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Lecture Outlines Chapter 17
Physics, 3rd Edition James S. Walker
Chapter 18 The Laws of
Thermodynamics
Units of Chapter 18
• The Zeroth Law of Thermodynamics
• The First Law of Thermodynamics
• Thermal Processes
• Specific Heats for an Ideal Gas: Constant Pressure, Constant Volume
• The Second Law of Thermodynamics
• Heat Engines and the Carnot Cycle
Units of Chapter 18
• Refrigerators, Air Conditioners, and Heat Pumps
• Entropy
• Order, Disorder, and Entropy
• The Third Law of Thermodynamics
18-1 The Zeroth Law of Thermodynamics
We have already discussed the zeroth law, and include it here for completeness:
If object A is in thermal equilibrium with object C,
and object B is separately in thermal equilibrium with
object C, then objects A and B will be in thermal
equilibrium if they are
placed in thermal contact.
18-2 The First Law of Thermodynamics
The first law of thermodynamics is a statement of the conservation of energy.
If a system’s volume is constant, and heat is added, its internal energy increases.
18-2 The First Law of Thermodynamics
If a system does work on the external world, and no heat is added, its internal energy decreases.
18-2 The First Law of Thermodynamics
Combining these gives the first law of
thermodynamics. The change in a system’s
internal energy is related to the heat Q and the work W as follows:
It is vital to keep track of the signs of Q and W.
18-2 The First Law of Thermodynamics
The internal energy of the system depends only on its temperature. The work done and the heat added, however, depend on the details of the process involved.
Example 18-1 Heat, Work, and Internal Energy
På en joggningstur på stranden utför en person W = 430 kJ och avger Q = - 380 kJ. Vad är ändringen i den inre energin? När personen övergår till promenad, avgår Q = - 120 kJ och ΔU = - 260 kJ. Vad blev W?
Example 18-1 Heat, Work, and Internal Energy
På en joggningstur på stranden utför en person arbetet 430 kJ och avger värmet 380 kJ. Vad är ändringen i den inre energin?
ΔU = Q - W = - 380 kJ - 430 kJ = - 810 kJ
När personen övergår till promenad, avgår
värmet 120 kJ och ändringen i inre energi blir - 260 kJ. Vad blir W?
ΔU = Q - W ger
W = Q - ΔU = -120 kJ – (- 260 kJ) = 140 kJ
18-3 Thermal Processes
We will assume that all processes we discuss are quasi-static – they are slow enough that the system is always in equilibrium.
We also assume they are reversible:
For a process to be reversible, it must be possible to return both the system and its surroundings to exactly the same states they were in before the process began.
18-3 Thermal Processes
This is an idealized reversible process. The gas is compressed; the temperature is constant, so heat leaves the gas. As the gas expands, it
draws heat from the reservoir, returning the gas and the reservoir to their initial states. The
piston is assumed frictionless.
18-3 Thermal Processes
Work done by an expanding gas, constant pressure (isobar process): (18-4)
Exercise 18-1
En gas med det konstanta trycket 150 kPa
expanderar från ursprungsvolymen 0,76 m3 till 0,92 m3. Hur stort arbete utför gasen?
W = PΔV = 150 kPa(0,92 – 0,76)m3 = 24 kJ
Example 18-2 Work Area
Example 18-2
En gas expanderar från ursprungsvolymen 0,40 m3 till 0,62 m3 under det att trycket ökar linjärt från 110 kPa till 230 kPa. Hur stort arbete utför gasen? (W = PΔV)
”Kvadraten” ger arbetet W1 = 110 kPa(0,62 – 0,40)m3 = 24,2 kJ
”Triangeln” ger arbetet
W2 = {(230 – 110)kPa(0,22 m3)}/2 = 13,2 kJ W = 37 kJ
18-3 Thermal Processes
If the volume stays constant (isochor process), nothing moves and no work is done.
Active Example 18-1 Find the Total Work
Active Example 18-1 Find the Total Work
En gas undergår en trestegs process enligt figur 18-7. Hur stort arbete utför gasen under
processen från A till B?
Längs 1 är arbetet W1 = 120kPa(0,40 m3 - 0,25 m3) = 18 kJ
W2 = 0 (isochor process)
Längs 3 ger ”kvadraten” arbetet W3k = 120 kPa(0,54 – 0,40)m3 = 16,8 kJ
och ”triangeln” ger arbetet
W3t = {(180 – 120)kPa(0,14 m3)}/2 = 4,2 kJ W = 39 kJ
18-3 Thermal Processes
If the temperature is constant (isoterm process), the pressure varies inversely with the volume.
[PV=NkT]
18-3 Thermal Processes
The work done is the area under the curve:
W = ∫P•dV = {P =NkT/V} = NkT∫dV/V=NkT•ln(Vf / Vi)
Example 18-3 Heat Flow
Example 18-3 Heat Flow
En cylinder innehåller 0,50 mol av en monoatomisk gas med temperaturen 310 K. När gasen expanderar isotermt, från ursprungsvolymen 0,31 m3 till 0,45 m3, utför gasen ett arbete. Hur stor mängd värme Q
måste tillföras gasen, för att temperaturen skall kunna hållas konstant? ΔU = Q – W (18-3)
Gasen utför arbetet (W > 0) W = NkT•ln(Vf/Vi) = nRT•ln(Vf / Vi) =
0,50 mol•8,31 J/(mol •K)•310K•ln(1,4516) = 480 J Ändringen i den inre energin ΔU = 0, eftersom den bara beror på temperaturen (U = 3NkT/2, 17-15) Q = ΔU + W = 480 J
Conceptual Checkpoint 18-1 Ideal or Not?
Inre energin för en viss gas ökar då den komprimeras isotermt. Är gasen ideal?
18-3 Thermal Processes
An adiabatic process is one in which no heat flows into or out of the system. The adiabatic P-V curve is similar to the isothermal one, but is steeper. One way to ensure that a process is adiabatic is to insulate the system.
Conceptual Checkpoint 18-2 Pressure versus Volume Vägen (i)i är en isoterm. Vilken väg är den adiabatiska?
Example 18-4 Work into Energy När en viss gas komprimeras adiabatiskt så görs arbetet (på gasen W<0) 640 J. Vad är ändringen i inre energin? ΔU = Q – W
Another way to ensure that a process is
effectively adiabatic is to have the volume
change occur very quickly. In this case, heat has no time to flow in or out of the system.
18-3 Thermal Processes
18-3 Thermal Processes
Here is a summary of the different types of thermal processes:
18-4 Specific Heats for an Ideal Gas:
Constant Pressure, Constant Volume
Specific heats for ideal gases must be quoted either at constant pressure or at constant
volume. For a constant-volume process,
18-4 Specific Heats for an Ideal Gas:
Constant Pressure, Constant Volume
At constant pressure,
Conceptual Checkpoint 18-3 Comparing Specific Heats
Hur är stort är det molära specifika värmet Cv (vid konstant volym), jämfört med det molära specifika värmet Cp (vid konstant tryck)? > = < ?
Hint: Se på figurerna 18-12 och 18-13 (som finns på de två föregående slides, eller i nödfall nästa)!
18-4 Specific Heats for an Ideal Gas:
Constant Pressure, Constant Volume
Both CV and CP can be calculated for a
monatomic ideal gas using the first law of thermodynamics.
18-4 Specific Heats for an Ideal Gas: Constant Pressure, Constant Volume
Både CV och CP kan beräknas för en enatomig ideal gas genom att använda första huvudsatsen (18-3) ΔU = Q – W För en process vid konstant volym är W = 0, så att Q(v) = ΔU = (17-15) = 3nRΔT/2 = (16-11)= nCvΔT Cv = 3R/2 (18-6) För en process vid konstant tryck är W = P ΔV = (ideal gas) = nR ΔT Då blir Q(p) = ΔU + W = 3nR ΔT/2 + nR ΔT = nCpΔT Cp = 5R/2 (18-7)
18-4 Specific Heats for an Ideal Gas:
Constant Pressure, Constant Volume
Although this calculation was done for an ideal, monatomic gas, it works well for real gases.
Exercise 18-2
Beräknade värmet som behövs för att värma 0,200 mol av en enatomisk, ideal gas 5,00° C a) under
konstant volym och b) under konstant tryck.
För en process vid konstant volym är W = 0, så att Q(v) = nCvΔT = (Cv = 3R/2) = = 0,200 mol•1,5•8,31 J/(mol•K) •5,00 K = 12,5 J
För en process vid konstant tryck är Q(p) = nCpΔT = (Cp = 5R/2) =
= 0,200 mol•2,5•8,31 J/(mol•K) •5,00 K = 20,8 J
18-4 Specific Heats for an Ideal Gas:
Constant Pressure, Constant Volume
The P-V curve for an adiabat is given by
where
Example 18-5 Hot Air
Example 18-5 Hot Air
En behållare med ursprungsvolymen 0,0625 m3
innehåller 2,50 mol av en enatomig ideal gas med temperaturen 315 K. Gasen sammanpressas
adiabatiskt till volymen 0,0350 m3. Vad blir då
gasens a) tryck b) temperatur?
Ideal gas, så PV = nRT gäller, således är
Pi=2,50 mol•8,31 J/(mol•K)•315 K/0,0625m3=104,7 kPa Adiabatisk process gäller dvs PVγ = konstant
PiViγ = PfVfγ
Pf = 104,7 kPa (0,0625/0,0350)5/3 = 275,2 kPa
T=275,2 kPa•0,0350 m /2,50 mol•8,31 J/(mol•K)=464 K
18-5 The Second Law of Thermodynamics
We observe that heat always flows
spontaneously from a warmer object to a
cooler one, although the opposite would not violate the conservation of energy. This
direction of heat flow is one of the ways of expressing the second law of
thermodynamics:
When objects of different temperatures are brought into thermal contact, the spontaneous flow of heat that results is always from the high temperature object to the low temperature object. Spontaneous heat flow never proceeds in the reverse direction.
18-6 Heat Engines and the Carnot Cycle
A heat engine is a device that converts heat into work. A classic example is the steam engine.
Fuel heats the water; the vapor expands and does work against the piston; the vapor
condenses back into water again and the cycle
repeats.
18-6 Heat Engines and the Carnot Cycle
All heat engines have:
• a high-temperature reservoir
• a low-temperature reservoir
• a cyclical engine
These are illustrated schematically here.
18-6 Heat Engines and the Carnot Cycle
An amount of heat Qh is supplied from the hot
reservoir to the engine during each cycle. Of that heat, some appears as work, and the rest, Qc, is given off as waste heat to the cold reservoir.
The efficiency is the fraction of the heat
supplied to the engine that appears as work.
18-6 Heat Engines and the Carnot Cycle
The efficiency can also be written:
In order for the engine to run, there must be a temperature difference; otherwise heat will not be transferred.
Example 18-6 Heat into Work
Example 18-6 Heat into Work
En värmemaskin med verkningsgraden 24,0 % utför 1250 J arbete. Bestäm Qh och Qc.
Ekvation 18-12 e = W/Qh = (Qh – Qc)/Qh = 1 – Qc/Qh
Qh = W/0,24 = 5210 J (5208 J) Qc = Qh(1 –e) = 3960 J (3958 J)
18-6 Heat Engines and the Carnot Cycle
The maximum-efficiency heat engine is described in Carnot’s theorem:
If an engine operating between two constant- temperature reservoirs is to have maximum
efficiency, it must be an engine in which all processes are reversible. In addition, all reversible engines
operating between the same two temperatures, Tc and Th, have the same efficiency.
This is an idealization; no real engine can be perfectly reversible.
18-6 Heat Engines and the Carnot Cycle
If the efficiency depends only on the two
temperatures, the ratio of the temperatures must be the same as the ratio of the transferred heats.
Therefore, the maximum efficiency of a heat engine can be written:
18-6 Heat Engines and the Carnot Cycle
The maximum work a heat engine can do is then:
If the two reservoirs are at the same
temperature, the efficiency is zero; the
smaller the ratio of the cold temperature to the hot temperature, the closer the efficiency will be to 1.
Conceptual Checkpoint 18-4 Comparing Efficiencies
Anta att en värmemaskin kan arbeta på två olika sätt.
I det ena processen arbeter den mellan
temperaturerna 400 K/200K och i det andra mellan 600 K/400K. Är verkningsgraden i den första
processen > = < än i det andra?
Ekvation 18-14 Wmax = emaxQh= (1 – Tc/Th )Qh
Active Example 18-2 Find the Temperature
Active Example 18-2 Find the Temperature
Om värmemaskinen i exemple 18-6 arbeter med maximal effektivitet och dess kalla reservoar hålls vid temperaturen 295 K, vad är då temperaturen på den varma reservoaren?
Värmemaskinens verkningsgraden 24,0 % är
maximal. Ekvation 18-14 ger då e = 1 – Tc/Th
Th = Tc/(1 – e) = 388 K
18-7 Refrigerators, Air Conditioners, and Heat Pumps
While heat will flow spontaneously only from a higher temperature to a lower one, it can be
made to flow the other way if work is done on the system. Refrigerators, air conditioners,
and heat pumps all use work to transfer heat from a cold object to a hot object.
Figure 18-17
Cooling and heating devices
18-7 Refrigerators, Air Conditioners, and Heat Pumps
If we compare the heat engine and the refrigerator, we see that the refrigerator is basically a heat engine running
backwards – it uses work to extract heat from the cold
reservoir (the inside of the refrigerator) and exhausts to the kitchen. Note that
- more heat is exhausted to the kitchen than is removed from the refrigerator.
18-7 Refrigerators, Air Conditioners, and Heat Pumps
An ideal refrigerator would remove the most heat from the interior while requiring the
smallest amount of work. This ratio is called the coefficient of performance, COP:
Typical refrigerators have COP values between 2 and 6. Bigger is better!
Exercise 18-3
Ett kylskåp har ett COP-värde av 2,5. Hur mycket arbete måste tillföras kylskåpet för att
värmemängden 225 J skall föras bort?
Ekvation 18-15 ger W = Qc/COP = 225 J/2,5 = 90 J
(dvs 225 J + 90 J värme i köket)
18-7 Refrigerators, Air Conditioners, and Heat Pumps
An air conditioner is
essentially identical to a
refrigerator; the cold reservoir is the interior of the house or other space being cooled, and the hot reservoir is outdoors.
Exhausting an air conditioner within the house will result in the house becoming warmer, just as keeping the refrigerator door open will result in the
kitchen becoming warmer.
Conceptual Checkpoint 18-5 Room Temperature
Tidsbrist har gjort att en ”air conditioner” tillfälligt ställs på matsalsbordet i väntan på en placering på sin rätta plats. Blir nu matsalen
varmare/kallare/ingen skillnad?
(Jämför med ”kylskåpstalet”)
18-7 Refrigerators, Air Conditioners, and Heat Pumps
Finally, a heat pump is the same as an air conditioner, except with the reservoirs reversed. Heat is removed from the cold reservoir
outside, and exhausted into the house, keeping it warm. Note that the work the pump does actually contributes to the desired result (a warmer house) in this case.
Example 18-7 Pumping Heat
18-7 Refrigerators, Air Conditioners, and Heat Pumps
In an ideal heat pump with two operating
temperatures (cold and hot), the Carnot relationship holds; the work needed to add heat Qh to a room is:
The COP for a heat pump:
Example 18-7 Pumping Heat + Exercise 18-4
En ideal värmepump som uppfyller Carnotrelationen W = Qh – Qc = Qh(1 - Qc/Qh) = Qh(1 – Tc/Th) = 275 J
används för att värma ett rum till 293 K. Hur mycket värme tillförs rummet om utomhustemperaturen är a) 273 k?
Qh = W/(1-Tc / Th) = 4,03 kJ (COP ≈ 15) b) 263 K?
Qh = p.s.s. = 2,69 kJ (COP ≈ 10)
Exercise 18-4 Värmepump med COP ≡ Qh/W = 3,5 tillför 2,5 kJ. W = ? (714 J)
18-8 Entropy
A reversible engine has the following relation between the heat transferred and the reservoir temperatures
(Carnotrelationen):
Rewriting,
This quantity, Q/T, is the same for both reservoirs, and is defined as the change in entropy.
18-8 Entropy
For this definition to be valid, the heat transfer must be reversible.
In a reversible heat engine, it can be shown that the entropy does not change.
Example 18-8 Melts in Your Hand Beräkna entropiändringen då 0,125 kg is av 0°C smälter. Anta att smältningen är reversibel.
Q = mL = 0,125 kg•335 kJ/kg ΔS=Q/T=41875 J/273 K= 153 J/K
18-8 Entropy
A real engine will operate at a lower efficiency than a reversible engine; this means that less heat is converted to work. Therefore,
Any irreversible process results in an increase of entropy.
18-8 Entropy
To generalize:
• The total entropy of the universe increases whenever an irreversible process occurs.
• The total entropy of the universe is unchanged whenever a reversible process occurs.
Since all real processes are irreversible, the
entropy of the universe continually increases. If entropy decreases in a system due to work
being done on it, a greater increase in entropy occurs outside the system.
Example 18-9 Entropy Is Not Conserved! Beräkna entropiändringen nedan ΔSh = - Q/Th = - 1,82 J/K; ΔSc = Q/Tc = 3,44 J/K; ΔS = ΔSh + ΔSc = 1,62 J/K
18-8 Entropy
As the total entropy of the universe
increases, its ability to do work decreases.
The excess heat exhausted during an
irreversible process cannot be recovered;
doing that would require a decrease in entropy, which is not possible.
Active Example 18-3 Find the Work
Active Example 18-3 Find the Work
En reversibel värmemaskin arbetar mellan
temperaturerna Th och Tc enligt figur. Bestäm W då Qh= 1050 J.
W = Qh(1 – Tc/Th) = 494 J
Lägg märke till att ΔS = ΔSh + ΔSc = 0 (som väntat?!) ΔSh = -1050 J/576 K = - 1,82 J/K
ΔSc = 556 J/305 K = 1,82 J/K
18-9 Order, Disorder, and Entropy
Entropy can be thought of as the increase in disorder in the universe. In this diagram, the end state is less ordered than the initial state – the separation between low and high
temperature areas has been lost.
18-9 Order, Disorder, and Entropy
If we look at the ultimate fate of the universe in light of the continual increase in entropy, we might envision a future in which the entire universe would have come to the same
temperature. At this point, it would no longer be possible to do any work, nor would any type of life be possible. This is referred to as the “heat death” of the universe.
18-9 Order, Disorder, and Entropy
So if entropy is continually increasing, how is life possible? How is it that species can evolve into ever more complex forms? Doesn’t this
violate the second law of thermodynamics?
No – life and increasing complexity can exist because they use energy to drive their
functioning. The overall entropy of the universe is still increasing. When a living entity stops
using energy, it dies, and its entropy can increase rather quickly.
18-10 The Third Law of Thermodynamics
Absolute zero is a temperature that an object can get arbitrarily close to, but never attain.
Temperatures as low as 2.0 x 10-8 K have been achieved in the laboratory, but absolute zero will remain ever elusive – there is simply nowhere to
“put” that last little bit of energy.
This is the third law of thermodynamics:
It is impossible to lower the temperature of an object to absolute zero in a finite number of steps.
Summary of Chapter 18
• When two objects have the same temperature, they are in thermal equilibrium.
• The first law of thermodynamics is a statement of energy conservation that includes heat.
•
• The internal energy of a system depends only on its temperature, pressure, and volume.
• A quasi-static process is one in which the
system may be considered to be in equilibrium at all times.
Summary of Chapter 18
• In a reversible process it is possible to return the system and its surroundings to their initial states.
• Irreversible processes cannot be undone.
• The work done during a process is equal to the area under the curve in the PV plot.
• The work done at constant pressure is
• The work done at constant volume is zero.
• The work done in an isothermal expansion is
Summary of Chapter 18
• An adiabatic process is one where no heat transfer occurs.
• The value of the specific heat depends on
whether it is at constant pressure or at constant volume.
• Molar specific heat is defined by:
• For a monatomic gas at constant volume:
• For a monatomic gas at constant pressure:
Summary of Chapter 18
• In a PV plot, is constant, where
• For a monatomic ideal gas,
• The spontaneous flow of heat between objects in thermal equilibrium is always from the hotter one to the colder one.
• A heat engine converts heat into work.
• Efficiency of a heat engine:
Summary of Chapter 18
• A reversible engine has the maximum possible efficiency,
• The maximum possible work:
• Refrigerators, air conditioners, and heat pumps use work to transfer heat from a cold region to a hot region.
Summary of Chapter 18
• Coefficient of performance of a refrigerator:
• Work done by an ideal heat pump:
• Coefficient of performance for a heat pump:
Summary of Chapter 18
• Change of entropy during a reversible heat exchange:
• Total entropy of the universe increases
whenever an irreversible process occurs; total entropy is unchanged after an ideal reversible process.
• Entropy is a measure of disorder.
• The heat death of the universe will occur when everything is the same temperature and no more work can be done.
Summary of Chapter 18
• It is impossible to lower the temperature of an object to absolute zero in a finite number of
steps.
