Termodinamika

Thermodynamic Systems andThermodynamic Systems and
Their SurroundingsTheir Surroundings
ThermodynamicsThermodynamics is the branch of physicsis the branch of physics
that is built upon the fundamental laws thatthat is built upon the fundamental laws that
heat and work obey.heat and work obey.

ThermodynamicsThermodynamics
 In thermodynamics the collection of objectsIn thermodynamics the collection of objects
upon which attention is being focused isupon which attention is being focused is
called thecalled the system,system, while everything else in thewhile everything else in the
environment is called theenvironment is called the surroundings.surroundings.
 The system and its surroundings are separatedThe system and its surroundings are separated
by walls of some kind. Walls that permit heatby walls of some kind. Walls that permit heat
to flow through them, such as those of theto flow through them, such as those of the
engine block, are calledengine block, are called diathermal walls.diathermal walls.
Perfectly insulating walls that do not permitPerfectly insulating walls that do not permit
heat to flow between the system and itsheat to flow between the system and its
surroundings are calledsurroundings are called adiabatic walls.adiabatic walls.

State of a SystemState of a System
 To understand what the laws ofTo understand what the laws of
thermodynamics have to say about thethermodynamics have to say about the
relationship between heat and work, it isrelationship between heat and work, it is
necessary to describe the physical conditionnecessary to describe the physical condition
oror state of a system.state of a system.
 The state of the system would be specifiedThe state of the system would be specified
by giving values for the pressure, volume,by giving values for the pressure, volume,
temperature, and mass of the hot air.temperature, and mass of the hot air.

The Zeroth Law ofThe Zeroth Law of
 (a) Systems A and B are(a) Systems A and B are
surrounded by adiabaticsurrounded by adiabatic
walls and register the samewalls and register the same
temperature on thetemperature on the
thermometer. (b) When A isthermometer. (b) When A is
put into thermal contact withput into thermal contact with
B through diathermal walls,B through diathermal walls,
no net flow of heat occursno net flow of heat occurs
between the systems.between the systems.

Two systems individually in thermalTwo systems individually in thermal
equilibrium with a third systemequilibrium with a third system are in thermalare in thermal
equilibrium with each other.equilibrium with each other.

Temperature is the indicator of thermalTemperature is the indicator of thermal
equilibrium in the sense that there is noequilibrium in the sense that there is no
net flow of heat between two systems innet flow of heat between two systems in
thermal contact that have the samethermal contact that have the same
temperature.temperature.
There can be no flow of heat within aThere can be no flow of heat within a
system in thermal equilibrium.system in thermal equilibrium.

The First Law ofThe First Law of

 The internal energy of a system changesThe internal energy of a system changes
from an initial valuefrom an initial value UUii to a final value ofto a final value of UUff
due to heatdue to heat QQ and workand work WW::
∆∆UU == UUff -- UUii == Q – WQ – W
 Q is positive when the system gains heatQ is positive when the system gains heat
and negative when it loses heat. W isand negative when it loses heat. W is
positive when work is done by the systempositive when work is done by the system
and negative when work is done on theand negative when work is done on the
system.system.
The First Law ofThe First Law of

The Internal EnergyThe Internal Energy
 The internal energy depends only on theThe internal energy depends only on the
state of a system, not on the method bystate of a system, not on the method by
which the system arrives at a given state.which the system arrives at a given state.

EXAMPLE : An Ideal GasEXAMPLE : An Ideal Gas
 The temperature of three moles of aThe temperature of three moles of a
monatomic ideal gas is reduced frommonatomic ideal gas is reduced from TTii ==
540 K to540 K to TTff = 350 K by two different= 350 K by two different
methods. In the first method 5500 J of heatmethods. In the first method 5500 J of heat
flows into the gas, while in the second,flows into the gas, while in the second,
1500 J of heat flows into it. In each case1500 J of heat flows into it. In each case
find (a) the change in the internal energyfind (a) the change in the internal energy
and (b) the work done by the gas.and (b) the work done by the gas.

SolutionSolution
 Since the internal energy of a monatomic ideal gasSince the internal energy of a monatomic ideal gas
isis UU = (= (33
//22)) nRTnRT and since the number of molesand since the number of moles nn
is fixed, only a change in temperatureis fixed, only a change in temperature TT can altercan alter
the internal energy.the internal energy.
 Since the change inSince the change in TT is the same in both methods,is the same in both methods,
the change inthe change in UU is also the same. From the givenis also the same. From the given
temperatures, the changetemperatures, the change ∆∆UU in internal energyin internal energy
can be determined. Then, the first law ofcan be determined. Then, the first law of
thermodynamics can be used withthermodynamics can be used with ∆∆UU and theand the
given heat values to calculate the work.given heat values to calculate the work.

(a)(a) Using EquationUsing Equation UU = (= (33
//22)) nRTnRT for thefor the
internal energy of a monatomic ideal gas,internal energy of a monatomic ideal gas,
we find for each method of adding heat thatwe find for each method of adding heat that
(b)(b) SinceSince ∆∆UU is now known and the heat isis now known and the heat is
given in each method,given in each method, ∆∆UU == Q – WQ – W can becan be
used to determine the work:used to determine the work:
SolutionSolution

Thermal ProcessesThermal Processes
 A system can interact with its surroundingsA system can interact with its surroundings
in many ways, and the heat and work thatin many ways, and the heat and work that
come into play always obey the first law ofcome into play always obey the first law of
thermodynamics.thermodynamics.
 thermal process is assumed to bethermal process is assumed to be quasi-quasi-
static,static, which means that it occurs slowlywhich means that it occurs slowly
enough that a uniform pressure andenough that a uniform pressure and
temperature exist throughout all regions oftemperature exist throughout all regions of
the system at all times.the system at all times.

An Isobaric ProcessAn Isobaric Process
The substance in the chamberThe substance in the chamber
is expanding isobaricallyis expanding isobarically
because the pressure is heldbecause the pressure is held
constant by the externalconstant by the external
atmosphere and the weight ofatmosphere and the weight of
the piston and the block.the piston and the block.
An isobaric process is oneAn isobaric process is one
that occurs at constantthat occurs at constant
pressure.pressure.

 The Figure shows a substance (solid, liquid, orThe Figure shows a substance (solid, liquid, or
gas) contained in a chamber fitted with agas) contained in a chamber fitted with a
frictionless piston.frictionless piston.
 The pressureThe pressure PP experienced by the substance isexperienced by the substance is
always the same and is determined by the externalalways the same and is determined by the external
atmosphere and the weight of the piston and theatmosphere and the weight of the piston and the
block resting on it.block resting on it.
 Heating the substance makes it expand and doHeating the substance makes it expand and do
workwork WW in lifting the piston and block through thein lifting the piston and block through the
displacementdisplacement ss..

 The work can be calculated fromThe work can be calculated from W = FsW = Fs, where, where FF
is the magnitude of the force andis the magnitude of the force and ss is theis the
magnitude of the displacement.magnitude of the displacement.
 The force is generated by the pressureThe force is generated by the pressure PP acting onacting on
the bottom surface of the piston (area =the bottom surface of the piston (area = AA),),
according toaccording to F = PAF = PA..
 With this substitution forWith this substitution for FF, the work becomes, the work becomes WW
= (= (PAPA))ss. But the product. But the product AA ·· ss is the change inis the change in
volume of the material,volume of the material, ∆∆V = VV = Vff -- VVii, where, where VVff andand
VVii are the final and initial volumes, respectively.are the final and initial volumes, respectively.

 Thus, the expression for the work isThus, the expression for the work is
Isobaric processIsobaric process
W = PW = P∆∆V = P(VV = P(Vff – V– Vii ))
 A positive value for the work done by aA positive value for the work done by a
system when it expands isobarically (Vsystem when it expands isobarically (Vff
exceeds Vexceeds Vii).).
 That Equation also applies to an isobaricThat Equation also applies to an isobaric
compression (Vcompression (Vff less than Vless than Vii). Then, the). Then, the
work is negative, since work must be donework is negative, since work must be done
on the system to compress it.on the system to compress it.

For an isobaric process, aFor an isobaric process, a
pressure-versus-volumepressure-versus-volume
plot is a horizontal straightplot is a horizontal straight
line, and the work doneline, and the work done
[[WW == PP((VVff
22
–– VVii
22
)] is the)] is the
colored rectangular areacolored rectangular area
under the graph.under the graph.

An Isochoric ProcessAn Isochoric Process
a) The substance in the chambera) The substance in the chamber
is being heated isochoricallyis being heated isochorically
because the rigid chamber keepsbecause the rigid chamber keeps
the volume constant. (b) Thethe volume constant. (b) The
pressure–volume plot for anpressure–volume plot for an
isochoric process is a verticalisochoric process is a vertical
straight line. The area under thestraight line. The area under the
graph is zero, indicating that nograph is zero, indicating that no
work is done.work is done.
an isochoric process, one thatan isochoric process, one that
occurs at constant volume.occurs at constant volume.

 FigureFigure aa illustrates an isochoric process inillustrates an isochoric process in
which a substance (solid, liquid, or gas) iswhich a substance (solid, liquid, or gas) is
heated. The substance would expand if itheated. The substance would expand if it
could, but the rigid container keeps thecould, but the rigid container keeps the
volume constant, so the pressure–volumevolume constant, so the pressure–volume
plot shown in Figureplot shown in Figure bb is a vertical straightis a vertical straight
lineline
 Because the volume is constant, theBecause the volume is constant, the
pressure inside rises, and the substancepressure inside rises, and the substance
exerts more and more force on the walls.exerts more and more force on the walls.

 While enormous forces can be generated inWhile enormous forces can be generated in
the closed container, no work is done, sincethe closed container, no work is done, since
the walls do not move. Consistent with zerothe walls do not move. Consistent with zero
work being done, the area under the verticalwork being done, the area under the vertical
straight line in Figurestraight line in Figure bb is zero.is zero.
 Since no work is done, the first law ofSince no work is done, the first law of
thermodynamics indicates that the heat inthermodynamics indicates that the heat in
an isochoric process serves only to changean isochoric process serves only to change
the internal energy:the internal energy: ∆∆U = Q - W = QU = Q - W = Q..

An Isothermal ProcessAn Isothermal Process
The colored area gives theThe colored area gives the
work done by the gas for thework done by the gas for the
process fromprocess from XX toto YY..
 A third important thermalA third important thermal
process is anprocess is an isothermalisothermal
process, one that takes place atprocess, one that takes place at
constant temperature.constant temperature.

The Adiabatic ProcessThe Adiabatic Process
 Last, there is theLast, there is the adiabatic process, oneadiabatic process, one
that occurs without the transfer of heat.that occurs without the transfer of heat.
 Since there is no heat transfer,Since there is no heat transfer, QQ equalsequals
zero, and the first law indicates thatzero, and the first law indicates that
∆∆U = Q - W = -WU = Q - W = -W..
 Thus, when work is done by a systemThus, when work is done by a system
adiabatically, the internal energy of theadiabatically, the internal energy of the
system decreases by exactly the amount ofsystem decreases by exactly the amount of
the work done.the work done.

 Thus, when work is done by a systemThus, when work is done by a system
adiabatically, the internal energy of theadiabatically, the internal energy of the
system decreases by exactly the amount ofsystem decreases by exactly the amount of
the work done. When work is done on athe work done. When work is done on a
system adiabatically, the internal energysystem adiabatically, the internal energy
increases correspondingly.increases correspondingly.
The Adiabatic ProcessThe Adiabatic Process

Thermal Processes ThatThermal Processes That
Utilize an Ideal GasUtilize an Ideal Gas

Isothermal Expansion orIsothermal Expansion or
CompressionCompression
(a)(a) The ideal gas in the cylinder is expandingThe ideal gas in the cylinder is expanding
isothermally at temperatureisothermally at temperature TT. The force holding. The force holding
the piston in place is reduced slowly, so thethe piston in place is reduced slowly, so the
expansion occurs quasi-statically.expansion occurs quasi-statically. (b)(b) The workThe work
done by the gas is given by the colored area.done by the gas is given by the colored area.

 When a system performs work isothermally,When a system performs work isothermally,
the temperature remains constant.the temperature remains constant.
 In FigureIn Figure aa, for instance, a metal cylinder, for instance, a metal cylinder
containscontains nn moles of an ideal gas, and themoles of an ideal gas, and the
large mass of hot water maintains thelarge mass of hot water maintains the
cylinder and gas at a constant Kelvincylinder and gas at a constant Kelvin
temperaturetemperature TT. The piston is held in place. The piston is held in place
initially so the volume of the gas isinitially so the volume of the gas is VVii..

 As the external force applied to the piston is reducedAs the external force applied to the piston is reduced
quasi-statically, the gas expands to the final volumequasi-statically, the gas expands to the final volume
VVff ..
 Figure b gives a plot of pressure (Figure b gives a plot of pressure (P = nRTP = nRT // VV) versus) versus
volume for the process.volume for the process.
 The solid red line in the graph is called an isothermThe solid red line in the graph is called an isotherm
(meaning “constant temperature”) because it(meaning “constant temperature”) because it
represents the relation between pressure and volumerepresents the relation between pressure and volume
when the temperature is held constant.when the temperature is held constant.

 The workThe work WW done by the gas isdone by the gas is notnot given bygiven by
W = PW = P∆∆V = PV = P((VVff -- VVii) because the pressure) because the pressure
is not constant.is not constant.
 Nevertheless, the work is equal to the areaNevertheless, the work is equal to the area
under the graph. The techniques of integralunder the graph. The techniques of integral
calculus lead to the following resultcalculus lead to the following result forfor WW::






=
i
f
V
V
nRTW ln

 Where does the energy for this workWhere does the energy for this work
originate?originate?
 Since the internal energy of any ideal gas isSince the internal energy of any ideal gas is
proportional to the Kelvin temperature (proportional to the Kelvin temperature (UU==
((33
//22)) nRTnRT for a monatomic ideal gas, forfor a monatomic ideal gas, for
example), the internal energy remainsexample), the internal energy remains
constant throughout an isothermal process,constant throughout an isothermal process,
and the change in internal energy is zero.and the change in internal energy is zero.

 The first law of thermodynamics becomesThe first law of thermodynamics becomes
∆∆UU = 0 == 0 = Q - WQ - W. In other words,. In other words, Q = WQ = W,,
and the energy for the work originates inand the energy for the work originates in
the hot water.the hot water.

Adiabatic Expansion orAdiabatic Expansion or
a)a) The ideal gas in the cylinder is expanding adiabatically.The ideal gas in the cylinder is expanding adiabatically.
The force holding the piston in place is reduced slowly, soThe force holding the piston in place is reduced slowly, so
the expansion occurs quasi-statically.the expansion occurs quasi-statically. (b)(b) A plot ofA plot of
pressure versus volume yields the adiabatic curve shown inpressure versus volume yields the adiabatic curve shown in
red, which intersects the isotherms (blue) at the initialred, which intersects the isotherms (blue) at the initial
temperaturetemperature TTii and the final temperatureand the final temperature TTff . The work done. The work done
by the gas is given by the colored area.by the gas is given by the colored area.

 When a system performs work adiabatically,When a system performs work adiabatically,
no heat flows into or out of the system.no heat flows into or out of the system.
 FigureFigure aa shows an arrangement in whichshows an arrangement in which nn
moles of an ideal gas do work undermoles of an ideal gas do work under
adiabatic conditions, expanding quasi-adiabatic conditions, expanding quasi-
statically from an initial volumestatically from an initial volume VVii to a finalto a final
volumevolume VVff..
 The arrangement is similar to that in FigureThe arrangement is similar to that in Figure
for isothermal expansion.for isothermal expansion.

 However, a different amount of work is done here,However, a different amount of work is done here,
because the cylinder is now surrounded by insulatingbecause the cylinder is now surrounded by insulating
material that prevents the flow of heat, somaterial that prevents the flow of heat, so QQ = 0.= 0.
According to the first law of thermodynamics, theAccording to the first law of thermodynamics, the
change in internal energy ischange in internal energy is ∆∆U = Q - W = -WU = Q - W = -W..
 Since the internal energy of an ideal monatomic gas isSince the internal energy of an ideal monatomic gas is
UU= (= (33
//22))nRTnRT , it follows that, it follows that ∆∆U =U = ((33
//22))nR(TnR(Tff – T– Tii)) ,,
wherewhere TTii andand TTff are the initial and final Kelvinare the initial and final Kelvin
temperatures. With this substitution, the relationtemperatures. With this substitution, the relation ∆∆UU
= -= -WW becomesbecomes

 When an ideal gas expands adiabatically, it doesWhen an ideal gas expands adiabatically, it does
positive work, sopositive work, so WW is positive.is positive.
 Therefore, the termTherefore, the term TTii -- TTff is also positive, and theis also positive, and the
final temperature of the gas must be less than thefinal temperature of the gas must be less than the
initial temperature.initial temperature.
 The internal energy of the gas is reduced toThe internal energy of the gas is reduced to
provide the necessary energy to do the work, andprovide the necessary energy to do the work, and
because the internal energy is proportional to thebecause the internal energy is proportional to the
Kelvin temperature, the temperature decreases.Kelvin temperature, the temperature decreases.

 FigureFigure bb shows a plot of pressure versusshows a plot of pressure versus
volume for the adiabatic process. Thevolume for the adiabatic process. The
adiabatic curve (red) intersects theadiabatic curve (red) intersects the
isotherms (blue) at the higher initialisotherms (blue) at the higher initial
temperature [temperature [TTii == PPiiVVii/(/(nRnR)] and the lower)] and the lower
final temperature [final temperature [TTff == PPffVVff/(/(nRnR)]. The)]. The
colored area under the adiabatic curvecolored area under the adiabatic curve
represents the work done.represents the work done.

 The equation that gives the adiabatic curveThe equation that gives the adiabatic curve
(red) between the initial pressure and(red) between the initial pressure and
volume (volume (PPii,, VVii) and the final pressure and) and the final pressure and
volume (volume (PPff,, VVff) in Figure) in Figure bb can be derivedcan be derived
using integral calculus. The result isusing integral calculus. The result is
 where the exponentwhere the exponent γγ is the ratio of theis the ratio of the
specific heat capacities at constant pressurespecific heat capacities at constant pressure
and constant volume,and constant volume, γγ == ccPP//ccVV..

THE SECOND LAW OFTHE SECOND LAW OF
THERMODYNAMICSTHERMODYNAMICS

THERMODYNAMICS: THE HEAT FLOWTHERMODYNAMICS: THE HEAT FLOW
STATEMENTSTATEMENT
Heat flows spontaneously from a substance at aHeat flows spontaneously from a substance at a
higher temperature to a substance at a lowerhigher temperature to a substance at a lower
temperature and does not flow spontaneously intemperature and does not flow spontaneously in
the reverse directionthe reverse direction

It is important to realize that the second lawIt is important to realize that the second law
of thermodynamics deals with a differentof thermodynamics deals with a different
aspect of nature than does the first law ofaspect of nature than does the first law of
thermodynamics.thermodynamics.
The second law is a statement about theThe second law is a statement about the
natural tendency of heat to flow from hot tonatural tendency of heat to flow from hot to
cold, whereas the first law deals with energycold, whereas the first law deals with energy
conservation and focuses on both heat andconservation and focuses on both heat and
work.work.

AA heat engineheat engine is any device that uses heat to performis any device that uses heat to perform
work. It has three essential features:work. It has three essential features:
1
.
Heat is supplied to the engine at a relatively highHeat is supplied to the engine at a relatively high
temperature from a place called thetemperature from a place called the hot reservoir.hot reservoir.
2
.
Part of the input heat is used to perform work by thePart of the input heat is used to perform work by the
working substanceworking substance of the engine, which is the materialof the engine, which is the material
within the engine that actually does the work (e.g., thewithin the engine that actually does the work (e.g., the
gasoline–air mixture in an automobile engine).gasoline–air mixture in an automobile engine).
3
.
The remainder of the input heat is rejected at aThe remainder of the input heat is rejected at a
temperature lower than the input temperature to a placetemperature lower than the input temperature to a place
called thecalled the cold reservoircold reservoir

This schematic representationThis schematic representation
of a heat engine shows the inputof a heat engine shows the input
heat (magnitude =heat (magnitude = QQHH) that) that
originates from the hotoriginates from the hot
reservoir, the work (magnitudereservoir, the work (magnitude
== WW) that the engine does, and) that the engine does, and
the heat (magnitude =the heat (magnitude = QQCC ) that) that
the engine rejects to the coldthe engine rejects to the cold
reservoir.reservoir.

To be highly efficient, a heat engine must produceTo be highly efficient, a heat engine must produce
a relatively large amount of work from as littlea relatively large amount of work from as little
input heat as possible. Thus, theinput heat as possible. Thus, the efficiency eefficiency e of aof a
heat engine is defined as the ratio of the workheat engine is defined as the ratio of the work WW
done by the engine to the input heatdone by the engine to the input heat QQHH::
If the input heat were converted entirely intoIf the input heat were converted entirely into
work, the engine would have an efficiency ofwork, the engine would have an efficiency of
1.00, since1.00, since W = QW = QHH; such an engine would be; such an engine would be
100% efficient.100% efficient.

What is it that allows a heat engine toWhat is it that allows a heat engine to
operate with maximum efficiency?operate with maximum efficiency?
The French engineer Sadi Carnot (1796–1832)The French engineer Sadi Carnot (1796–1832)
proposed that a heat engine has maximumproposed that a heat engine has maximum
efficiency when the processes within the engineefficiency when the processes within the engine
are reversible.are reversible. A reversible process is one inA reversible process is one in
which both the system and its environmentwhich both the system and its environment
can be returned to exactly the states they werecan be returned to exactly the states they were
in before the process occurred.in before the process occurred.

In a reversible process,In a reversible process, bothboth the system andthe system and
its environment can be returned to their initialits environment can be returned to their initial
states.states.
A process that involves an energy-dissipatingA process that involves an energy-dissipating
mechanism, such as friction, cannot bemechanism, such as friction, cannot be
reversible because the energy wasted due toreversible because the energy wasted due to
friction would alter the system or thefriction would alter the system or the
environment or both.environment or both.

CARNOT'S PRINCIPLE: ANCARNOT'S PRINCIPLE: AN
ALTERNATIVE STATEMENT OF THEALTERNATIVE STATEMENT OF THE
SECOND LAW OF THERMODYNAMICSSECOND LAW OF THERMODYNAMICS
No irreversible engine operating between twoNo irreversible engine operating between two
reservoirs at constant temperatures can have a greaterreservoirs at constant temperatures can have a greater
efficiency than a reversible engine operating betweenefficiency than a reversible engine operating between
the same temperatures. Furthermore, all reversiblethe same temperatures. Furthermore, all reversible
engines operating between the same temperaturesengines operating between the same temperatures
have the same efficiency.have the same efficiency.

Natural Limits on the Efficiency of aNatural Limits on the Efficiency of a
Heat EngineHeat Engine
Consider a hypothetical engine that receives 1000 JConsider a hypothetical engine that receives 1000 J
of heat as input from a hot reservoir and deliversof heat as input from a hot reservoir and delivers
1000 J of work, rejecting no heat to a cold reservoir1000 J of work, rejecting no heat to a cold reservoir
whose temperature is above 0 K. Decide whetherwhose temperature is above 0 K. Decide whether
this engine violates the first or the second law ofthis engine violates the first or the second law of
thermodynamics, or both.thermodynamics, or both.

The first law of thermodynamics is an expressionThe first law of thermodynamics is an expression
of energy conservation. From the point of view ofof energy conservation. From the point of view of
energy conservation, nothing is wrong with anenergy conservation, nothing is wrong with an
engine that converts 1000 J of heat into 1000 J ofengine that converts 1000 J of heat into 1000 J of
work.work.
Energy has been neither created nor destroyed; itEnergy has been neither created nor destroyed; it
has only been transformed from one form (heat) tohas only been transformed from one form (heat) to
another (work).another (work).

This engine does, however, violate the secondThis engine does, however, violate the second
law of thermodynamics. Since all of the inputlaw of thermodynamics. Since all of the input
heat is converted into work, the efficiency of theheat is converted into work, the efficiency of the
engine is 1, or 100%.engine is 1, or 100%.
Since we know thatSince we know that TTCC is above 0 K, it is clearis above 0 K, it is clear
that the ratiothat the ratio TTCC//TTHH is greater than zero, so theis greater than zero, so the
maximum possible efficiency is less than 1, ormaximum possible efficiency is less than 1, or
less than 100%.less than 100%.

A Heat PumpA Heat Pump
An ideal or Carnot heat pump is used toAn ideal or Carnot heat pump is used to
heat a house to a temperature ofheat a house to a temperature of TTHH = 294= 294
K (21 °C). How much work must be doneK (21 °C). How much work must be done
by the pump to deliverby the pump to deliver QQHH = 3350 J of heat= 3350 J of heat
into the house when the outdoorinto the house when the outdoor
temperaturetemperature TTCC is (a) 273 K (0 °C) and (b)is (a) 273 K (0 °C) and (b)
252 K (-21 °C)?252 K (-21 °C)?

The conservation of energy (The conservation of energy (QQHH == WW ++ QQCC))
applies to the heat pump. Thus, the work can beapplies to the heat pump. Thus, the work can be
determined fromdetermined from W = QW = QHH -- QQCC, provided we can, provided we can
obtain a value forobtain a value for QQCC, the heat taken by the pump, the heat taken by the pump
from the outside. To determinefrom the outside. To determine QQCC, we use the fact, we use the fact
that the pump is a Carnot heat pump and operatesthat the pump is a Carnot heat pump and operates
reversibly. Therefore, the relationreversibly. Therefore, the relation QQCC//QQHH == TTCC//TTHH
applies. Solving it forapplies. Solving it for QQCC, we obtain, we obtain QQCC == QQHH((TTCC//TTHH).).
Using this result, we find thatUsing this result, we find that
ReasoningReasoning

SolutionSolution
(a)(a) At an indoor temperature ofAt an indoor temperature of TTHH = 294 K and an= 294 K and an
outdoor temperature ofoutdoor temperature of TTCC = 273 K, the work needed= 273 K, the work needed
isis
(b)(b) This solution is identical to that in part (a),This solution is identical to that in part (a),
except that it is now cooler outside, soexcept that it is now cooler outside, so TTCC = 252 K.= 252 K.
The necessary work is , which is more thanThe necessary work is , which is more than
in part (a).in part (a).

To introduce the idea of entropy we recall theTo introduce the idea of entropy we recall the
relationrelation QQCC//QQHH == TTCC//TTHH that applies to a Carnotthat applies to a Carnot
engine. This equation can be rearranged asengine. This equation can be rearranged as QQCC//TTCC ==
QQHH//TTHH, which focuses attention on the heat, which focuses attention on the heat QQ divideddivided
by the Kelvin temperatureby the Kelvin temperature TT. The quantity. The quantity QQ//TT isis
called the change in the entropycalled the change in the entropy ∆∆SS::
ENTROPYENTROPY
the SI unit for entropy is a joule per kelvin (J/K).

THERMODYNAMICS STATED INTHERMODYNAMICS STATED IN
TERMS OF ENTROPYTERMS OF ENTROPY
The total entropy of the universe does notThe total entropy of the universe does not
change when a reversible process occurschange when a reversible process occurs
((∆∆SSuniverseuniverse = 0 ) and increases when an= 0 ) and increases when an
irreversible process occurs (irreversible process occurs (∆∆SSuniverseuniverse 0 ).0 ).

The Entropy of the Universe IncreasesThe Entropy of the Universe Increases
This Figure shows 1200 J ofThis Figure shows 1200 J of
heat flowing spontaneouslyheat flowing spontaneously
through a copper rod from a hotthrough a copper rod from a hot
reservoir at 650 K to a coldreservoir at 650 K to a cold
reservoir at 350 K. Determinereservoir at 350 K. Determine
the amount by which thisthe amount by which this
irreversible process changes theirreversible process changes the
entropy of the universe,entropy of the universe,
assuming that no other changesassuming that no other changes
occur.occur.

ReasoningReasoning
The hot-to-cold heat flow is irreversible, soThe hot-to-cold heat flow is irreversible, so
the relationthe relation ∆∆SS = (= (QQ//TT))RR is applied to ais applied to a
hypothetical process whereby the 1200 J ofhypothetical process whereby the 1200 J of
heat is taken reversibly from the hotheat is taken reversibly from the hot
reservoir and added reversibly to the coldreservoir and added reversibly to the cold
reservoir.reservoir.

SolutionSolution
The total entropy change of the universe is theThe total entropy change of the universe is the
algebraic sum of the entropy changes for eachalgebraic sum of the entropy changes for each
reservoir:reservoir:

Order to DisorderOrder to Disorder
Find the change in entropy that results when aFind the change in entropy that results when a
2.3-kg block of ice melts slowly (reversibly)2.3-kg block of ice melts slowly (reversibly)
at 273 K (0 °C).at 273 K (0 °C).

ReasoningReasoning
Since the phase change occurs reversibly at a constantSince the phase change occurs reversibly at a constant
temperature, the change in entropy can be found bytemperature, the change in entropy can be found by
using,using, ∆∆SS = (= (QQ//TT))RR, where, where QQ is the heat absorbed byis the heat absorbed by
the melting ice. This heat can be determined by usingthe melting ice. This heat can be determined by using
the relationthe relation Q = mLQ = mLff, where, where mm is the mass andis the mass and LLff ==
3.35 × 103.35 × 1055
J/kg is the latent heat of fusion of water.J/kg is the latent heat of fusion of water.
& Solution& Solution

THE THIRD LAW OF THERMODYNAMICSTHE THIRD LAW OF THERMODYNAMICS
It is not possible to lower the temperature of anyIt is not possible to lower the temperature of any
system to absolute zero in a finite number ofsystem to absolute zero in a finite number of
steps.steps.

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