HW 4 Ch 18, 19 Due: 11:59pm 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 P needed for it to operate and the maximum heat energy that can be removed per second Q C,max /Δt 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 device that uses work to remove heat energy from a cold 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 coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input.
Hint 2. Definition of the performance coefficient The performance coefficient k of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir Q C to the work W input to the refrigerator: k=
Recall that power is defined as work per unit time.
QC W
.
ANSWER:
Typesetting math: 95%
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 the
room. Rank from largest to smallest. To rank items as equivalent, overlap them.
Hint 1. Temperature and Conservation of energy A refrigerator uses electrical energy to remove heat from its interior and expel it into the environment. By conservation of energy, the energy expelled into the room must be the sum of the energy extracted from the interior of the refrigerator and the energy input to run the refrigerator. Notice 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
± PSS 19.1 HeatEngine Problems Learning Goal: To practice ProblemSolving Strategy 19.1 for heatengine problems. A heat engine uses the closed cycle shown in the diagram below. The working substance is n moles of monatomic ideal gas. Find the efficiency η of such a cycle. Use the values for pressure and volume shown in the diagram, and assume that the process between points 1 and 3 is isothermal.
PROBLEMSOLVING STRATEGY 19.1 Heatengine problems MODEL: Identify each process in the cycle.
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 n , p, V , and T at one point in the cycle. Use the ideal gas law and equations for specific gas processes to determine p, V , and T at the beginning and end of each process. Calculate Q , W s , and ΔEth for each process. Find W out by adding W s for each process in the cycle. If the geometry is simple, you can confirm this value by finding the area enclosed within the pV curve. Add just the positive values of Q to find Q H . Verify that (ΔEth ) net = 0. This is a selfconsistency check to verify that you haven't made any mistakes. Calculate the thermal efficiency η and any other quantities you need to complete the solution. ASSESS: Is (ΔEth ) net
= 0
? Do all the signs of W s and Q make sense? Does η 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 volume between 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 and 3, respectively? ANSWER: isochoric and isobaric isobaric and isochoric isobaric and isothermal isothermal and isochoric
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 efficiency η 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 η of a heat engine is given by the ratio of the work done by the engine (work output, W out ) to the energy transferred into the engine (heat input, Q H ): η = W out /Q
H
.
Hint 2. How to find the work done in one cycle There are two ways to compute the net work W out done by the engine during one full cycle. Using a geometrical approach, calculate the area enclosed by the pV curve for the cycle. W out is equal to the value of this area. Alternatively, identify all the processes in one full cycle and compute the work W s done by the engine in each process. Find W out by adding W s 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 A between the graphs of two functions f (x) and g(x) over a certain interval [a , b] is given by b
(f − g) dx . For this particular problem, take the function whose graph is the curve for the process 3→1 as the function f , and the function whose graph is A= ∫
a
the curve for the process 1→2 as g . Use the ideal gas law to find the correct mathematical expressions for these functions. Perform the integration over the V interval [3, 9] (10 2 cm3 ).
Hint 4. Second method: Find the work done during process 1→2 How much work W 12 is done by the gas during process 1→2 ? Express your answer in joules.
Hint 1. Work in isobaric processes In an isobaric process at pressure p, the work done by the system in which the volume changes by an amount ΔV is given by W s = pΔV
.
ANSWER: W 12
=
J
Hint 5. Find the work done during process 2→3 How much work W 23 is done by the gas during process 2→3 ? Express your answer in joules.
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: W 23
=
J
Hint 6. Find the work done during process 3→1 How much work W 31 is done by the gas during process 3→1 ? Express your answer in joules.
Hint 1. Work in isothermal processes In an isothermal process at temperature T , the work done by the system when the volume changes from Vi to Vf is given by W s = nRT ln(Vf /Vi )
,
where n is the number of moles in the gas and R is the gas constant.
Hint 2. How to solve this problem when moles and temperature are not given Leave n (the number of moles of gas) and T (the temperature of the gas during the process 3→1 ) as symbols in your calculation. You will find in the end that you need a numerical value for the product nRT . Use the ideal gas law to compute this product in terms of the pressure and volume at either point 1 or point 3. ANSWER: W 31
=
J
Hint 7. Find which processes contribute to the heat input During which processes in the cycle is heat delivered to the engine? 1. 1→2 2. 2→3 3. 3→1 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.
ANSWER:
Hint 8. Compute the heat input during process 2→3 What amount of heat Q 23 is delivered to the engine during process 2→3 ? Express your answer in joules.
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 ΔT is Q = nCV ΔT , where n is the number of moles in the gas, and CV is the heat capacity at constant volume.
Hint 2. The properties of monatomic gases For monatomic gases, the heat capacity at constant volume is CV =
3R
where R is the gas constant.
2
,
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 T can be expressed in terms of some known quantities using the ideal gas law, pV = nRT . Thus, use this law to find the temperature at points 2 and 3 in the pV diagram, and then compute ΔT = T3 − T2 . Note that you won’t need to know how many moles are in the gas because when you calculate the amount of heat transferred in the process, the number of moles n will cancel out. ANSWER: Q
23
=
J
Hint 9. Compute the heat input during process 3→1 What amount of heat Q 31 is delivered to the engine during process 3→1 ? Express your answer in joules.
Hint 1. Heat transfer during isothermal processes Since in an isothermal process the thermal energy of the gas does not change, ΔEth is zero. From the first law of thermodynamics, it follows that the heat transfer during an isothermal process is equal to the work done by the system, or Q = W s . In other words, all of the heat input during an isothermal process is converted into work done by the gas. ANSWER: Q 31
=
J
ANSWER: η
= 0.206
Correct If you prefer to express the efficiency as a percentage, in this case you would say that the engine is 20.6% efficient.
Assess Part C
Here is a table of energy transfers for this problem, with some entries missing: Process
Ws
(J)
Q
(J)
ΔEth
(J)
1→2
−120
?
?
2→3
0
180
180
3→1
198
198
0
What must be the heat input Q 12 in process 1→2 to satisfy the condition that (ΔEth ) net
= 0
?
Express your answer in joules.
Hint 1. How to approach the problem To compute the heat transfer Q 12 occurring during the isobaric process, you can either use the generic expression for heat transfer in isobaric processes, Q = nCp ΔT , where Cp is the heat capacity at constant pressure, or you can apply directly the first law of thermodynamics, ΔEth = Q − W s , to the process 1→2 . To do that, you'll need to know the change in thermal energy of the gas during the isobaric process. Read from the table the changes in thermal energy during the isochoric and the isothermal processes, and then calculate ΔEth for the isobaric process using the fact that ΔEth = 0 for a complete cycle. ANSWER: Q
12
= 300 J
Correct You can easily verify that your results make sense. The net heat transfer per cycle is Q net = Q 12 + Q 23 + Q 31 = 78 J , which is equal to the work W out done by the engine in one cycle. By the first law of thermodynamics, it follows that ΔEth = 0.0 J , which is what we would expect for a cyclic process.
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 vaporliquid phase equilibrium. The compressor takes in lowpressure, lowtemperature vapor and compresses it adiabatically to highpressure, hightemperature 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 highpressure, high temperature 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 lowpressure, lowtemperature 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 kg 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 Q c is absorbed by the refrigerant while it is in the evaporator? The heat of vaporization of the refrigerant is 1.50×105 J/kg . Express your answer numerically in joules.
Hint 1. How to approach the problem 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 the evaporator, it is already at boiling temperature and all the absorbed heat is used in the liquidvapor phase change. Note that the pressure is kept constant in the evaporator. Also, a small fraction of refrigerant is still liquid when it leaves the evaporator, so the refrigerant must still be in phase equilibrium at that point and no change in temperature has occurred in the evaporator.
Hint 2. Find the percentage of refrigerant transformed to vapor Consider a refrigerator where, in each cycle, 54.0 % of refrigerant enters the evaporator as liquid in liquidvapor phase equilibrium and 95% leaves as vapor. Assume that the pressure of the refrigerant is kept constant while it is in the evaporator. What is the percentage m% of the total mass of refrigerant transformed to vapor in the evaporator per cycle? ANSWER: m%
=
%
ANSWER: Qc
= 5.15×105 J
Correct
Part B In each cycle, the change in internal energy of the refrigerant when it leaves the compresser is 1.20×105 J . What is the work W done by the motor of the compressor? Express your answer in joules.
Hint 1. Adiabatic compression Recall that the compressor performs an adiabatic compression; that is, no heat exchange occurs while the refrigerant is in the compressor. Then use the first law of thermodynamics to determine W . ANSWER: W
= 1.20×105 J
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 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 Q h 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. Express your answer numerically in joules.
Hint 1. How to approach the problem When the air conditioner is operated in the cooling mode, in each cycle it absorbs an amount of heat Q c from the air inside the room, as you calculated in Part A, and gives off a certain amount of heat Q h 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, the same quantity Q c is now absorbed from the air outside the room and the amount of heat Q h 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 Q h .
Hint 2. Find the right expresssion for the first law of thermodynamics Suppose you have an ideal system that absorbs heat Q c from a cold reservoir and releases heat Q h to a hot reservoir. The net input of work required by this system for operation is W in . 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 to the starting point.
ANSWER: Q Q Q
c
h
h
= W in − Q = Q = Q
h
c
− W in
c
+ W in
ANSWER: Qh
=
4
J
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 enginesdevices, 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. If a heat engine is run backward (i.e., with every input and output reversed), it becomes a heat pump (as pictured schematically ). Work W in must be put into a heat pump, and it then pumps heat from a colder temperature Tc to a hotter temperature Th , 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 Q h of a heat pump or the heat going in to the cold side Q c 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, K . In a refrigerator, this is the ratio of heat removed from the cold side Q c to work put in: K f rig =
Qc Win
.
In a heat pump the coefficient of performance is the ratio of heat exiting the hot side Q h to the work put in: K pump =
Qh Win
.
Take Q h , and Q c to be the magnitudes of the heat emitted and absorbed respectively.
Part A What is the relationship of W in to the work W done by the system? Express W in in terms of W and other quantities given in the introduction.
Hint 1. Note the differences in wording Recall that W is the work done by the system; W in is the work done on the system. ANSWER: W in
=
−W
Correct
Part B Find Q h , the heat pumped out by the ideal heat pump. Express Q h in terms of Q c and W in .
Hint 1. Conservation of energy and the first law Apply conservation of energy. If you think in terms of the first law of thermodynamics, remember the sign conventions for heat and work and note that the internal energy does not change in an engine after one cycle. ANSWER: Q
h
=
W in + Q
c
Correct
Part C A heat pump is used to heat a house in winter; the inside radiators are at Th and the outside heat exchanger is at Tc . If it is a perfect (i.e., Carnot cycle) heat pump, what is Kpump , its coefficient of performance? Give your answer in terms of Th and Tc .
Hint 1. Heat pump efficiency in terms of Qh and Qc What is the efficiency of a heat pump Kpump in terms of the heats in and out? Use the expression for the efficiency of the heat pump and the expression that you found involving the work done on the system, W in , and the outside heats, Q h and Q c . Give your answer in terms of Q h and Q c . ANSWER: Kpump
=
Hint 2. Relation between Qh and Qc in a Carnot cycle Recall that in a Carnot cycle, Qh Qc
=
Th Tc
.
ANSWER:
Kpump
=
Th T h −T c
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 sides hold true. Indicate the statement that must apply for the heat pump to work. ANSWER: Q Q Q Q
h
h
h
h
< Q > Q < Q > Q
c
and Th
< Tc
.
c
and Th
< Tc
.
c
and Th
> Tc
.
and Th c
> Tc
.
Correct
Part E Assume that you heat your home with a heat pump whose heat exchanger is at Tc = 2∘ C, and which maintains the baseboard radiators at ∘ Th = 47 C. 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 the heat generation the less work needs to be done and the lower the heating bill. You are given that the cost of Q h is $1000. You also have an equation for Kpump in terms of the temperatures: Kactual = 75
% of
K pump =
3
Th
4
T h −T c
.
Set this equal to Q h /W in and solve for the monetary value of W in , the amount of external energy input the pump requires. You can measure energies in units of currency for this calculation.
Hint 2. Units of T h and T c Keep in mind that when calculating an efficiency of a thermodynamic device you need to use temperature in kelvins. That is, 0∘ C
= 273K
.
ANSWER: Cost = 187.5 dollars
Correct 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 Th , 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 Q c can be obtained from a groundwater that is abundant or flowing (e.g., an underground stream).
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? Give the letter(s) of the design(s) in alphabetical order, without commas or spaces (e.g., ACD).
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: CF
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 (e.g., 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 Thot and Tcold has a maximum thermal efficiency e max given by e max = 1 −
T cold T hot
.
An engine with thermal efficiency greater than that given by the above equation violates the second law of thermodynamics. ANSWER: BD
Correct
Part C Which of the remaining designs has the highest thermal efficiency? ANSWER: device A device E
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 n 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 e = W /Q h , where W is the net work done by the engine and Q h is the quantity of heat put into the engine at a (high) temperature Th . Furthermore, he showed that the engine must necessarily put an amount of heat Q c back into a heat reservoir at a lower temperature Tc . 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 expands isothermally from state A to state B, absorbing heat Q h from a reservoir at temperature Th . The gas then expands adiabatically until it reaches a temperature Tc , in state C. The gas is compressed isothermally to state D, giving off heat Q c . Finally, the gas is adiabatically compressed to state A, its original state.
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: 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 W done by the gas after it completes a single Carnot cycle. Express the work in terms of any or all of the quantities Q h , Th , Q c , and Tc .
Hint 1. How to approach the problem 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 the first law of thermodynamics: dQ = dU + dW .
Hint 2. Compute the change in internal energy What is the net change in the gas's internal energy U after one complete cycle? ANSWER: change in U =
ANSWER: W
=
Q
h
Correct
−Q
c
Part C Suppose there are n moles of the ideal gas, and the volumes of the gas in states A and B are, respectively, VA and VB . Find Q h , the heat absorbed by the gas as it expands from state A to state B. Express the heat absorbed by the gas in terms of n , VA , VB , the temperature of the hot reservoir, Th , and the gas constant R.
Hint 1. General method of finding Qh First, find the work W AB done by the gas as it expands from state A to state B. Then, use the first law of thermodynamics to relate W AB to both Q h and the change in the gas's internal energy. Finally, recall that this process is isothermal; what does this tell you about the change in the gas's internal energy?
Hint 2. Find the work done by the gas What is W AB , the work done by the gas as it expands from state A to state B? Express the work in terms of n , VA , VB , the temperature of the hot reservoir Th , and the gas constant R.
Hint 1. How to find the work done by the gas To find W AB , the net work done by the gas as it proceeds from state A to state B, you need to integrate dW state B.
= p dV
from state A to
Hint 2. Express p in terms of V Use the ideal gas equation of state to find an expression for the pressure p in terms of n , R, Th , and V . ANSWER: p(V )
=
Hint 3. Express the integral over dV Given that p(V )
∝ 1/V
and that W AB = ∫
VB
what is the integral ∫ V
dV /V
VB VA
p(V ) dV
,
?
A
Express your answer in terms of VA and VB . ANSWER:
∫
VB VA
dV V
=
ANSWER: W AB
=
Hint 3. Relation between Qh and WAB 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 heat flow into the gas: Q h = W AB . ANSWER:
Q
h
=
nRT h (ln(
Correct
VB VA
))
Part D The volume of the gas in state C is VC , and its volume in state D is VD . Find Q c , the magnitude of the heat that flows out of the gas as it proceeds from state C to state D. Express your answer in terms of n , VC , VD , Tc (the temperature of the cold reservoir), and R.
Hint 1. How to approach the problem First, find the work W CD done by the gas as it expands from state C to state D. Then, use the first law of thermodynamics to relate W CD to both Q c and the change in the gas's internal energy. Finally, recall that this process is isothermal; what does this tell you about the change in the gas's internal energy? ANSWER:
Q
c
=
nRT c (ln(
VC VD
))
Correct Observe that the three parts together imply that W BC + W DA = 0 . This is because BC and DA are adiabatic processes. So using the first law, W BC = −ΔUBC = −nCV (Th − Tc ) , whereas W DA = −ΔUDA = −nCV (Tc − Th ) . So W BC = −W DA , or W BC + W DA = 0 . 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 VC /VD in terms of VA and VB .
Hint 1. How to approach the problem Suppose the ratio of the gas's specific heats is denoted by γ = Cp /Cv . Along adiabatic curves, you know that pV γ = C , where C is some constant. Rewrite this equation in terms of T and V instead of p and V . Then use this new equation to relate the temperature and volume at the end points of the two adiabatic legs of the Carnot cycle. This will give you two equations that you can solve for VC /VD .
Hint 2. Rewrite pV
γ
in terms of T and V
Use the ideal gas equation of state to eliminate p from the expression pV γ . Express your answer in terms of γ , n , R, and the temperature T . ANSWER: pV
γ
=
Hint 3. Express T h and V B in terms of T c and V C (γ−1)
States B and C are connected by an adiabatic expansion. Use the result found in the previous hint to find an expression for Th VB Express your answer in terms of Tc , VC , and γ . ANSWER: Th V
(γ−1)
B
=
Hint 4. Express T h and V A in terms of T c and V D (γ−1)
States D and A are connected by an adiabatic expansion. Use the result found in Hint 2 to find an expression for Th VA Express your answer in terms of Tc , VD , and γ . ANSWER:
.
.
(γ−1)
Th V
A
=
Hint 5. Solving for V C /V D in terms of V A and V B Combine the results of the previous two hints, eliminating the temperatures Th and Tc . (One way to do this is to divide one equation by the other.) This should allow you to solve for VC /VD in terms of VA and VB . ANSWER: =
VC /VD
VB VA
Correct
Part F Using your expressions for Q h and Q c (found in Parts C and D), and your result from Part E, find a simplified expression for Q c /Q h . No volume variables should appear in your expression, nor should any constants (e.g., n or R). ANSWER:
Q c /Q h
=
Tc Th
Correct
Part G The efficiency of any engine is, by definition, e Find the efficiency e Carnot of a Carnot engine.
= W /Q
h
. Carnot proved that no engine can have an efficiency greater than that of a Carnot engine.
Express the efficiency in terms of Th and Tc .
Hint 1. Express the efficiency in terms of Qh and Qc Using your result from Part B, find the efficiency e (of any engine) in terms of the engine's heat input and output. Express your answer in terms of Q h and Q c . ANSWER: e
=
ANSWER:
e Carnot
=
T h −T c Th
Correct Because Tc 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 temperature Th . This explains why (historically) there were some spectacular explosions of boilers used for steam power.
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 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 Q h at a temperature Th and does work W . In the process of doing this it generates waste heat Q c at a cooler temperature Tc . Take Q h and Q c to be the magnitudes of the heat absorbed and emitted, respectively; therefore both quantities are positive.
Part A A heat engine is designed to do work. This is possible only if certain relationships between the heats and temperatures at the input and output hold true. Which of the following sets of statements must apply for the heat engine to do work? ANSWER: Qh < Qc
and Th
< Tc
Qh > Qc
and Th
< Tc
Qh < Qc
and Th
> Tc
Qh > Qc
and Th
> Tc
Correct
Part B Find the work W done by the "ideal" heat engine. Express W in terms of Q h and Q c . ANSWER: W
= Q_{h}Q_{c}
Correct
Part C The thermal efficiency \texttip{e}{e} of a heat engine is defined as follows: e = W/Q_{\rm h}. Express the efficiency in terms of \texttip{Q_{\rm h}}{Q_h} and \texttip{Q_{\rm c}}{Q_c}. ANSWER:
\texttip{e}{e} = \large{\frac{Q_{h}Q_{c}}{Q_{h}}}
Correct
Problem 19.10 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 indicate whether the quantity increases, decreases, or stays the same during that process. ANSWER:
Correct
Problem 19.63 A heat engine with 0.100 {\rm mol} of a monatomic ideal gas initially fills a 3000 \;{\rm cm}^{3} cylinder at 600 {\rm K} . The gas goes through the following closed cycle:
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? Express your answer with the appropriate units. ANSWER:
Score Summary: Your score on this assignment is 81.5%. You received 7.33 out of a possible total of 9 points.
