression ratio of an Otto-cycle engine from 6.0 to 12, what happens to the

engine's thermal efficiency?

ANSWER:

it decreases

it remains the same

it increases, but by a factor less than 2

it increases by a factor greater than 2

it increases by a factor of 2

it increases, but not enough information is given to decide by what factor

The thermal efficiency of an Otto-cycle engine with compression ratio is

where is the ratio of heat capacities for air. For we find

and for we find

Thus the efficiency increases, but by a factor less than 2.

Test Your Understanding 20.8: Microscopic Interpretation of Entropy

Description: [[This problem tests student understanding of the microscopic interpretation of

entropy. Multiple-choice problem based on Test Your Understanding 20.8.]] A thermodynamic

system has a large number of possible microscopic states. (a) If you double the...

A thermodynamic system has a large number of possible microscopic states.

Part A

If you double the number of possible microscopic states, what happens to the entropy of the

system?

ANSWER:

it decreases

it remains the same

it increases, but by a factor less than 2

it increases by a factor of 2

it increases by a factor greater than 2

it increases, but not enough information is given to decide by what factor

If a thermodynamic system has possible microscopic states, the entropy of that system is

where is the Boltzmann constant. If the number of possible microscopic states is doubled to

, the ratio of the final entropy to the initial entropy is

This is greater than 1, so the entropy increases. We are told that the initial number of possible

microscopic states is a large number. Therefore is much greater than 2, ln is much greater

than ln 2, and the ratio (ln 2)/(ln ) is a small positive number. This tells us that is equal to

1 plus a small positive number, so the ratio of entropies is between 1 and 2.

± A Carnot Refrigerator

Description: ± Includes Math Remediation. 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: two isothermal 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 outside temperatures.

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.

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 equations

Recall that in a Carnot cycle,

,

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 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 the temperatures, 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:

=

1.38×10−2

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:

=

145

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

also very low, and the refrigerator requires a great deal of energy input to do the job. This

partially explains why it is so expensive to produce and 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 by the 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:

=

14.0

Express your answer in joules to three significant figures.

ANSWER:

=

0.357

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.