The Second Law of Thermodynamics Mp

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The Second Law of Thermodynamics

Due: 8:00pm on Monday, December 12, 2011

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

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Carnot Cycle: Just How Ideal Is It?

Learning Goal: To understand the quantitative relationships related to ideal (Carnot) engines and the limitations of such devices 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, the most 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 that operates between the same two reservoirs.

The Carnot engine operates cyclically, just like any real engine.The Carnot cycle includes four reversible steps: two isothermal processes 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 good mathematical representation of efficiency of any heat engine?

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ANSWER :

Correct Part B

During the Carnot cycle, the overall entropy ________. Hint B.1 Some useful equations

Hint not displayed ANSWER : increases decreases remains constant Correct Part C

Which of the following gives the efficiency of the Carnot engine? Hint C.1 Some useful equations

Hint not displayed ANSWER

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Correct 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 : 0 0.455 0.545 1 infinity Correct Part E

We have stressed that the Carnot engine does not exist in real life: It is a purely theoretical device, useful for understanding the limitations of heat engines. Real engines never operate on the Carnot cycle; their efficiency is hence lower than that of the Carnot 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. Correct

The Carnot cycle contains only reversible processes. To be reversible, a process must allow the system to equilibrate with 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 the cold 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 most efficient one.

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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

. Express your answer in terms of , , and .

ANSWER

: ₌

Correct

Heat Engines Introduced

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

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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?

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ANSWER : and and and and Correct Part B

Find the work done by the "ideal" heat engine. Express in terms of and .

ANSWER

: = _Correct

Part C

The thermal efficiency of a heat engine is defined as follows: . Express the efficiency in terms of and .

ANSWER

: ₌

Correct

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 should understand the following:

1. Heat engines must be cyclical; that is, they must return to their original state some time after having absorbed some heat and 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 engine is reversible, another result of the laws of thermodynamics.

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heat pump (as pictured schematically ). 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 is called 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 ratio of the heat to the work in heat pumps and refrigerators is called the coefficient 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

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

: = _Correct

Part B

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

B.1 Conservation of energy and the first law Hint not displayed Express in terms of and .

ANSWER

: = _Correct

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 Hint not displayed Hint

C.2 Relation between and in a Carnot cycle Hint not displayed Give your answer in terms of and .

ANSWER

: ₌

Correct Part D

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

ANSWER : and . and . and . and . Correct 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, the more 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

Hint not displayed Express the cost in dollars.

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ANSWER

: Cost = 187.5Correct dollars

This savings is accompanied by more initial capital costs, both for the heat pump and for the generous area of baseboard heaters 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 outside air is humid and not much above zero degrees Celsius. Therefore heat pumps are most useful in temperate climates or where the heat can be obtained from a groundwater that is abundant or flowing (e.g., an underground stream).

Second Law of Thermodynamics

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 the types of final states of the system that naturally evolve from a given initial state. The second law has many practical applications. For example it explains the limits of efficiency for heat engines and refrigerators. To develop a better understanding of this law, try these conceptual questions.

Part A

The thermodynamic processes that occur in nature ____________. ANSWER

: convert thermal energy into mechanical energy lead to a more ordered state

cannot be reversed do not conserve energy Correct

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 it constant. For irreveresible processes, the entropy increases; for the reversible ones, the entropy remains constant.

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

: heat energy to flow from a colder body to a hotter body an ideal heat engine to have the efficiency of 99% an ideal heat engine to have non-zero power.

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

The ideal engine follows a reversible cycle--therefore, an infinitely slow one. If the work is being done at the infinitely slow 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 heat energy 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 to construct a machine in which a heat flow from a colder to a hotter object is accompanied by another process, such as work 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 Hint not displayed

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.

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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 Hint not displayed ANSWER

: the amount of heat consumed per second the temperature of the cold reservoir the temperature of the hot reservoir the size of the cold reservoir the size of the hot reservoir Correct

Part E

How would you increase the coefficient of performance of an ideal refrigerator? Hint E.1 Graphical approach to the problem

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

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

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.

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. Correct

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 hot reservoir 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 statements is important. However, it is not a trivial matter: Your textbook and discussions should help you to get a better grasp of this equivalency.

± From Hot to Cool: The Second Law of Thermodynamics

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.

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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. Correct

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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.

Part C

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

: It increases.

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

Correct

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

: = -50Correct

Part F

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

: = 62.5Correct

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.

: = 12.5Correct

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.

± PSS 20.1 Heat Engines

Learning Goal: To practice Problem-Solving Strategy 20.1 Heat Engines.

Steam at a temperature = 280 and = 1.00 enters a heat engine at an unknown flow rate. After passing through the heat engine, it is released at a temperature = 100 and = 1.00 . The measured power output of the engine is 600 , and the exiting steam has a heat transfer rate of = 3450 . Find the efficiency of the engine and the molar flow rate of steam through the engine. The constant pressure molar heat capacity

for steam is 37.47 .

Problem Solving Strategy: Heat engines

IDENTIFY the relevant concepts:

A heat engine is any device that converts heat partially to work.

SET UP the problem using the following steps:

1. Carefully define what the thermodynamic system is.

2. For multi-step processes with more than one step, identify the initial and final states for each step.

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3. Identify the known quantities and the target variables.

4. The first law, , can be applied just once to each step in a thermodynamic process, so you will often need additional equations. The equation

is useful in situations for which the thermal efficiency of the engine is relevant. It's helpful to sketch an energy-flow diagram.

EXECUTE the solution as follows:

1. Be very careful with the sign conventions for and the various 's. is positive when the system expands and does work; is negative when the system is compressed. Each is positive if it represents heat entering the system and is negative if it represents heat leaving the system.

2. Power is work per unit time ( ), and heat current is heat transfer per unit time ( ).

3. Keeping steps 1 and 2 in mind, solve for the target variables.

EVALUATE your answer:

Use the first law of thermodynamics to check your results, paying particular attention to algebraic signs.

IDENTIFY the relevant concepts

This heat engine partially converts heat from the incoming steam into work, so the problem solving strategy for heat engines is applicable.

SET UP the problem using the following steps Part A

Which of the following quantities are known? Check all that apply.

ANSWER

: The power output of the engine, The molar flow rate of steam,

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The temperature of steam as it leaves the engine, The heat transfer rate for steam entering the engine, The temperature of the steam as it enters the engine, The constant pressure molar heat capacity of steam, The efficiency of the engine,

Correct

The efficiency and the molar flow rate of steam through the engine are both target variables in this problem. Even though the heat transfer rate of the steam entering the heat engine is not given in the problem statement, it is not a target variable.

The energy-flow diagram for this system is shown. Heat flows into the engine at a rate of as steam at a temperature . Work leaves the engine at a rate . The remaining heat leaves the engine at a rate of as steam at temperature .

EXECUTE the solution as follows Part B

What is the efficiency of the heat engine? Hint

B.1 How to approach the problem

You are asked to find the efficiency of the heat engine.

To find the efficiency of the heat engine, first find the heat transfer rate of the incoming steam from the relationships between work and power and between heat and heat transfer

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rate. Once you know the incoming heat transfer rate, you can then calculate the efficiency of the heat engine using the power delivered by the engine.

Hint

B.2 Find the heat transfer rate of the incoming steam What is the heat transfer rate of the incoming steam ?

Hint

B.2.1 Find an expression for the heat transfer rate of the incoming steam Hint not displayed

Express your answer numerically in joules per second. ANSWER

: = 4050_Correct

Hint

B.3 Find an expression for the efficiency of the heat engine Hint not displayed

Express the efficiency numerically to three significant figures. ANSWER

: = 0.148Correct

An efficiency of 0.148 indicates that 14.8 of the heat from the incoming steam is converted to work. The remaining 85.2 is expelled from the heat engine as heat in the outlet steam.

Part C

What is the molar flow rate of steam into the engine? Hint

C.1 How to approach the problem

Hint not displayed Hint

C.2 The net heat flowing into the heat engine

The heat required for the temperature change of moles of steam can be expressed in terms of the difference in temperatures of the entering and exciting steam as

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.

The heat lost by the steam is equal to the heat gained by the steam engine, so the net heat flowing into the steam engine can be expressed as

. Hint

C.3 Find an expression for the molar flow rate of steam through the heat engine Express the molar flow rate of steam through the heat engine as a function of the generated power , the constant pressure molar heat capacity , and the change in temperature . Keep in mind that is related to , and since the energy of the heat engine is conserved, .

Express the molar flow rate in terms of , , and . ANSWER

: ₌ Answer not displayed

Express the molar flow rate in moles per second to three significant figures. ANSWER

: = 8.90×10

−2

Correct

Knowing the flow rate of steam through the heat engine is important for a few reasons. In the design stage, an estimate of the flow rate is used the size the heat engine. During the operation of the heat engine, enough steam must flow through the heat engine to produce the required amount of power.

EVALUATE your answer Part D

Which of the following changes, if made individually, would cause an increase in the amount of power produced by the heat engine?

Check all that apply. ANSWER

: smaller value of

larger value of

higher molar flow rate, smaller value of lower molar flow rate,

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larger value of Correct

A heat engine can produce more power either by improving its efficiency or by increasing its throughput. If either the entering heat transfer rate increases or the exiting heat transfer rate decreases, the power generated by the engine will increase. Dividing both sides of

by time makes the relationship between power and heat transfer rates clearer:

.

By either increasing the entering heat transfer rate or decreasing the exiting heat transfer rate, the efficiency of the heat engine is increased. If the molar flow rate increases, the amount of heat that the engine converts into work per unit time increases, but the engine is not necessarily more efficient.

A Three-Step Gas Cycle

A monatomic ideal gas has pressure and temperature . It is contained in a cylinder of volume with a movable piston, so that it can do work on the outside world.

Consider the following three-step transformation of the gas:

1. The gas is heated at constant volume until the pressure reaches (where ). 2. The gas is then expanded at constant temperature until the pressure returns to . 3. The gas is then cooled at constant pressure until the volume has returned to . It may be helpful to sketch this process on the pV plane.

Part A

How much heat is added to the gas during step 1 of the process? Hint A.1 First law of thermodynamics

The heat added to a gas equals the change in energy of the gas plus the work done by the

gas: .

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Hint not displayed Hint A.3 How to find

The energy of a monatomic ideal gas is . The number of gas particles and the Boltzmann constant do not change during step 1, so depends only on . Hint A.4 How to find

To find , use the ideal gas law . Solve for and take the difference between the value of at the beginning and that at the end of step 1.

Express the heat added in terms of , , and . ANSWER

: = _Correct

Part B

How much work is done by the gas during step 2? Hint

B.1 How to approach this problem

B.2 Find

B.3 Find the initial and final volumes

Hint not displayed Express the work done in terms of , , and .

ANSWER

: = _Correct

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How much work is done by the gas during step 3?

If you've drawn a graph of the process, you won't need to calculate an integral to answer this question.

Hint

C.1 The easy way to solve this problem

C.2 Find the formula for work done

Hint not displayed Express the work done in terms of , , and .

ANSWER

: = _Correct

An Air Conditioner: Refrigerator or 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 the right contain a fluid (the working substance) called refrigerant, which is typically in vapor-liquid phase equilibrium. The compressor takes in low-pressure, low-temperature vapor and compresses it adiabatically to high-pressure, high-temperature vapor, which then reaches the condenser. Here the refrigerant is at a higher temperature than that of the air surrounding the condenser coils and it releases heat by undergoing a phase change. The refrigerant leaves the condenser coils as a high-pressure, high-temperature

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liquid and expands adiabatically at a controlled rate in the expansion valve. As the fluid expands, it cools down. Thus, when it enters the evaporator coils, the refrigerant is at a lower temperature than its surroundings and it absorbs heat. The air surrounding the evaporator cools down and most of the refrigerant in the evaporator coils vaporizes. It then reaches the compressor as a 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 1.50×105 _.

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Find the percentage of refrigerant transformed to vapor Hint not displayed

Express your answer numerically in joules. ANSWER

: = 5.15×10

5

Correct

Part B

In each cycle, the change in internal energy of the refrigerant when it leaves the compresser is 1.20×105_{. What is the work done by the motor of the compressor?}

Hint

B.1 Adiabatic compression

Hint not displayed Express your answer in joules.

ANSWER

: = 1.20×10

5

Correct

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If the direction of the refrigerant flow is inverted in an air conditioner, the air conditioning unit turns into a heat pump and it can be used for heating rather than cooling. In this case, the coils where the refrigerant would condense in the air conditioner become the evaporator coils when the unit is operated as a heat pump, and, vice versa, the evaporator coils of the air conditioner become the condenser coils in the heat pump. Suppose you operate the air conditioner described in Parts A and B as 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 processes during the cycle.

Hint

C.1 How to approach the problem

C.2 Find the right expresssion for the first law of thermodynamics Hint not displayed

Express your answer numerically in joules. ANSWER : = 6.35×10 5 Correct

Carnot Cycle

After Count Rumford (Benjamin Thompson) and James Prescott Joule had shown the equivalence of mechanical energy and heat, it was natural that engineers believed it possible to make a "heat engine" (e.g., a steam engine) that would convert heat completely into mechanical energy. Sadi Carnot considered a hypothetical piston engine that contained moles of an ideal gas, showing first that it was reversible, and most importantly that— regardless of the specific heat of the gas—it had limited efficiency, defined as , where is the net work done by the engine and is the quantity of heat put into the engine at a (high) temperature . Furthermore, he showed that the engine must necessarily put an amount of heat back into a heat reservoir at a lower temperature .

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Carnot cycle is shown in the figure. The working gas first expands isothermally from state A to state B, absorbing heat from a reservoir at temperature . The gas then expands adiabatically until it reaches a temperature

, in state C. The gas is compressed isothermally to state D, giving off heat . Finally, the gas is adiabatically compressed to state A, its original state.

Part A

Which of the following statements are true? Hint A.1 Heat flow in an adiabatic process

Hint not displayed Check all that apply.

ANSWER

: For the gas to do positive work, the cycle must be traversed in a _{clockwise manner.} Positive heat is added to the gas as it proceeds from state C to state D. The net work done by the gas is proportional to the area inside the closed curve.

The heat transferred as the gas proceeds from state B to state C is greater than the heat transferred as the gas proceeds from state D to state A.

Correct Part B

Find the total work done by the gas after it completes a single Carnot cycle. Hint How to approach the problem

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B.1

B.2 Compute the change in internal energy Hint not displayed

Express the work in terms of any or all of the quantities , , , and . ANSWER

: = _Correct

Part C

Suppose there are moles of the ideal gas, and the volumes of the gas in states A and B are, respectively, and . Find , the heat absorbed by the gas as it expands from state A to state B.

Hint

C.1 General method of finding

C.2 Find the work done by the gas

C.3 Relation between and

Express the heat absorbed by the gas in terms of , , , the temperature of the hot reservoir, , and the gas constant .

ANSWER

: ₌

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Part D

The volume of the gas in state C is , and its volume in state D is . Find , the magnitude of the heat that flows out of the gas as it proceeds from state C to state D.

Hint

D.1 How to approach the problem

Express your answer in terms of , , , (the temperature of the cold reservoir), and .

ANSWER

: ₌

Correct

Observe that the three parts together imply that . This is because BC and DA are adiabatic processes. So using the first law, ,

whereas . So , or . This is a

general result: Any two adiabatic processes operating between the same two temperatures result in the same amount of work, regardless of the pressure and volume differences. Part E

Now, by considering the adiabatic processes (from B to C and from D to A), find the ratio in terms of and .

Hint

E.1 How to approach the problem

E.2 Rewrite in terms of and

E.3 Express and in terms of and Hint not displayed

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Hint

E.4 Express and in terms of and Hint not displayed Hint E.5

Solving for in terms of and Hint not displayed ANSWER

: ₌

Correct Part F

Using your expressions for and (found in Parts C and D), and your result from Part E, find a simplified expression for .

No volume variables should appear in your expression, nor should any constants (e.g., or ).

ANSWER

: ₌

Correct Part G

The efficiency of any engine is, by definition, . Carnot proved that no engine can have an efficiency greater than that of a Carnot engine. Find the efficiency of a Carnot engine.

Hint

G.1 Express the efficiency in terms of and Hint not displayed Express the efficiency in terms of and .

ANSWER

: ₌

Correct

Because is generally fixed (e.g., the cold reservoir for power plants is often a river or a lake), engineers, trying to increase efficiency, have always sought to raise the upper

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temperature . This explains why (historically) there were some spectacular explosions of boilers used for steam power.

Carnot Heat Engine Pressure versus Volume Graph Conceptual Question

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

Hint not displayed Hint A.2 Carnot cycle

: positive

zero negative Correct

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.

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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 difference between 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 Hint not displayed Complete the sentence above.

ANSWER

: greater than

less than Correct

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 Hint not displayed Hint

C.2 Comparing the entropy changes

Hint not displayed Complete the sentence above.

ANSWER

: greater than

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less than Correct

Part D

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

Hint

D.1 Total change in entropy

Hint not displayed Complete the sentence above.

ANSWER

: positive

negative zero Correct

Because there is no heat flow during an adiabatic process, there is no entropy change for either the gas or the surroundings. (This is an approximation, but we can wrap the gas sample in lots of insulation to isolate it from the surroundings.)

During the isothermal compression, an amount of heat energy flows out of the gas at a temperature . This results in a decrease in the gas's entropy of magnitude . However, heat energy will naturally flow out of the gas into the surroundings only if there is a (slight) temperature difference between the gas and the surroundings.

Part E

In a real isothermal compression, the temperature of the surroundings must be ___________ the temperature of the gas.

Complete the sentence above. ANSWER

: greater than

less than Correct

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

Complete the sentence above. ANSWER : greater than equal to less than Correct Part G

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

Complete the sentence above. ANSWER

: positive

negative zero Correct

Therefore, although the net entropy change of the gas is still zero after a real Carnot-like cycle, the net entropy change of the entire system is positive, because during both the isothermal expansion and compression phases the net change in entropy of the system was positive. This is unavoidable and a manifestation of the second law of thermodynamics.

Does Entropy Really Always Increase?

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 not displayed Hint A.2 Find the heat absorbed by the lake

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Hint A.3 Entropy change in an isothermal process Hint not displayed Express your answer numerically in joules per kelvin

ANSWER

: = 1780Correct

Part B

Has the entropy of the aluminum bar decreased or increased? Hint B.1 How to approach the question

: Since the entropy change of a system is always positive, we can deducethat 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.

Correct 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

Hint not displayed Express your answer numerically in joules per Kelvin

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ANSWER

: = 1710Correct

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 Correct

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

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

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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?

Hint not displayed Hint A.2 Isothermal process

Hint not displayed ANSWER : increases decreases remains constant cannot be determined Correct 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

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ANSWER

: yes

no Correct

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

: yes

no Correct

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}

: yes

no Correct

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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 Hint not displayed

ANSWER

: yes

no Correct

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

: yes

no Correct

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.

Internal-Combustion Engine Prototypes Ranking Task

Six new prototypes for internal-combustion engines are tested in the laboratory. For each engine, the heat energy input and output per cycle, and the designed number of cycles per second are measured.

Part A

Rank these engines on the basis of the work they perform per cycle. Hint A.1 How to approach the problem

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Hint A.2 First law of thermodynamics applied to an engine cycle Hint not displayed

Hint A.3 Net heat transfer

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

:

View Correct Part B

Rank these engines on the basis of their designed power output. Hint

B.1 Calculating power

:

View Correct Part C

Rank these engines on the basis of their thermal efficiency. Hint

C.1 Definition of thermal efficiency

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ANSWER :

View Correct

Irreversible versus Reversible Processes

Part A

Which of the following conditions should be met to make a process perfectly reversible? Hint A.1 Reversible processes

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 or pressure gradients.

The system should not be close to equilibrium. Correct

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

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 . Correct

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Refrigerator Light

The inside of an ideal refrigerator is at a temperature , while the heating coils on the back of the refrigerator are at a temperature . Owing to a malfunctioning switch, the light bulb within the refrigerator remains on when the the door is closed. The power of the light bulb is

; assume that all of the energy generated by the light bulb goes into heating the inside of the refrigerator.

For all parts of this problem, you must assume that the refrigerator operates as an ideal Carnot engine in reverse between the respective temperatures.

Part A

If the temperatures inside and outside of the refrigerator do not change, how much extra power does the refrigerator consume as a result of the malfunction of the switch?

Hint A.1 Find the coefficient of performance Find the refrigerator's coefficient of performance .

Hint

A.1.1 Relate the heats exchanged to the work done

The coefficient of performance is defined as , where is the heat absorbed from the cold reservoir (the inside of the refrigerator) and is the work done by the refrigerator's motor. If is the heat given off by the refrigerator, find an expression for

.

Give your answer in terms of and . ANSWER

: = _Correct

Hint

A.1.2 Relating the heats and the work to the appropriate temperatures The absolute temperature scale may be defined by the following ratio:

,

where and are the respective heats expelled to a cold reservoir and absorbed from a hot reservoir for an ideal Carnot engine. Recall that we are assuming the refrigerator to be a Carnot engine operating in reverse, so that heat is aborbed from the cold reservoir and expelled to the hot reservoir. That's what a good refrigerator does, of course. Give your answer in terms of and .

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ANSWER

: ₌

Correct

Hint A.2 How much heat is absorbed from the cold reservoir for the properly working refrigerator?

Suppose that, before the light was on, the refrigerator absorbed an amount of heat from the cold compartment during every fixed time interval . (We could say, for instance,

, but the fixed time interval will cancel in the final answer.) After the light comes on, the temperature of the cold compartment should not change (according to the problem statement). For this to be true, how much heat will the refrigerator have to absorb in a time while the light remains on?

Express your answer in terms of , , and . ANSWER

: ₌ _{Answer not displayed}

Hint A.3 How much work is done by the malfunctioning refrigerator? Hint not displayed

Hint A.4 Convert heat into power

Hint A.5 How much power was consumed by the properly working refrigerator? Hint not displayed

Hint A.6 Putting it all together

Hint not displayed Express the extra power in terms of , , and .

ANSWER

: ₌

Correct

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of , the extra power needed to run the refrigerator with the light on is much smaller than . The next part of this problem shows this for a specific case.

Part B

Suppose the refrigerator has a 25-W light bulb, the temperature inside the refrigerator is , and the temperature of the heat dissipation coils on the back of the refrigerator is . Find the extra power consumed by the refrigerator. Keep in mind that you will need to use absolute units of temperature (i.e., kelvins).

Express your answer numerically in watts to three significant figures. ANSWER

: = 3.20Correct

Six Carnot Engines with Varying Reservoirs Ranking Task

Six Carnot engines operating between different hot and cold reservoirs are described below. The heat energy transferred to the gas 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

.

:

View Correct Part B

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

Hint

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B.2 Does the second law of thermodynamics apply? Hint not displayed

:

View Correct

Six New Heat Engines Conceptual Question

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

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

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

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ANSWER

: CFCorrect 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 Hint not displayed

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

ABD).

ANSWER

: BDCorrect Part C

Which of the remaining designs has the highest thermal efficiency? ANSWER

: device A

device E Correct

± From Hot to Cool: A Change in Entropy

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 the process. Note also that the ice starts out at the melting point, and you are asked about the change in entropy by the time it just melts. In other words, you can assume that the temperature of the "ice water" remains constant as well.

Hint not displayed Hint A.2 Description of entropy

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Hint not displayed Hint A.3 Heat needed to melt the ice

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

ANSWER

: = 8.35×10

−2

Correct

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

Melting Ice with a Carnot 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 6.00×10−2_.

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

Hint not displayed Hint A.2 Temperature conversion

Hint not displayed Hint A.3 Calculate the heat rejected

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Hint A.4 Calculate the heat absorbed

Hint not displayed Hint A.5 Using the first law of thermodynamics

: = 7340Correct

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.

± Entropy Change of an Expanding Gas

Two moles of an ideal gas undergo a reversible isothermal expansion from 2.57×10−2 _to

4.86×10−2 _{at a temperature of 29.1 .}

Part A

What is the change in entropy of the gas? Hint A.1 How to approach the problem

Hint not displayed Hint A.2 Calculate the work done by the gas

Hint not displayed Hint A.3 Calculating the change in entropy

Hint not displayed Express your answer numerically in joules per kelvin.

ANSWER

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Correct

As the gas expands, the greater volume allows the molecules of the gas to explore a greater range of positions, so the disorder of the gas (the molecules of which can have an increased randomness of position in an increased volume) will therefore be increased as well.

Problem 20.51

A Carnot engine operates between two heat reservoirs at temperatures and . An inventor proposes to increase the efficiency by running one engine between and an intermediate temperature and a second engine between and using as input the heat expelled by the first engine.

Part A

Compute the efficiency of this composite system.

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

: ₌

Correct Part B

Compare the efficiency of this composite system to that of the original engine. ANSWER

: overall efficiency has been reduced overall efficiency has been increased overall efficiency hasn`t been changed Correct

Exercise 20.20

An ideal Carnot engine operates between 505 and 180 with a heat input of 300 per cycle.

Part A

How much heat is delivered to the cold reservoir in each cycle? ANSWER

: = 175_Correct

Part B

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through a height of 105 ? ANSWER

: 3700Correct cycles

Problem 20.58

A 1.60×10−2_{- cube of ice at an initial temperature of -18.0 is placed in 0.700 of}

water at 42.0 in an insulated container of negligible mass. Part A

Calculate the change in entropy of the system. ANSWER

: = 3.50Correct

Multiple Choice Question - 20.3

Part A

Two Carnot heat engines operate in tandem as follows: engine A takes in 13.0 kJ per cycle from a heat reservoir at a temperature of 470 K. The heat rejected by engine A is received by engine B, which performs 4.3 kJ of net work per cycle. Engine B, in turn, rejects heat at a temperature of 320 K. The temperature at which engine A rejects heat to engine B, in SI units, is closest to:

ANSWER : 398 449 395 475 422 Correct

Multiple Choice Question - 20.26

Part A

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A diesel engine operates reversibly on the cycle abcda, using 9.0 moles of an ideal gas. Paths bc and da are adiabatic processes. The operating temperatures of points a, b, c, and d of the cycle are 375 K, 450 K, 432 K, and 250 K, respectively. The adiabatic constant of the gas is 1.50.

In Fig. 20.4, the heat intake during the isobaric expansion, in kJ, is closest to: ANSWER : 29 25 37 33 41 Correct

Short Answer Question - 20.5

An ideal Carnot engine operates between reservoirs having temperatures of 125°C and -20.0°C. Each cycle the heat expelled by this engine is used to melt 37.0 g of ice at 0.00°C. The heat of fusion of water is 3.34 × 105_{J/kg and the heat of vaporization of water is 2.25 ×}

106_J/kg.

Part A

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ANSWER

: 7080Correct J

Part B

How much heat per cycle does this engine absorb at the hot reservoir? ANSWER

: 1.94×10

4

Correct J

Score Summary:

Your score on this assignment is 98.9%.

You received 222.56 out of a possible total of 225 points. [ Print ]