Download ME12001 Thermodynamics T7

Thermodynamics T7
Irreversible versus Reversible Processes
Part A
Which of the following conditions should be met to make a process perfectly reversible?
Check all that apply.
Hint 1. Reversible processes
A reversible process is a transition from one state of a thermodynamic system to another, during which the system is always very close to a state
of mechanical and thermal equilibrium. This implies that the process should be slow and smooth to allow temperature and pressure to change
only by infinitesimal amounts.
ANSWER:
Any mechanical interactions taking place in the process should be frictionless.
Any thermal interactions taking place in the process should occur across infinitesimal temperature or pressure gradients.
The system should not be close to equilibrium.
Correct
Part B
Based on the results found in the previous part, which of the following processes are not reversible?
Check all that apply.
Hint 1. How to approach the problem
As you found in the previous part, a process that takes place through infinitesimal changes in the conditions of a system is reversible. Do any of
the processes listed involve nonnegligible changes in temperature and pressure?
ANSWER:
Melting of ice in an insulated ice-water mixture at 0∘ C.
Lowering a frictionless piston in a cylinder by placing a bag of sand on top of the piston.
Lifting the piston described in the previous statement by removing one grain of sand at a time.
Freezing water originally at 5∘ C.
Correct
A Three-Step Gas Cycle
A monatomic ideal gas has pressure p1 and temperature T1 . It is contained in a cylinder of volume V1 with a movable piston, so that it can do work on the
outside world.
Consider the following three-step transformation of the gas:
1. The gas is heated at constant volume until the pressure reaches Ap1 (where A > 1).
2. The gas is then expanded at constant temperature until the pressure returns to p1 .
3. The gas is then cooled at constant pressure until the volume has returned to V1 .
It may be helpful to sketch this process on the pV plane.
Part A
How much heat Q1 is added to the gas during step 1 of the process?
Express the heat added in terms of p1 , V1 , and A.
Hint 1. First law of thermodynamics
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The heat added to a gas equals the change in energy of the gas plus the work done by the gas: Q = ΔU + W .
Hint 2. Think about W
The work done by a gas is
V
final
W = ∫ Vinitial
p dV .
Think about the volume of the gas during step 1. If you were going to compute the work done during step 1, what would Vinitial and Vfinal be?
Hint 3. How to find ΔU
The energy of a monatomic ideal gas is U = (3/2)N kB T . The number of gas particles N and the Boltzmann constant kB do not change during
step 1, so ΔU depends only on ΔT .
Hint 4. How to find ΔT
To find ΔT , use the ideal gas law pV = N kB T . Solve for T and take the difference between the value of T at the beginning and that at the end
of step 1.
ANSWER:
Q1 = 3 (A − 1)p1 V 1
2
Correct
Part B
How much work W2 is done by the gas during step 2?
Express the work done in terms of p1 , V1 , and A.
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Hint 1. How to approach this problem
The work done by a gas is
V
final
W = ∫ Vinitial
p dV .
Since the pressure is not constant during step 2, you'll need to use the ideal gas law to express p as a function of V to solve this integral. You'll
also need to determine the appropriate initial and final volumes of the gas.
Hint 2. Find p(V )
What is the pressure p of the gas during step 2, as a function of volume V ?
Express your answer in terms of V , A, p1 , and V1 .
Hint 1. Find the temperature during step 2
What is the temperature T2 during step 2? (That is, what is the temperature at the end of step 1?)
Give your answer in terms of T1 and other variables given in the introduction.
ANSWER:
T2 = AT1
ANSWER:
p(V ) =
Ap1 V1
V
Hint 3. Find the initial and final volumes
Since step 1 of the process took place at constant volume, you know that the initial volume of the gas (at the beginning of step 2) is V1 . What is
V2 , the volume of the gas at the end of step 2?
Give your answer in terms of V1 and other variables given in the introduction.
ANSWER:
V2 = AV1
ANSWER:
W2 = Ap1 V1 ln(A)
Part C
How much work W3 is done by the gas during step 3?
If you've drawn a graph of the process, you won't need to calculate an integral to answer this question.
Express the work done in terms of p1 , V1 , and A.
Hint 1. The easy way to solve this problem
Recall that an integral is a calculation of the area under a curve. In the case of work done by a gas, the integral is equal to the area under the plot
of pressure versus volume. If this plot is a horizontal line, the area is a rectangle. If the plot is a vertical line, the area is zero (rectangle of zero
width). If the plot is any other straight line, the area is a triangle. Only resort to calculus when the area under the curve cannot be solved with
simple geometry (as was the case for W2 ).
Hint 2. Find the formula for work done
What is the work W done by a gas held at constant pressure p as its volume changes from V2 to V1 ?
ANSWER:
W = p(V2 − V1 )
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ANSWER:
W3 = −(A − 1)p1 V1
Exercise 20.3
A gasoline engine takes in 1.44×104 J and delivers 4000 J of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.60×104
J/g .
Part A
What is the thermal efficiency?
ANSWER:
e = 0.278
Correct
Part B
How much heat is discarded in each cycle?
ANSWER:
1.04×104
J
Correct
Part C
What mass of fuel is burned in each cycle?
ANSWER:
m = 0.313 g
Correct
Part D
If the engine goes through 70.0 cycles per second, what is its power output in kilowatts?
ANSWER:
280
kW
Correct
Part E
If the engine goes through 70.0 cycles per second, what is its power output in horsepower?
ANSWER:
376
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hp
Correct
Exercise 20.5
The pV diagram in the figure shows a cycle of a heat engine that uses 0.250 mole of an ideal gas
having γ = 1.40. The curved part ab of the cycle is adiabatic.
Part A
Find the pressure of the gas at point a.
ANSWER:
Pa = 12.3 atm
Correct
Part B
How much heat enters this gas per cycle?
ANSWER:
Qin = 5470 J
Correct
Part C
Where does the entering of heat happen?
ANSWER:
a->b.
b->c.
c->a.
Correct
Part D
How much heat leaves this gas in a cycle?
ANSWER:
Qleav = -3730 J
Part E
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Where does the leaving of heat of this gas occur?
ANSWER:
a->b.
b->c.
c->a.
Correct
Part F
How much work does this engine do in a cycle?
ANSWER:
A = 1750 J
Part G
What is the thermal efficiency of the engine?
ANSWER:
e = 31.9 %
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 QH at
a temperature TH and does work W . In
the process of doing this it generates
waste heat QC at a cooler temperature
TC .
Take QH to be positive and QC is
negative, since they are respectively the
heat absorbed and emitted by the heat
engine.
Part A
A heat engine is designed to do work. This is possible only if certain relationships between the heats and temperatures at the input and output hold true.
Which of the following sets of statements must apply for the heat engine to do work?
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Recall that |QH | and |QC | are respectively the magnitudes of the heat aborbed QH and the heat emitted QC by the heat engine.
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 QH and QC .
ANSWER:
W = QH + QC
Correct
Part C
The thermal efficiency e of a heat engine is defined as follows: e = W /QH .
Express the efficiency in terms of QH and QC .
ANSWER:
e=
QH +QC
QH
Correct
± PSS 20.1 Heat Engines
Learning Goal:
To practice Problem-Solving Strategy 20.1 Heat Engines.
Steam at a temperature TH = 155 ∘ C and p = 1.00 atm enters a heat engine at an unknown flow rate. After passing through the heat engine, it is released at
a temperature TC = 100 ∘ C and p = 1.00 atm . The measured power output P of the engine is 570 J/s , and the exiting steam has a heat transfer rate of
HC = 3150 J/s . Find the efficiency e of the engine and the molar flow rate n/t of steam through the engine. The constant pressure molar heat capacity Cp
for steam is 37.47 J/(mol ⋅ K) .
Problem Solving Strategy: Heat engines
IDENTIFY the relevant concepts:
A heat engine is any device that converts heat partially to work.
SET UP the problem using the following steps:
1. Carefully define what the thermodynamic system is.
2. For multi-step processes with more than one step, identify the initial and final states for each step.
3. Identify the known quantities and the target variables.
4. The first law, ΔU = Q − W , can be applied just once to each step in a thermodynamic process, so you will often need additional equations.
The equation
Q
Q
W
= 1 + QC = 1 − ∣∣ QC ∣∣
QH
H
H
is useful in situations for which the thermal efficiency of the engine is relevant. It's helpful to sketch an energy-flow diagram.
e=
EXECUTE the solution as follows:
1. Be very careful with the sign conventions for W and the various Q's. W is positive when the system expands and does work; W is negative
when the system is compressed. Each Q is positive if it represents heat entering the system and is negative if it represents heat leaving the
system.
2. Power is work per unit time (P = W /t), and heat current H is heat transfer per unit time (H = Q/t).
3. Keeping steps 1 and 2 in mind, solve for the target variables.
EVALUATE your answer:
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Use the first law of thermodynamics to check your results, paying particular attention to algebraic signs.
IDENTIFY the relevant concepts
This heat engine partially converts heat from the incoming steam into work, so the problem solving strategy for heat engines is applicable.
SET UP the problem using the following steps
Part A
Which of the following quantities are known?
Check all that apply.
ANSWER:
The heat transfer rate for steam leaving the engine, HC
The constant pressure molar heat capacity of steam, Cp
The temperature of steam as it leaves the engine, TC
The efficiency of the engine, e
The heat transfer rate for steam entering the engine, HH
The molar flow rate of steam, n/t
The temperature of the steam as it enters the engine, TH
The power output of the engine, P
The efficiency and the molar flow rate of steam through the engine are both target variables in this problem. Even though the heat transfer rate of
the steam entering the heat engine HH is not given in the problem statement, it is not a target variable.
The energy-flow diagram for this system is shown. Heat flows into the engine at a rate of HH as steam at a temperature TH . Work leaves the
engine at a rate P . The remaining heat leaves the engine at a rate of HC as steam at temperature TC .
EXECUTE the solution as follows
Part B
What is the efficiency of the heat engine?
Express the efficiency numerically to three significant figures.
Hint 1. How to approach the problem
You are asked to find the efficiency of the heat engine.
To find the efficiency of the heat engine, first find the heat transfer rate of the incoming steam from the relationships between work and power and
between heat and heat transfer rate. Once you know the incoming heat transfer rate, you can then calculate the efficiency of the heat engine
using the power delivered by the engine.
Hint 2. Find the heat transfer rate of the incoming steam
What is the heat transfer rate of the incoming steam HH ?
Express your answer numerically in joules per second.
Hint 1. Find an expression for the heat transfer rate of the incoming steam
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Find an expression for HH .
Start with the equation W = QH + QC and replace W , QC , and QH with expressions involving the power P , the heat transfer rates
HC , HH , and the time t. Refer to EXECUTE step 2 if you need help recalling the relationships between these variables.
Express your answer in terms of HC and P .
Hint 1. How to put the expression in terms of power and heat transfer rates
Work is related to power by W = P t . Heat is related to the heat transfer rate by Q = Ht. Replace all work and heat expressions
in
W = QH + QC
with power and heat transfer rates, simplify the expression, and solve for HH .
ANSWER:
HH = P − HC
ANSWER:
HH = 3720 J/s
Hint 3. Find an expression for the efficiency of the heat engine
Find an expression for the efficiency of the heat engine e.
Start with the expression
e=
W
,
QH
and replace W and QH with expressions involving the power P generated by the engine, the heat transfer rate of the steam entering the engine
HH , and time t. Refer to EXECUTE step 2 if you need help recalling the relationships between these variables.
Express e in terms of P and HH .
Hint 1. How to put the efficiency equation into terms of power and heat transfer rates
Work is related to power by W = P t . Heat is related to the heat transfer rate by QH = HH t. Replace all work and heat expressions in
e=
with power and heat transfer rates and simplify the expression.
W
QH
ANSWER:
P
e= H
H
ANSWER:
e = 0.153
An efficiency of 0.153 indicates that 15.3 % of the heat from the incoming steam is converted to work. The remaining 84.7 % is expelled from the
heat engine as heat in the outlet steam.
Part C
What is the molar flow rate of steam into the engine?
Express the molar flow rate in moles per second to three significant figures.
Hint 1. How to approach the problem
The molar flow rate n/t of steam can be calculated by relating the temperature change and molar heat capacity to the power (P = W /t)
delivered by the steam engine.
From the first law of thermodynamics, you know that W = Q. That is, the energy of the heat engine is conserved and the net heat flowing into
the engine equals the net work done by the engine. You can use the fact that the steam is a gas held at a constant pressure to find an expression
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for Q involving n.
Hint 2. The net heat flowing into the heat engine
The heat required for the temperature change of n moles of steam can be expressed in terms of the difference in temperatures of the entering
and exciting steam as
Q = nCp (ΔT ).
The heat lost by the steam is equal to the heat gained by the steam engine, so the net heat flowing into the steam engine can be expressed as
Q = −nCp (ΔT ).
Hint 3. Find an expression for the molar flow rate of steam through the heat engine
Express the molar flow rate n/t of steam through the heat engine as a function of the generated power P , the constant pressure molar heat
capacity Cp , and the change in temperature ΔT . Keep in mind that P is related to W , and since the energy of the heat engine is conserved,
W = Q.
Express the molar flow rate in terms of P , Cp , and ΔT .
ANSWER:
−P
n/t = (C )(ΔT )
p
ANSWER:
n/t = 0.277 mol/s
Knowing the flow rate of steam through the heat engine is important for a few reasons. In the design stage, an estimate of the flow rate is used the
size the heat engine. During the operation of the heat engine, enough steam must flow through the heat engine to produce the required amount of
power.
EVALUATE your answer
Part D
Which of the following changes, if made individually, would cause an increase in the amount of power produced by the heat engine?
Check all that apply.
ANSWER:
lower molar flow rate, n/t
smaller value of HC
higher molar flow rate, n/t
larger value of HC
larger value of HH
smaller value of HH
A heat engine can produce more power either by improving its efficiency or by increasing its throughput. If either the entering heat transfer rate
increases or the exiting heat transfer rate decreases, the power generated by the engine will increase. Dividing both sides of W = |QH | − |QC | by
time makes the relationship between power and heat transfer rates clearer:
P = |HH | − |HC |.
By either increasing the entering heat transfer rate or decreasing the exiting heat transfer rate, the efficiency of the heat engine is increased. If the
molar flow rate increases, the amount of heat that the engine converts into work per unit time increases, but the engine is not necessarily more
efficient.
± Isothermal Engine
Consider a thermal engine filled with N molecules of a monatomic ideal gas. The gas undergoes a cyclic transformation, as shown. Processes AB and CD
are isothermal (constant temperature), and processes BC and DA are isochoric (constant volume).
The quantities p1 , p2 , V1 , and V2 are defined in the figure. For example, when the gas is in state
A, it has pressure p2 and volume V1 ; when the gas is in state B, it has volume V2 , but the pressure
is not given.
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Work done by the gas
Part A
Find WAB , the work done by the gas during process AB.
Express the work in terms of p1 , p2 , V1 , and/or V2 .
Hint 1. How to approach the problem
To find the work done by the gas, you need to integrate
V
final
W = ∫ Vinitial
p(V ) dV .
To do this integral, you first need to find an expression for p(V ), the pressure of the gas as a function of its volume.
Hint 2. Find p(V ) for process AB
Find an expression for p(V ), the pressure of the gas along the AB segment of the cycle, as a function of the volume V .
Express p(V ) in terms of the given quantities ( p1 , p2 , V1 , and/or V2 ) and V .
Hint 1. Find p(V )V for process AB
The ideal gas law is pV = N kB T , where kB is Boltzmann's constant. Therefore, for an isothermal transformation, p(V )V is constant.
Along the segment AB, what is this constant?
Express your answer in terms of p2 and V1 .
ANSWER:
(p(V )V )AB = p2 V1
ANSWER:
p(V ) =
p2 V1
V
Hint 3. Integrating to find the work
You should have found that the pressure p(V ) is proportional to 1/V . Find the integral of dV /V from initial volume V1 to final volume V2 .
Express your answer in terms of V1 and V2 .
Hint 1. Help with the integral
The indefinite integral of dV /V is ln(V ) + constant. When you do the integral, however, you will have definite limits of integration: V1 to
V2 .
ANSWER:
V
∫ V12 dV /V
=
ln( VV2 )
1
ANSWER:
WAB = p2 V 1 ln(
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V2
)
V1
Part B
Find WBC , the work done by the gas during process BC.
Express your answer in terms of p1 , p2 , V1 , and/or V2 .
Hint 1. Work done in an isochoric process
Recall that the infinitesimal amount of work done is given by dW = P dV . Along segment BC, however, the volume doesn't change. (The
process is isochoric.) In other words, dV = 0 along the entire segment. What, then, is WBC ?
ANSWER:
WBC = 0
Part C
Find WCD , the work done by the gas during process CD.
Express your answer in terms of p1 , p2 , V1 , and/or V2 .
Hint 1. How to approach the problem
If you need help with this part, see the hints for Part A.
ANSWER:
WCD = −p1 V 2 ln(
V2
)
V1
Part D
Find WDA , the work done by the gas during process DA.
Express your answer in terms of p1 , p2 , V1 , and/or V2 .
Hint 1. How to approach the problem
If you need help with this part, see the hint for Part B.
ANSWER:
WDA = 0
The internal energy
Now, you will investigate the change in the gas's internal energy during each of the four segments of its cycle. You may find it helpful to recall that, for a
monatomic ideal gas, Cv = (3/2)N kB , where Cv is the heat capacity of the gas at constant volume.
Part E
What is the change in internal energy UB − UA during the process AB? What is the change in internal energy UD − UC during the process CD?
Express the change in internal energies in terms of p1 , p2 , V1 , and/or V2 . Separate your answers with a comma.
Hint 1. How to approach the problem
The segment AB of the gas's cycle is an isothermal process. The temperature is constant throughout this process. Now, recall that the internal
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energy of an ideal gas only depends on the temperature of the gas. If the gas's temperature doesn't change during process AB, can the gas's
internal energy change?
The segment CD is also an isothermal process.
ANSWER:
UB − UA , UD − UC = 0,0
Part F
What is the change in internal energy UC − UB during the process BC? What is the change in internal energy UA − UD during the process DA?
Express the change in internal energies in terms of p1 , p2 , V1 , and/or V2 . Separate your answers with a comma.
Hint 1. How to approach the problem
First, find TC and TB , the temperatures at states C and B respectively, in terms of given quantities (pressure and volume). Then compute
UC − UB using the expression U = (3/2)N kB T for the internal energy of an ideal monatomic gas.
You should find the change in internal energy of process DA in the same manner.
Hint 2. Find the temperature of the gas in state B
Using the ideal gas equation of state, find TB , the temperature of the gas when it is in state B.
Express the temperature in terms of N , kB , and other given quantities ( p1 , p2 , V1 , and/or V2 ).
Hint 1. Relationship between temperature in states A and B
As the gas proceeds from state A to state B, its temperature does not change. Therefore, TA = TB . You may find it easier to calculate the
temperature when the gas is in state A, since you know the pressure and volume of the gas in state A.
ANSWER:
TB =
p2 V1
NkB
Hint 3. Find the temperature of the gas in state C
Using the ideal gas equation of state, find TC , the temperature of the gas when it is in state C.
Express your answer in terms of N , kB , and other given quantities ( p1 , p2 , V1 , and/or V2 ).
ANSWER:
TC =
p1 V2
NkB
ANSWER:
UC − UB , UA − UD = 32 (p1 V 2 − p2 V 1 ),− 32 (p1 V 2 − p2 V 1 )
Part G
What is the total change in the gas's internal energy after it has completed a cycle?
ANSWER:
Uend of
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cycle
− Ubeginning of
cycle =
0
After completing an entire cycle, the gas has returned to its initial state. Therefore the total change in internal energy, after completing a cycle, is
zero.
Heat
Finally, consider the heat absorbed by the gas during its cycle. Define QAB to be heat absorbed by the gas in process AB, etc. If a process is exothermic
(i.e., heat is given off by the gas during the process), then the quantity of heat absorbed will be negative.
Part H
Indicate the processes in this cycle during which heat is flowing into the gas. That is, indicate the segments where Q is positive.
Hint 1. How to approach the problem
Find an expression for QAB in terms of UB − UA and WAB . You already have expressions for UB − UA and WAB from Parts E and A, so you
should be able to determine whether QAB is a positive or a negative number. If it's positive, then heat flows into the gas during segment AB. You
need to repeat this for the rest of the segments of the cycle: BC, CD, and DA.
Hint 2. Identify the formula for heat in terms of work and internal energy
Consider segment AB of the cycle. Define QAB to be the heat absorbed by the gas as it proceeds from state A to state B. To determine whether
QAB is positive or negative, use the first law of thermodynamics.
Which of the following is the correct expression for QAB in terms of the change in internal energy UB − UA and the work done by the gas WAB
during segment AB?
Hint 1. Sign convention for the first law of thermodynamics
Here's one way to figure out the proper signs when writing down the first law of thermodynamics. Consider the internal energy of the gas.
If heat is absorbed by the gas, then its internal energy should increase. If work is done by the gas, then its internal energy should
decrease. Therefore, dU = dQ − dW . (That is, the change in internal energy dU equals the heat absorbed dQ minus the work done
dW .) You can now solve this equation for dQ .
ANSWER:
QAB = (UB − UA ) + WAB
QAB = (UB − UA ) − WAB
QAB = −(UB − UA ) + WAB
QAB = −(UB − UA ) − WAB
ANSWER:
DA only
CD and DA
DA and AB
AB and BC
An Air Conditioner: Refrigerator or Heat Pump?
The typical operation cycle of a common refrigerator is shown schematically in the figure . Both the
condenser coils to the left and the evaporator coils to the right contain a fluid (the working
substance) called refrigerant, which is typically in vapor-liquid phase equilibrium. The compressor
takes in low-pressure, low-temperature vapor and compresses it adiabatically to high-pressure,
high-temperature vapor, which then reaches the condenser. Here the refrigerant is at a higher
temperature than that of the air surrounding the condenser coils and it releases heat by undergoing
a phase change. The refrigerant leaves the condenser coils as a high-pressure, high-temperature
liquid and expands adiabatically at a controlled rate in the expansion valve. As the fluid expands, it
cools down. Thus, when it enters the evaporator coils, the refrigerant is at a lower temperature than
its surroundings and it absorbs heat. The air surrounding the evaporator cools down and most of
the refrigerant in the evaporator coils vaporizes. It then reaches the compressor as a low-pressure,
low-temperature vapor and a new cycle begins.
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02/04/16 01:02
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 Qc 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 liquid-vapor 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 liquid-vapor phase equilibrium and 95%
leaves as vapor. Assume that the pressure of the refrigerant is kept constant while it is in the evaporator. What is the percentage m% of the total
mass of refrigerant transformed to vapor in the evaporator per cycle?
ANSWER:
m% = 49.0 %
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 Qh 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 Qc from the air inside the room, as you
calculated in Part A, and gives off a certain amount of heat Qh 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
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quantity Qc is now absorbed from the air outside the room and the amount of heat Qh 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 Qh .
Hint 2. Find the right expresssion for the first law of thermodynamics
Suppose you have an ideal system that absorbs heat Qc from a cold reservoir and releases heat Qh to a hot reservoir. The net input of work
required by this system for operation is Win . 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:
Qc = Win − Qh
Qh = Qc − Win
Qh = Qc + Win
Correct
ANSWER:
Qh = 6.35×105 J
Exercise 20.11
A refrigerator has a coefficient of performance of 2.25, runs on an input of 95.0 W of electrical power, and keeps its inside compartment at 5.00 ∘ C.
Part A
If you put a dozen 1.00 L plastic bottles of water at 31.0 ∘ C this refrigerator, how long will it take for them to be cooled down to 5.00 ∘ C? (Ignore any
heat that leaves the plastic.)
ANSWER:
t = 102 minutes
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 QC,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 QC to the work W input
to the refrigerator:
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k=
Recall that power is defined as work per unit time.
QC
.
W
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 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:
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