Mastering Physics HW 4 Ch 18, 19

7242019 Mastering Physics HW 4 Ch 18 19

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HW 4 Ch 18 19

Due 1159pm on Thursday October 1 2015

To understand how points are awarded read the Grading Policy for this assignment

Refrigerator Prototypes 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 heatenergy that can be removed per second from its interior are given

Part A

Rank these refrigerators on the basis of their performance coefficient

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

Hint 1 How to approach the problem

A refrigerator is a devic e t hat uses work to remove heat energy from a c old reservoir and deposit it into a hot reservoir By conservation of energy the energy deposited in the hot reservoir is the sum of the work done on the refrigerator and the energy removed from the cold reservoir

A good refrigerator (with a large performance coeffic ient) will remove a large amount of heat energy from the cold reservoir for a s mall amount of work input

Hint 2 Definition of the performance coefficient

The performance coeffic ient 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

ANSWER

Correct

Part B

The six refrigerators are placed in six identical sealed rooms Rank the refrigerators on the basis of the rate at which they raise the temperature of th

P

Δ 983156 Q

C m a x

983147 Q

C

W

983147 =

Q

C

W

Typesetting math 95



room


Hint 1 Temperature and Conservation of energy

A refrigerator uses electrical energy to remove heat from its interior and expel it into the environment B y conservation of energy the energyexpelled into the room must be the sum of the energy extracted from the interior of the refrigerator and the energy input to run the refrigeratorNotice that the rate at which the temperature of the room rises is directly proportional to the rate at which energy is expelled into the room

ANSWER

Correct

plusmn PSS 191 Heat-Engine Problems

Learning Goal

To practice Problem-Solving Strategy 191 for heat-engine problems

A heat engine uses the closed cyc le s hown in the diagram below The working substance ismoles of monatomic ideal gas Find the efficiency of such a cycle Use the values for pressureand volume shown in the diagram and assume that the process between points 1 and 3 isisothermal

PROBLEM-SOLVING STRAT EGY 191 Heat-engine problems

MODEL Identify each process in the cycle

983150

η



VISUALIZE Draw the pV diagram of the cycle

SOLVE There are several steps in the mathematical analysis

Use the ideal gas law to complete your knowledge of and at one point in the cycleUse the ideal gas law and equations for specific gas processes to determine and at the beginning and end of each processCalculate and for each processFind by adding for each process in the cycle If the geometry is simple you can confirm this value by finding the area enclosedwithin the pV curve

Add just the positive values of to find

Verify that This is a self-consist ency check to verify that you havent made any mistakes

Calculate the thermal effic iency and any other quantities you need to complete the solution

ASSESS Is Do all the signs of and make sense Does have a reasonable value Have you answered the question

Model

Part A

The cycle used by the engine is composed of three processes a process at constant pressure between point 1 and 2 a process at constant volumebetween points 2 and 3 and an isothermal process between points 3 and 1 What are the processes between points 1 and 2 and between points 2 an3 respectively

ANSWER

Correct

Visualize

The pV diagram is already given in the problem introduction You may want to make a copy of the diagram in your notes so that you can add further information to it as you work through the next part

Solve

Part B

Find the effic iency of the heat engine

Express your answer as a decimal number to three significant figures

Hint 1 The efficiency of a heat engine

The efficiency of a heat engine is given by the ratio of the work done by the engine (work output ) to the energy transferred into theengine (heat input )

Hint 2 How to find the work done in one cycle

There are two ways to compute the net work done by the engine during one full cycle Using a geometrical approach calculate the areaenclosed by the pV curve for the cycle is equal to the value of this area Alternatively identify all the processes in one full cyc le andcompute the work done by the engine in each process Find by adding for each process in the cycle

Hint 3 First method Use calculus to compute areas

Because the region enclosed by the pV curve shown in this problem does not have a simple shape you need to use your knowledge of calculus to compute its area Recall that the area between the graphs of two functions and over a certain interval [ ] is given

by

For this particular problem take the function whose graph is the curve for the process as the function and the function whose graph is

983150 983152 V T

983152 V T

Q W

s

Δ E

t h

W

o u t

W

s

Q Q

H

( Δ = 0 E

t h

)

n e t

η

( Δ = 0 E

t h

)

n e t

W

s

Q η

isochoric and isobaric

isobaric and isochoric

isobaric and isothermal

isothermal and isochoric

η

η W

o u t

Q

H

η = W

o u t

Q

H

W

o u t

W

o u t

W

s

W

o u t

W

s

A 983142 ( 983160 ) 983143 ( 983160 )

a b

A = ( 983142 minus 983143 ) 983140 983160 int

b

a

3 rarr 1 983142



the curve for the process as Use the ideal gas law to find the correct mathematical expressions for these functions Perform the

integration over the V interval [3 9]

Hint 4 Second method Find the work done during process

How much work is done by the gas during process

Express your answer in joules

Hint 1 Work in isobaric processes

In an isobaric process at pressure the work done by the syst em in which the volume changes by an amount is given by

ANSWER

Hint 5 Find the work done during process



Hint 1 Work in isochoric processes

Recall that an isochoric process is a process in which the volume does not change Consequently no work is done on or by the system

ANSWER




Hint 1 Work in isothermal processes

In an isothermal process at temperature the work done by the sys tem when the volume changes from to is given by

where is the number of moles in the gas and is the gas constant

Hint 2 How to solve this problem when moles and temperature are not given

Leave (the number of moles of gas) and (the temperature of the gas during the process ) as symbols in your calculation Youwill find in the end that you need a numerical value for the product Use the ideal gas law to compute this product in terms of thepressure and volume at either point 1 or point 3

ANSWER

Hint 7 Find which processes contribute to the heat input

During which processes in the cycle is heat delivered to the engine

123

Enter the letter(s) of all the correct answers in alphabetical order Do not use commas For example if you think all three processes

deliver heat to the engine enter ABC

1 rarr 2 983143

( c ) 1 0

2

m

3

1 rarr 2

W

1 2

1 rarr 2

983152 Δ V

= 983152 Δ V W

983155

=W

1 2

J

2 rarr 3

W

2 3

2 rarr 3

=W

2 3

J

3 rarr 1

W

3 1

3 rarr 1

T V

i

V

f

= 983150 R T l n ( ) W

983155

V

f

V

i

983150 R

983150 T 3 rarr 1

983150 R T

=W

3 1

J

1 rarr 2

2 rarr 3

3 rarr 1



ANSWER

Hint 8 Compute the heat input during process

What amount of heat is delivered to the engine during process


Hint 1 Heat transfer during isochoric processes

The heat transfer during an isochoric process in which the temperature of the gas changes by an amount is

where is the number of moles in the gas and is the heat capacity at constant volume

Hint 2 The properties of monatomic gases

For monatomic gases the heat capacity at constant volume is

where is the gas constant

Hint 3 How to compute the change in temperature

Even though the temperature of the gas is not given in the problem at certain points in the cycle can be expressed in terms of someknown quantities using the ideal gas law Thus use this law to find the temperature at points 2 and 3 in the pV diagramand then compute Note that you wonrsquot need to know how many moles are in the gas because when you calculate theamount of heat transferred in the process the number of moles will cancel out

ANSWER




Hint 1 Heat transfer during isothermal processes

Since in an isothermal process the thermal energy of the gas does not change is zero From the first law of thermodynamics itfollows that the heat transfer during an isothermal process is equal to the work done by the syst em or In other words all of the heat input during an isothermal process is converted into work done by the gas

ANSWER

ANSWER

Correct

If you prefer to express the effic iency as a percentage in this case you would say that the engine is 206 effic ient

Assess

Part C

2 rarr 3

Q

2 3

2 rarr 3

Δ T

Q = 983150 Δ T C

V

983150 C

V

= C

V

3 R

2

R

T

983152 V =

983150 R T

Δ T = minus T

3

T

2

983150

=Q

2 3

J

3 rarr 1

Q

3 1

3 rarr 1

Δ E

t h

Q = W

s

=Q

3 1

J

= 0206η



Here is a table of energy transfers for this problem with some entries missing

Process (J) (J) (J)

What must be the heat input in process to satis fy the condition that



To compute the heat transfer occurring during the isobaric process you can either use the generic expression for heat transfer in isobaric

processes where is the heat capacity at constant pressure or you can apply directly the first law of thermodynamics

to the process To do that youll need to know the change in thermal energy of the gas during the isobaric processRead from the table the changes in thermal energy during the isochoric and the isothermal processes and then calculate for the isobaricprocess using the fact that for a complete cyc le

ANSWER

Correct

You can easily verify that your results make sense The net heat transfer per cyc le is which is equal to the

work done by the engine in one cyc le By the first law of thermodynamics it follows that which is what we would expect foa cyclic process

An Air Conditioner Refrigerator or Heat Pump

The typical operation cycle of a common refrigerator is shown schematically in the figure Boththe condenser coils to the left and the evaporator coils to the right contain a fluid (the workingsubstance) 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-pressurehigh-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 liquid and expands adiabatically at a controlled rate in the expansion valve As thefluid expands it cools down Thus when it enters the evaporator coils the refrigerant is at alower 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 thecompressor 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 700 of refrigerant flowing through it s circuiteach cycle The refrigerant enters the evaporator coils in phase equilibrium with 540 of its mass as liquid and the rest as vapor It flows throughthe 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 150times10 5

Express your answer numerically in joules


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

W

s

Q Δ E

t h

1 rarr 2 minus 1 2 0

2 rarr 3 0 1 8 0 1 8 0

3 rarr 1 1 9 8 1 9 8 0

Q

1 2

1 rarr 2 ( Δ = 0 E

t h

)

n e t

Q

1 2

Q = 983150 Δ T C

983152

C

p

Δ = Q minus E

t h

W

s

1 rarr 2

Δ E

t h

Δ = 0 E

t h

= -300Q

1 2

J

= + + = 7 8 J Q

n e t

Q

1 2

Q

2 3

Q

3 1

W

o u t

Δ = 0 0 J E

t h

k g

Q

c

J k g



Hint 2 Find the percentage of refrigerant transformed to vapor

Consider a refrigerator where in each cycle 540 of refrigerant enters the evaporator as liquid in liquid-vapor phase equilibrium and 95leaves 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

ANSWER

Correct

Part B

In each cycle the change in internal energy of the refrigerant when it leaves the compresser is 120times105 What is the work done by the motorthe compressor


Hint 1 Adiabatic compression

Recall that the compressor performs an adiabatic compression that is no heat exchange occurs while the refrigerant is in the compressorThen use the first law of thermodynamics to determine

ANSWER

Correct

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 it can be used for heatingrather than cooling In this case the coils where the refrigerant would condense in the air conditioner become the evaporator coils when the unit isoperated as a heat pump and vice versa the evaporator coils of the air conditioner become the condenser coils in the heat pump Suppose youoperate 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



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 operation cycle remains the same In the heating mode then thesame 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 2 Find the right expresssion for the first law of thermodynamics

Suppose you have an ideal sys tem 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 1 Cyclic processes

Remember that the net internal energy change in a cyclic process is zero since the system has the same temperature when it returns tothe starting point

983149

=983149

= 515times105 Q

c

J

J W

W

= 120times105 W J

Q

h

Q

c

Q

h

Q

c

Q

h

Q

h

Q

c

Q

h

W

i n



ANSWER

ANSWER

All attempts used correct answer withheld by instructor

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 cy clical that is they must return to their original state some time aft er having absorbed some heat and done somework)

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

If a heat engine is run backward (ie with every input and output reversed) it becomes a heat pump (as pictured schematic ally ) Work must be puinto 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 heatpump)

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 theratio 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 workput in

In a heat pump the coeffic ient 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 sys tem

Express in terms of and other quantities given in the introduction

Hint 1 Note the differences in wording

Recall that is the work done by the syst em is the work done on the system

ANSWER

Correct

= minus Q

c

W

i n

Q

h

= minus Q

h

Q

c

W

i n

= + Q

h

Q

c

W

i n

=Q

h

4

J

W

i n

T

c

T

h

Q

h

Q

c

K Q

c

= K

f r i g

Q

c

W

i n

Q

h

= K

p u m p

Q

h

W

i n

Q

h

Q

c

W

i n

W

W

i n

W

W W

i n

=W

i n

minus W



Part B

Find the heat pumped out by the ideal heat pump

Express in terms of and

Hint 1 Conservation of energy and the first law

Apply conservation of energy If you think in t erms of the first law of thermodynamics remember the sign conventions for heat and work andnote that the internal energy does not change in an engine after one cycle

ANSWER

Correct

Part C

A heat pump is used t o heat a house in winter t he inside radiators are at and the outside heat ex changer is at If it is a perfect (ie Carnotcyc le) heat pump what is its coeffic ient of performance

Give your answer in terms of and

Hint 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 sys tem and the outside heats and


ANSWER

Hint 2 Relation between and in a Carnot cycle

Recall that in a Carnot cycle

ANSWER

Correct

Part D

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 sideshold true Indicate the statement that must apply for the heat pump to work

ANSWER

Q

h

Q

h

Q

c

W

i n

=Q

h

+ W

i n

Q

c

T

h

T

c

K

p u m p

T

h

T

c

Q

h

Q

c

K

p u m p

W

i n

Q

h

Q

c

Q

h

Q

c

=K

p u m p

Q

h

Q

c

=

Q

h

Q

c

T

h

T

c

=K

p u m p

T

983144

minus T

983144

T

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c



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

Express the cost in dollars

Hint 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 theheat generation the less work needs to be done and the lower the heating billYou 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 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

ANSWER

Correct

This savings is accompanied by more initial capital costs both for the heat pump and for the generous area of baseboard heaters needed totransfer enough heat to the house without raising which would reduce the coeffic ient of performance An additional problem is icing of theoutside heat exchanger which is very difficult to avoid if the outside air is humid and not much above zero degrees Celsius Therefore heatpumps are most useful in temperate climates or where the heat can be obtained from a groundwater that is abundant or flowing (eg an

underground stream)

Six New Heat Engines Conceptual Question

As part of your job at the patent office you are asked to evaluate t he s ix designs s hown in the figure for innovativ e new heat engines

Part A

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

Give the letter(s) of the design(s) in alphabetical order without commas or spaces (eg ACD)

= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c



Hint 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

ANSWER

Correct

Part B

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

Give the letter(s) of the design(s) in alphabetical order without commas or spaces (eg ABD)

Hint 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 effic iencygiven by

An engine with thermal efficiency greater than t hat given by the above equation v iolates the second law of thermodynamics

ANSWER

Correct

Part C

Which of the remaining designs has the highest thermal efficiency

ANSWER

Correct

Carnot Cycle

After Count Rumford (Benjamin Thompson) and James Prescott Joule had shown t he equivalence of mechanical energy and heat it was natural that

engineers believed it possible to make a heat engine (eg a steam engine) that would convert heat completely into mechanical energy Sadi Carnotconsidered a hypothetical piston engine that contained moles of an ideal gas showing first that it was reversible and most importantly thatmdashregardlesof the specific heat of the gasmdashit had limited effic iency 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 hea

reservoir at a lower temperature

The cycle associated with a Carnot engine is known as a Carnot cycle A pV plot of the Carnot cycle is shown in the figure The working gas first expandisothermally 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 i

original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A

Which of the following statements are true

Check all that apply

Hint 1 Heat flow in an adiabatic process

An adiabatic process is one in which heat does not flow into or out of the gas

ANSWER

Correct

Part B

Find the total work done by the gas after it completes a single Carnot cycle

Express the work in terms of any or all of the quantities and


Find the total amount of heat added during the entire cycle and the change in internal energy of the gas over the entire cycle Then apply thefirst law of thermodynamics

Hint 2 Compute the change in internal energy

What is the net change in the gass internal energy after one complete cycle

ANSWER

ANSWER

Correct

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 tostate A

W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



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 absorb

by the gas as it expands from state A to state B

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

Hint 1 General method of finding

First find the work done by the gas as it expands from state A to state B Then use the first law of thermodynamics to relate to

both and the change in the gass internal energy Finally recall that this process is isothermal what does this tell you about the change inthe gass internal energy

Hint 2 Find the work done by the gas

What is the work done by the gas as it expands from state A to state B

Express the work in terms of the temperature of the hot reservoir and the gas constant

Hint 1 How to find the work done by the gas

To find the net work done by the gas as it proceeds from state A to state B you need to integrate from state A tostate B

Hint 2 Express in terms of

Use the ideal gas equation of state to find an expression for the pressure in terms of and

ANSWER

Hint 3 Express the integral over

Given that and that

what is the integral

Express your answer in terms of and

ANSWER

ANSWER

Hint 3 Relation between and

Because the process is isothermal the internal energy of the gas does not change Therefore the work done by the gas will equal the net heatflow into the gas

ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



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 proceed

from state C to state D

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


First find the work done by the gas as it expands from state C to state D Then use the first law of thermodynamics to relate toboth and the change in the gass internal energy Finally recall that this process is isothermal what does this tell you about the change in

the gass internal energy

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 thepressure 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


Suppose the ratio of the gass specific heats is denoted by Along adiabatic curves you know that where is some

constant Rewrite this equation in terms of and instead of and Then use this new equation to relate the temperature and volume atthe end points of the two adiabatic legs of the Carnot cyc le This will give you two equations that you can solve for

Hint 2 Rewrite in terms of and

Use the ideal gas equation of state to eliminate from the expression

Express your answer in terms of and the temperature

ANSWER

Hint 3 Express and in terms of and

States B and C are connected by an adiabatic expansion Use the result found in the previous hint to find an expression for


ANSWER


States D and A are connected by an adiabatic expansion Use the result found in Hint 2 to find an expression for


ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A

= minus Δ = minus 983150 ( minus ) W

B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ



Hint 5 Solving for in terms of and

Combine the results of the previous two hints eliminating the temperatures and (One way to do this is to divide one equation by theother) This should allow you to solve for in terms of and

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 (eg or )

ANSWER

Correct

Part G

The efficiency of any engine is by definition Carnot proved that no engine can have an effic iency greater than that of a Carnot engine

Find the effic iency of a Carnot engine

Express the efficiency in terms of and

Hint 1 Express the efficiency in terms of and

Using your result from Part B find the efficiency (of any engine) in terms of the engines heat input and output


ANSWER

ANSWER

Correct

Because is generally fixed (eg the cold reservoir for power plants is often a river or a lake) engineers trying to increase effic iency havealways sought to raise the upper temperature This explains why (historically ) there were some spectacular explosions of boilers used for steam power

Heat Engines Introduced

Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h



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 andmake the lid jump up and down why cant heat be made to do useful work

A heat engine is a devic e designed to convert heat into work The heat engines we will study will be c yclic The working s ubstance eventually returns to ioriginal state sometime after having absorbed a quantity of heat and done some work A cyclic heat engine cannot convert heat into work withoutgenerating some waste heat in the process Although by no means intuitively obvious this is an important fact of nature since it dramatically affects thetechnology 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 aremore than 100 efficient

Consequently the impossible heat engine pictured schematically here cannot exist even intheory 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 heatengine one that obeys the laws of thermodynamics It takes in heat

at a temperature and does work In the process of doing this it generateswaste heat at a cooler temperature

Take and to be the magnitudes

of the heat absorbed and emittedrespectively therefore both quantitiesare positive

Part A

A heat engine is designed to do work This is possible only if certain relationships between t he heats and temperatures at the input and output holdtrue Which of the following sets of statements must apply for the heat engine to do work

ANSWER

Correct

Part B

Find the work done by the ideal heat engine


ANSWER

Correct

Part C

The thermal efficiency texttipee of a heat engine is defined as follows e = WQ_rm h

Express the efficiency in terms of texttipQ_rm hQ_h and texttipQ_rm cQ_c

ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910

The cycle of the figure consists of three processes

Part A

Make a chart with rows labeled A - C and columns labeled Delta E_rm th W_rm s and Q Fill each box in the chart with + - or 0 to indicatewhether the quantity increases decreases or stays the same during that process

ANSWER

Correct

Problem 1963

A heat engine with 0100 rm mol of a monatomic ideal gas initially fills a 3000 rm cm^3 cylinder at 600 rm K The gas goes through the followinclosed cycle

texttipee = largefracQ_h-Q_cQ_h



- Isothermal expansion to 4000 rm cm^3- Isochoric cooling to 400 rm K - Isothermal compression to 3000 rm cm^3- Isochoric heating to 600 rm K

Part A

How much work does this engine do per cycle

Express your answer with the appropriate units

ANSWER

Completed correct answer withheld by instructor

Part B

What is its thermal efficiency


ANSWER

Score Summary

Your score on this assignment is 815You received 733 out of a possible total of 9 points



room


Hint 1 Temperature and Conservation of energy

A refrigerator uses electrical energy to remove heat from its interior and expel it into the environment B y conservation of energy the energyexpelled into the room must be the sum of the energy extracted from the interior of the refrigerator and the energy input to run the refrigeratorNotice that the rate at which the temperature of the room rises is directly proportional to the rate at which energy is expelled into the room

ANSWER

Correct

plusmn PSS 191 Heat-Engine Problems

Learning Goal

To practice Problem-Solving Strategy 191 for heat-engine problems

A heat engine uses the closed cyc le s hown in the diagram below The working substance ismoles of monatomic ideal gas Find the efficiency of such a cycle Use the values for pressureand volume shown in the diagram and assume that the process between points 1 and 3 isisothermal

PROBLEM-SOLVING STRAT EGY 191 Heat-engine problems

MODEL Identify each process in the cycle

983150

η










Model

Part A


ANSWER

Correct

Visualize


Solve

Part B









by


983150 983152 V T

983152 V T

Q W

s

Δ E

t h

W

o u t

W

s

Q Q

H

( Δ = 0 E

t h

)

n e t

η

( Δ = 0 E

t h

)

n e t

W

s

Q η





η

η W

o u t

Q

H

η = W

o u t

Q

H

W

o u t

W

o u t

W

s

W

o u t

W

s

A 983142 ( 983160 ) 983143 ( 983160 )

a b

A = ( 983142 minus 983143 ) 983140 983160 int

b

a

3 rarr 1 983142










ANSWER






ANSWER









ANSWER



123



1 rarr 2 983143

( c ) 1 0

2

m

3

1 rarr 2

W

1 2

1 rarr 2

983152 Δ V

= 983152 Δ V W

983155

=W

1 2

J

2 rarr 3

W

2 3

2 rarr 3

=W

2 3

J

3 rarr 1

W

3 1

3 rarr 1

T V

i

V

f

= 983150 R T l n ( ) W

983155

V

f

V

i

983150 R

983150 T 3 rarr 1

983150 R T

=W

3 1

J

1 rarr 2

2 rarr 3

3 rarr 1



ANSWER












ANSWER






ANSWER

ANSWER

Correct


Assess

Part C

2 rarr 3

Q

2 3

2 rarr 3

Δ T

Q = 983150 Δ T C

V

983150 C

V

= C

V

3 R

2

R

T

983152 V =

983150 R T

Δ T = minus T

3

T

2

983150

=Q

2 3

J

3 rarr 1

Q

3 1

3 rarr 1

Δ E

t h

Q = W

s

=Q

3 1

J

= 0206η




Process (J) (J) (J)







ANSWER

Correct






Part A






W

s

Q Δ E

t h


2 rarr 3 0 1 8 0 1 8 0

3 rarr 1 1 9 8 1 9 8 0

Q

1 2

1 rarr 2 ( Δ = 0 E

t h

)

n e t

Q

1 2

Q = 983150 Δ T C

983152

C

p

Δ = Q minus E

t h

W

s

1 rarr 2

Δ E

t h

Δ = 0 E

t h

= -300Q

1 2

J

= + + = 7 8 J Q

n e t

Q

1 2

Q

2 3

Q

3 1

W

o u t

Δ = 0 0 J E

t h

k g

Q

c

J k g






ANSWER

ANSWER

Correct

Part B





ANSWER

Correct

Part C













983149

=983149

= 515times105 Q

c

J

J W

W

= 120times105 W J

Q

h

Q

c

Q

h

Q

c

Q

h

Q

h

Q

c

Q

h

W

i n



ANSWER

ANSWER



Learning Goal












work put in


Part A





ANSWER

Correct

= minus Q

c

W

i n

Q

h

= minus Q

h

Q

c

W

i n

= + Q

h

Q

c

W

i n

=Q

h

4

J

W

i n

T

c

T

h

Q

h

Q

c

K Q

c

= K

f r i g

Q

c

W

i n

Q

h

= K

p u m p

Q

h

W

i n

Q

h

Q

c

W

i n

W

W

i n

W

W W

i n

=W

i n

minus W



Part B





ANSWER

Correct

Part C







ANSWER



ANSWER

Correct

Part D


ANSWER

Q

h

Q

h

Q

c

W

i n

=Q

h

+ W

i n

Q

c

T

h

T

c

K

p u m p

T

h

T

c

Q

h

Q

c

K

p u m p

W

i n

Q

h

Q

c

Q

h

Q

c

=K

p u m p

Q

h

Q

c

=

Q

h

Q

c

T

h

T

c

=K

p u m p

T

983144

minus T

983144

T

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c



Correct

Part E








Hint 2 Units of and


ANSWER

Correct


underground stream)



Part A



= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c





ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary











Model

Part A


ANSWER

Correct

Visualize


Solve

Part B









by


983150 983152 V T

983152 V T

Q W

s

Δ E

t h

W

o u t

W

s

Q Q

H

( Δ = 0 E

t h

)

n e t

η

( Δ = 0 E

t h

)

n e t

W

s

Q η





η

η W

o u t

Q

H

η = W

o u t

Q

H

W

o u t

W

o u t

W

s

W

o u t

W

s

A 983142 ( 983160 ) 983143 ( 983160 )

a b

A = ( 983142 minus 983143 ) 983140 983160 int

b

a

3 rarr 1 983142










ANSWER






ANSWER









ANSWER



123



1 rarr 2 983143

( c ) 1 0

2

m

3

1 rarr 2

W

1 2

1 rarr 2

983152 Δ V

= 983152 Δ V W

983155

=W

1 2

J

2 rarr 3

W

2 3

2 rarr 3

=W

2 3

J

3 rarr 1

W

3 1

3 rarr 1

T V

i

V

f

= 983150 R T l n ( ) W

983155

V

f

V

i

983150 R

983150 T 3 rarr 1

983150 R T

=W

3 1

J

1 rarr 2

2 rarr 3

3 rarr 1



ANSWER












ANSWER






ANSWER

ANSWER

Correct


Assess

Part C

2 rarr 3

Q

2 3

2 rarr 3

Δ T

Q = 983150 Δ T C

V

983150 C

V

= C

V

3 R

2

R

T

983152 V =

983150 R T

Δ T = minus T

3

T

2

983150

=Q

2 3

J

3 rarr 1

Q

3 1

3 rarr 1

Δ E

t h

Q = W

s

=Q

3 1

J

= 0206η




Process (J) (J) (J)







ANSWER

Correct






Part A






W

s

Q Δ E

t h


2 rarr 3 0 1 8 0 1 8 0

3 rarr 1 1 9 8 1 9 8 0

Q

1 2

1 rarr 2 ( Δ = 0 E

t h

)

n e t

Q

1 2

Q = 983150 Δ T C

983152

C

p

Δ = Q minus E

t h

W

s

1 rarr 2

Δ E

t h

Δ = 0 E

t h

= -300Q

1 2

J

= + + = 7 8 J Q

n e t

Q

1 2

Q

2 3

Q

3 1

W

o u t

Δ = 0 0 J E

t h

k g

Q

c

J k g






ANSWER

ANSWER

Correct

Part B





ANSWER

Correct

Part C













983149

=983149

= 515times105 Q

c

J

J W

W

= 120times105 W J

Q

h

Q

c

Q

h

Q

c

Q

h

Q

h

Q

c

Q

h

W

i n



ANSWER

ANSWER



Learning Goal












work put in


Part A





ANSWER

Correct

= minus Q

c

W

i n

Q

h

= minus Q

h

Q

c

W

i n

= + Q

h

Q

c

W

i n

=Q

h

4

J

W

i n

T

c

T

h

Q

h

Q

c

K Q

c

= K

f r i g

Q

c

W

i n

Q

h

= K

p u m p

Q

h

W

i n

Q

h

Q

c

W

i n

W

W

i n

W

W W

i n

=W

i n

minus W



Part B





ANSWER

Correct

Part C







ANSWER



ANSWER

Correct

Part D


ANSWER

Q

h

Q

h

Q

c

W

i n

=Q

h

+ W

i n

Q

c

T

h

T

c

K

p u m p

T

h

T

c

Q

h

Q

c

K

p u m p

W

i n

Q

h

Q

c

Q

h

Q

c

=K

p u m p

Q

h

Q

c

=

Q

h

Q

c

T

h

T

c

=K

p u m p

T

983144

minus T

983144

T

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c



Correct

Part E








Hint 2 Units of and


ANSWER

Correct


underground stream)



Part A



= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c





ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary











ANSWER






ANSWER









ANSWER



123



1 rarr 2 983143

( c ) 1 0

2

m

3

1 rarr 2

W

1 2

1 rarr 2

983152 Δ V

= 983152 Δ V W

983155

=W

1 2

J

2 rarr 3

W

2 3

2 rarr 3

=W

2 3

J

3 rarr 1

W

3 1

3 rarr 1

T V

i

V

f

= 983150 R T l n ( ) W

983155

V

f

V

i

983150 R

983150 T 3 rarr 1

983150 R T

=W

3 1

J

1 rarr 2

2 rarr 3

3 rarr 1



ANSWER












ANSWER






ANSWER

ANSWER

Correct


Assess

Part C

2 rarr 3

Q

2 3

2 rarr 3

Δ T

Q = 983150 Δ T C

V

983150 C

V

= C

V

3 R

2

R

T

983152 V =

983150 R T

Δ T = minus T

3

T

2

983150

=Q

2 3

J

3 rarr 1

Q

3 1

3 rarr 1

Δ E

t h

Q = W

s

=Q

3 1

J

= 0206η




Process (J) (J) (J)







ANSWER

Correct






Part A






W

s

Q Δ E

t h


2 rarr 3 0 1 8 0 1 8 0

3 rarr 1 1 9 8 1 9 8 0

Q

1 2

1 rarr 2 ( Δ = 0 E

t h

)

n e t

Q

1 2

Q = 983150 Δ T C

983152

C

p

Δ = Q minus E

t h

W

s

1 rarr 2

Δ E

t h

Δ = 0 E

t h

= -300Q

1 2

J

= + + = 7 8 J Q

n e t

Q

1 2

Q

2 3

Q

3 1

W

o u t

Δ = 0 0 J E

t h

k g

Q

c

J k g






ANSWER

ANSWER

Correct

Part B





ANSWER

Correct

Part C













983149

=983149

= 515times105 Q

c

J

J W

W

= 120times105 W J

Q

h

Q

c

Q

h

Q

c

Q

h

Q

h

Q

c

Q

h

W

i n



ANSWER

ANSWER



Learning Goal












work put in


Part A





ANSWER

Correct

= minus Q

c

W

i n

Q

h

= minus Q

h

Q

c

W

i n

= + Q

h

Q

c

W

i n

=Q

h

4

J

W

i n

T

c

T

h

Q

h

Q

c

K Q

c

= K

f r i g

Q

c

W

i n

Q

h

= K

p u m p

Q

h

W

i n

Q

h

Q

c

W

i n

W

W

i n

W

W W

i n

=W

i n

minus W



Part B





ANSWER

Correct

Part C







ANSWER



ANSWER

Correct

Part D


ANSWER

Q

h

Q

h

Q

c

W

i n

=Q

h

+ W

i n

Q

c

T

h

T

c

K

p u m p

T

h

T

c

Q

h

Q

c

K

p u m p

W

i n

Q

h

Q

c

Q

h

Q

c

=K

p u m p

Q

h

Q

c

=

Q

h

Q

c

T

h

T

c

=K

p u m p

T

983144

minus T

983144

T

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c



Correct

Part E








Hint 2 Units of and


ANSWER

Correct


underground stream)



Part A



= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c





ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary




ANSWER












ANSWER






ANSWER

ANSWER

Correct


Assess

Part C

2 rarr 3

Q

2 3

2 rarr 3

Δ T

Q = 983150 Δ T C

V

983150 C

V

= C

V

3 R

2

R

T

983152 V =

983150 R T

Δ T = minus T

3

T

2

983150

=Q

2 3

J

3 rarr 1

Q

3 1

3 rarr 1

Δ E

t h

Q = W

s

=Q

3 1

J

= 0206η




Process (J) (J) (J)







ANSWER

Correct






Part A






W

s

Q Δ E

t h


2 rarr 3 0 1 8 0 1 8 0

3 rarr 1 1 9 8 1 9 8 0

Q

1 2

1 rarr 2 ( Δ = 0 E

t h

)

n e t

Q

1 2

Q = 983150 Δ T C

983152

C

p

Δ = Q minus E

t h

W

s

1 rarr 2

Δ E

t h

Δ = 0 E

t h

= -300Q

1 2

J

= + + = 7 8 J Q

n e t

Q

1 2

Q

2 3

Q

3 1

W

o u t

Δ = 0 0 J E

t h

k g

Q

c

J k g






ANSWER

ANSWER

Correct

Part B





ANSWER

Correct

Part C













983149

=983149

= 515times105 Q

c

J

J W

W

= 120times105 W J

Q

h

Q

c

Q

h

Q

c

Q

h

Q

h

Q

c

Q

h

W

i n



ANSWER

ANSWER



Learning Goal












work put in


Part A





ANSWER

Correct

= minus Q

c

W

i n

Q

h

= minus Q

h

Q

c

W

i n

= + Q

h

Q

c

W

i n

=Q

h

4

J

W

i n

T

c

T

h

Q

h

Q

c

K Q

c

= K

f r i g

Q

c

W

i n

Q

h

= K

p u m p

Q

h

W

i n

Q

h

Q

c

W

i n

W

W

i n

W

W W

i n

=W

i n

minus W



Part B





ANSWER

Correct

Part C







ANSWER



ANSWER

Correct

Part D


ANSWER

Q

h

Q

h

Q

c

W

i n

=Q

h

+ W

i n

Q

c

T

h

T

c

K

p u m p

T

h

T

c

Q

h

Q

c

K

p u m p

W

i n

Q

h

Q

c

Q

h

Q

c

=K

p u m p

Q

h

Q

c

=

Q

h

Q

c

T

h

T

c

=K

p u m p

T

983144

minus T

983144

T

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c



Correct

Part E








Hint 2 Units of and


ANSWER

Correct


underground stream)



Part A



= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c





ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary





Process (J) (J) (J)







ANSWER

Correct






Part A






W

s

Q Δ E

t h


2 rarr 3 0 1 8 0 1 8 0

3 rarr 1 1 9 8 1 9 8 0

Q

1 2

1 rarr 2 ( Δ = 0 E

t h

)

n e t

Q

1 2

Q = 983150 Δ T C

983152

C

p

Δ = Q minus E

t h

W

s

1 rarr 2

Δ E

t h

Δ = 0 E

t h

= -300Q

1 2

J

= + + = 7 8 J Q

n e t

Q

1 2

Q

2 3

Q

3 1

W

o u t

Δ = 0 0 J E

t h

k g

Q

c

J k g






ANSWER

ANSWER

Correct

Part B





ANSWER

Correct

Part C













983149

=983149

= 515times105 Q

c

J

J W

W

= 120times105 W J

Q

h

Q

c

Q

h

Q

c

Q

h

Q

h

Q

c

Q

h

W

i n



ANSWER

ANSWER



Learning Goal












work put in


Part A





ANSWER

Correct

= minus Q

c

W

i n

Q

h

= minus Q

h

Q

c

W

i n

= + Q

h

Q

c

W

i n

=Q

h

4

J

W

i n

T

c

T

h

Q

h

Q

c

K Q

c

= K

f r i g

Q

c

W

i n

Q

h

= K

p u m p

Q

h

W

i n

Q

h

Q

c

W

i n

W

W

i n

W

W W

i n

=W

i n

minus W



Part B





ANSWER

Correct

Part C







ANSWER



ANSWER

Correct

Part D


ANSWER

Q

h

Q

h

Q

c

W

i n

=Q

h

+ W

i n

Q

c

T

h

T

c

K

p u m p

T

h

T

c

Q

h

Q

c

K

p u m p

W

i n

Q

h

Q

c

Q

h

Q

c

=K

p u m p

Q

h

Q

c

=

Q

h

Q

c

T

h

T

c

=K

p u m p

T

983144

minus T

983144

T

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c



Correct

Part E








Hint 2 Units of and


ANSWER

Correct


underground stream)



Part A



= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c





ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary







ANSWER

ANSWER

Correct

Part B





ANSWER

Correct

Part C













983149

=983149

= 515times105 Q

c

J

J W

W

= 120times105 W J

Q

h

Q

c

Q

h

Q

c

Q

h

Q

h

Q

c

Q

h

W

i n



ANSWER

ANSWER



Learning Goal












work put in


Part A





ANSWER

Correct

= minus Q

c

W

i n

Q

h

= minus Q

h

Q

c

W

i n

= + Q

h

Q

c

W

i n

=Q

h

4

J

W

i n

T

c

T

h

Q

h

Q

c

K Q

c

= K

f r i g

Q

c

W

i n

Q

h

= K

p u m p

Q

h

W

i n

Q

h

Q

c

W

i n

W

W

i n

W

W W

i n

=W

i n

minus W



Part B





ANSWER

Correct

Part C







ANSWER



ANSWER

Correct

Part D


ANSWER

Q

h

Q

h

Q

c

W

i n

=Q

h

+ W

i n

Q

c

T

h

T

c

K

p u m p

T

h

T

c

Q

h

Q

c

K

p u m p

W

i n

Q

h

Q

c

Q

h

Q

c

=K

p u m p

Q

h

Q

c

=

Q

h

Q

c

T

h

T

c

=K

p u m p

T

983144

minus T

983144

T

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c



Correct

Part E








Hint 2 Units of and


ANSWER

Correct


underground stream)



Part A



= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c





ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary




ANSWER

ANSWER



Learning Goal












work put in


Part A





ANSWER

Correct

= minus Q

c

W

i n

Q

h

= minus Q

h

Q

c

W

i n

= + Q

h

Q

c

W

i n

=Q

h

4

J

W

i n

T

c

T

h

Q

h

Q

c

K Q

c

= K

f r i g

Q

c

W

i n

Q

h

= K

p u m p

Q

h

W

i n

Q

h

Q

c

W

i n

W

W

i n

W

W W

i n

=W

i n

minus W



Part B





ANSWER

Correct

Part C







ANSWER



ANSWER

Correct

Part D


ANSWER

Q

h

Q

h

Q

c

W

i n

=Q

h

+ W

i n

Q

c

T

h

T

c

K

p u m p

T

h

T

c

Q

h

Q

c

K

p u m p

W

i n

Q

h

Q

c

Q

h

Q

c

=K

p u m p

Q

h

Q

c

=

Q

h

Q

c

T

h

T

c

=K

p u m p

T

983144

minus T

983144

T

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c



Correct

Part E








Hint 2 Units of and


ANSWER

Correct


underground stream)



Part A



= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c





ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary




Part B





ANSWER

Correct

Part C







ANSWER



ANSWER

Correct

Part D


ANSWER

Q

h

Q

h

Q

c

W

i n

=Q

h

+ W

i n

Q

c

T

h

T

c

K

p u m p

T

h

T

c

Q

h

Q

c

K

p u m p

W

i n

Q

h

Q

c

Q

h

Q

c

=K

p u m p

Q

h

Q

c

=

Q

h

Q

c

T

h

T

c

=K

p u m p

T

983144

minus T

983144

T

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c



Correct

Part E








Hint 2 Units of and


ANSWER

Correct


underground stream)



Part A



= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c





ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary




Correct

Part E








Hint 2 Units of and


ANSWER

Correct


underground stream)



Part A



= C T

c

2

∘

= C T

h

4 7

∘

Q

h

K

p u m p

= 7 5 K

a c t u a l

o f = K

p u m p

3

4

T

h

minus T

h

T

c

Q

h

W

i n

W

i n

T

h

T

c

C = 2 7 3 K 0

∘

Cost = 1875 dollars

T

h

Q

c





ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary






ANSWER

Correct

Part B






ANSWER

Correct

Part C


ANSWER

Correct

Carnot Cycle







original state

CF

T

h o t

T

c o l d

983141

m a x

= 1 minus 983141

m a x

T

c o l d

T

h o t

BD

device A

device E

983150

983141 = W Q

h

W Q

h

T

h

Q

c

T

c

Q

h

T

h

T

c

Q

c



Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary




Part A





ANSWER

Correct

Part B







ANSWER

ANSWER

Correct





W

Q

h

T

h

Q

c

T

c

983140 Q =

983140 U +

983140 W

U

change in =U

=W

minus Q

983144

Q

c



Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary




Part C














ANSWER


Given that and that



ANSWER

ANSWER



ANSWER

Correct

983150 V

A

V

B

Q

h

983150 V

A

V

B

T

h

R

Q

h

W

A B

W

A B

Q

h

W

A B

983150 V

A

V

B

T

h

R

W

A B

983140 W =

983152 983140 V

983152 V

983152 983150 R T

h

V

= 983152 ( V )

983140 V

983152 ( V ) prop 1 V

= 983152 ( V ) 983140 V

W

A B

int

V

B

V

A

983140 V V int

V

B

V

A

V

A

V

B

=int

V

B

V

A

983140 V

V

=W

A B

Q

h

W

A B

= Q

h

W

A B

=Q

h

983150 R

( l n

( 983081 983081

T

983144

V

B

V

A



Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary




Part D







ANSWER

Correct



Part E








ANSWER




ANSWER




ANSWER

V

C

V

D

Q

c

983150 V

C

V

D

T

c

R

W

C D

W

C D

Q

c

=Q

c

983150 R

( l n

( 983081 983081

T

c

V

C

V

D

+ = 0 W

B C

W

D A


B C

U

B C

C

V

T

h

T

c


D A

U

D A

C

V

T

c

T

h

= minus W

B C

W

D A

+ = 0 W

B C

W

D A

V

C

V

D

V

A

V

B

γ = C

983152

C

983158

983152 = C V

γ

C

T V 983152 V

V

C

V

D

983152 V

γ

T V

983152 983152 V

γ

γ 983150 R T

= 983152 V

γ

T

h

V

B

T

c

V

C

T

h

V

( γ minus 1 )

B

T

c

V

C

γ

=T

h

V

( γ minus 1 )

B

T

h

V

A

T

c

V

D

T

h

V

( γ minus 1 )

A

T

c

V

D

γ





ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary






ANSWER

Correct

Part F



ANSWER

Correct

Part G







ANSWER

ANSWER

Correct



Learning Goal

=T

h

V

( γ minus 1 )

A

V

C

V

D

V

A

V

B

T

h

T

c

V

C

V

D

V

A

V

B

= V

C

V

D

V

B

V

A

Q

h

Q

c

Q

c

Q

h

983150 R

= Q

c

Q

h

T

c

T

983144

983141 = W Q

h

983141

C a r n o t

T

h

T

c

Q

h

Q

c

983141

Q

h

Q

c

=983141

=983141

C a r n o t

minus T

983144

T

c

T

983144

T

c

T

h












Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary













Part A


ANSWER

Correct

Part B



ANSWER

Correct

Part C



ANSWER

Q

h

T

h

W

Q

c

T

c

Q

h

Q

c

and

and

and

and

lt Q

h

Q

c

lt T

h

T

c

gt Q

h

Q

c

lt T

h

T

c

lt Q

h

Q

c

gt T

h

T

c

gt Q

h

Q

c

gt T

h

T

c

W

W Q

h

Q

c

= Q_h-Q_cW



Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary




Correct

Problem 1910


Part A


ANSWER

Correct

Problem 1963






Part A



ANSWER


Part B



ANSWER

Score Summary





Part A



ANSWER


Part B



ANSWER

Score Summary


Mastering Physics HW 4 Ch 18, 19

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Transcript of Mastering Physics HW 4 Ch 18, 19