termodinamika

TERMODINAMIKATERMODINAMIKA

Thermodynamic Systems and Thermodynamic Systems and Their SurroundingsTheir Surroundings

ThermodynamicsThermodynamics is the branch of physics is the branch of physics that is built upon the fundamental laws that that is built upon the fundamental laws that heat and work obey.heat and work obey.

ThermodynamicsThermodynamics

In thermodynamics the collection of objects In thermodynamics the collection of objects upon which attention is being focused is upon which attention is being focused is called the called the system,system, while everything else in the while everything else in the environment is called the environment is called the surroundings.surroundings.

The system and its surroundings are separated The system and its surroundings are separated by walls of some kind. Walls that permit heat by walls of some kind. Walls that permit heat to flow through them, such as those of the to flow through them, such as those of the engine block, are called engine block, are called diathermal walls.diathermal walls. Perfectly insulating walls that do not permit Perfectly insulating walls that do not permit heat to flow between the system and its heat to flow between the system and its surroundings are called surroundings are called adiabatic walls.adiabatic walls.

State of a SystemState of a System

To understand what the laws of To understand what the laws of thermodynamics have to say about the thermodynamics have to say about the relationship between heat and work, it is relationship between heat and work, it is necessary to describe the physical condition necessary to describe the physical condition or or state of a system.state of a system.

The state of the system would be specified The 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 of The Zeroth Law of ThermodynamicsThermodynamics

(a) Systems A and B are (a) Systems A and B are surrounded by adiabatic surrounded by adiabatic walls and register the same walls and register the same temperature on the temperature on the thermometer. (b) When A is thermometer. (b) When A is put into thermal contact with put into thermal contact with B through diathermal walls, B through diathermal walls, no net flow of heat occurs no net flow of heat occurs between the systems.between the systems.

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


Temperature is the indicator of thermal Temperature is the indicator of thermal equilibrium in the sense that there is no equilibrium in the sense that there is no net flow of heat between two systems in net flow of heat between two systems in thermal contact that have the same thermal contact that have the same temperature.temperature.

There can be no flow of heat within a There can be no flow of heat within a system in thermal equilibrium.system in thermal equilibrium.


The First Law of The First Law of ThermodynamicsThermodynamics

The internal energy of a system changes The internal energy of a system changes from an initial value from an initial value UU ii to a final value of to a final value of UUff

due to heat due to heat QQ and work and work WW: :

∆∆UU = = UUff - - UU ii = = Q – WQ – W

Q is positive when the system gains heat Q is positive when the system gains heat and negative when it loses heat. W is and negative when it loses heat. W is positive when work is done by the system positive when work is done by the system and negative when work is done on the and negative when work is done on the system.system.

The First Law of The First Law of ThermodynamicsThermodynamics

The Internal EnergyThe Internal Energy

The internal energy depends only on the The internal energy depends only on the state of a system, not on the method by state 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 a The temperature of three moles of a monatomic ideal gas is reduced from monatomic ideal gas is reduced from TT ii = =

540 K to 540 K to TTff = 350 K by two different = 350 K by two different

methods. In the first method 5500 J of heat methods. 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 case 1500 J of heat flows into it. In each case find (a) the change in the internal energy find (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 gas Since the internal energy of a monatomic ideal gas

is is UU = ( = (33//22) ) nRTnRT and since the number of moles and since the number of moles nn

is fixed, only a change in temperature is fixed, only a change in temperature TT can alter can alter the internal energy.the internal energy.

Since the change in Since the change in TT is the same in both methods, is the same in both methods, the change in the change in UU is also the same. From the given is also the same. From the given temperatures, the change temperatures, the change ∆∆UU in internal energy in internal energy can be determined. Then, the first law of can be determined. Then, the first law of thermodynamics can be used with thermodynamics can be used with ∆∆UU and the and the given heat values to calculate the work.given heat values to calculate the work.

(a)(a) Using Equation Using Equation UU = ( = (33//22) ) nRTnRT for the for the

internal energy of a monatomic ideal gas, internal energy of a monatomic ideal gas, we find for each method of adding heat that we find for each method of adding heat that

(b)(b) Since Since ∆∆UU is now known and the heat is is now known and the heat is given in each method, given in each method, ∆∆UU = = Q – WQ – W can be can be used to determine the work: used to determine the work:

SolutionSolution

Thermal ProcessesThermal Processes

A system can interact with its surroundings A system can interact with its surroundings in many ways, and the heat and work that in many ways, and the heat and work that come into play always obey the first law of come into play always obey the first law of thermodynamics. thermodynamics.

thermal process is assumed to be thermal process is assumed to be quasi-quasi-static,static, which means that it occurs slowly which means that it occurs slowly enough that a uniform pressure and enough that a uniform pressure and temperature exist throughout all regions of temperature exist throughout all regions of the system at all times.the system at all times.

An Isobaric ProcessAn Isobaric Process

The substance in the chamber The substance in the chamber is expanding isobarically is expanding isobarically because the pressure is held because the pressure is held constant by the external constant by the external atmosphere and the weight of atmosphere and the weight of the piston and the block.the piston and the block.An isobaric process is one An isobaric process is one that occurs at constant that occurs at constant pressure.pressure.

The Figure shows a substance (solid, liquid, or The Figure shows a substance (solid, liquid, or gas) contained in a chamber fitted with a gas) contained in a chamber fitted with a frictionless piston.frictionless piston.

The pressure The pressure PP experienced by the substance is experienced by the substance is always the same and is determined by the external always the same and is determined by the external atmosphere and the weight of the piston and the atmosphere and the weight of the piston and the block resting on it. block resting on it.

Heating the substance makes it expand and do Heating the substance makes it expand and do work work WW in lifting the piston and block through the in lifting the piston and block through the displacement displacement ss..


The work can be calculated from The work can be calculated from W = FsW = Fs, where , where FF is the magnitude of the force and is the magnitude of the force and ss is the is the magnitude of the displacement. magnitude of the displacement.

The force is generated by the pressure The force is generated by the pressure PP acting on acting on the bottom surface of the piston (area = the bottom surface of the piston (area = AA), ), according to according to F = PAF = PA..

With this substitution for With this substitution for FF, the work becomes , the work becomes WW = (= (PAPA))ss. But the product . But the product AA · · ss is the change in is the change in volume of the material, volume of the material, ∆∆V = VV = V

ff - - VVii, where , where VVff and and VVii are the final and initial volumes, respectively. are the final and initial volumes, respectively.


Thus, the expression for the work is Thus, the expression for the work is Isobaric processIsobaric process

W = PW = P∆∆ V = P(VV = P(Vff – V – V

i i )) A positive value for the work done by a A 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 isobaric That Equation also applies to an isobaric compression (Vcompression (V

ff less than V less than Vii). Then, the ). Then, the

work is negative, since work must be done work is negative, since work must be done on the system to compress it. on the system to compress it.



For an isobaric process, a For an isobaric process, a pressure-versus-volume pressure-versus-volume plot is a horizontal straight plot is a horizontal straight line, and the work done line, and the work done [[WW = = PP((VVff

22 – – VVii22)] is the )] is the

colored rectangular area colored rectangular area under the graph.under the graph.

An Isochoric ProcessAn Isochoric Process

a) The substance in the chamber a) The substance in the chamber is being heated isochorically is being heated isochorically because the rigid chamber keeps because the rigid chamber keeps the volume constant. (b) The the volume constant. (b) The pressure–volume plot for an pressure–volume plot for an isochoric process is a vertical isochoric process is a vertical straight line. The area under the straight line. The area under the graph is zero, indicating that no graph is zero, indicating that no work is done. work is done. an isochoric process, one that an isochoric process, one that occurs at constant volume. occurs at constant volume.

Figure Figure aa illustrates an isochoric process in illustrates an isochoric process in which a substance (solid, liquid, or gas) is which a substance (solid, liquid, or gas) is heated. The substance would expand if it heated. The substance would expand if it could, but the rigid container keeps the could, but the rigid container keeps the volume constant, so the pressure–volume volume constant, so the pressure–volume plot shown in Figure plot shown in Figure bb is a vertical straight is a vertical straight lineline

Because the volume is constant, the Because the volume is constant, the pressure inside rises, and the substance pressure 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 in While enormous forces can be generated in the closed container, no work is done, since the closed container, no work is done, since the walls do not move. Consistent with zero the walls do not move. Consistent with zero work being done, the area under the vertical work being done, the area under the vertical straight line in Figure straight line in Figure bb is zero. is zero.

Since no work is done, the first law of Since no work is done, the first law of thermodynamics indicates that the heat in thermodynamics indicates that the heat in an isochoric process serves only to change an 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 the The colored area gives the work done by the gas for the work done by the gas for the process from process from XX to to YY. .

A third important thermal A third important thermal process is an process is an isothermal isothermal process, one that takes place at process, one that takes place at constant temperature.constant temperature.

The Adiabatic ProcessThe Adiabatic Process

Last, there is the Last, there is the adiabatic process, one adiabatic 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 equals equals zero, and the first law indicates that zero, and the first law indicates that ∆∆U = Q - W = -WU = Q - W = -W..

Thus, when work is done by a system Thus, when work is done by a system adiabatically, the internal energy of the adiabatically, the internal energy of the system decreases by exactly the amount of system decreases by exactly the amount of the work done. the work done.

Thus, when work is done by a system Thus, when work is done by a system adiabatically, the internal energy of the adiabatically, the internal energy of the system decreases by exactly the amount of system decreases by exactly the amount of the work done. When work is done on a the work done. When work is done on a system adiabatically, the internal energy system adiabatically, the internal energy increases correspondingly.increases correspondingly.

The Adiabatic ProcessThe Adiabatic Process

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

Isothermal Expansion or Isothermal Expansion or CompressionCompression

(a)(a) The ideal gas in the cylinder is expanding The ideal gas in the cylinder is expanding isothermally at temperature isothermally at temperature TT. The force holding . The force holding the piston in place is reduced slowly, so the the piston in place is reduced slowly, so the expansion occurs quasi-statically. expansion occurs quasi-statically. (b)(b) The work The 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 Figure In Figure aa, for instance, a metal cylinder , for instance, a metal cylinder contains contains nn moles of an ideal gas, and the moles of an ideal gas, and the large mass of hot water maintains the large mass of hot water maintains the cylinder and gas at a constant Kelvin cylinder and gas at a constant Kelvin temperature temperature TT. The piston is held in place . The piston is held in place initially so the volume of the gas is initially so the volume of the gas is VVii. .


As the external force applied to the piston is reduced As the external force applied to the piston is reduced quasi-statically, the gas expands to the final volume quasi-statically, the gas expands to the final volume VVff . .

Figure b gives a plot of pressure (Figure b gives a plot of pressure (P = nRT P = nRT / / VV) versus ) versus volume for the process. volume for the process.

The solid red line in the graph is called an isotherm The 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 volume represents the relation between pressure and volume when the temperature is held constant. when the temperature is held constant.


The work The work WW done by the gas is done by the gas is notnot given by given 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 area Nevertheless, the work is equal to the area

under the graph. The techniques of integral under the graph. The techniques of integral calculus lead to the following resultcalculus lead to the following result for for WW: :


=

i

f

V

VnRTW ln

Where does the energy for this work Where does the energy for this work originate? originate?

Since the internal energy of any ideal gas is Since the internal energy of any ideal gas is proportional to the Kelvin temperature (proportional to the Kelvin temperature (UU= = ((33//22) ) nRT nRT for a monatomic ideal gas, for for a monatomic ideal gas, for

example), the internal energy remains example), 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 becomes The 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 in and the energy for the work originates in the hot water. the hot water.


Adiabatic Expansion or Adiabatic Expansion or CompressionCompression

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, so The force holding the piston in place is reduced slowly, so the expansion occurs quasi-statically. the expansion occurs quasi-statically. (b)(b) A plot of A plot of pressure versus volume yields the adiabatic curve shown in pressure versus volume yields the adiabatic curve shown in red, which intersects the isotherms (blue) at the initial red, which intersects the isotherms (blue) at the initial temperature temperature TTii and the final temperature and 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.

Figure Figure aa shows an arrangement in which shows an arrangement in which nn moles of an ideal gas do work under moles of an ideal gas do work under adiabatic conditions, expanding quasi-adiabatic conditions, expanding quasi-statically from an initial volume statically from an initial volume VVii to a final to a final volume volume VVff. .

The arrangement is similar to that in Figure The 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 insulating because the cylinder is now surrounded by insulating material that prevents the flow of heat, so material that prevents the flow of heat, so QQ = 0. = 0. According to the first law of thermodynamics, the According to the first law of thermodynamics, the change in internal energy is change in internal energy is ∆∆U = Q - W = -WU = Q - W = -W. .

Since the internal energy of an ideal monatomic gas is Since the internal energy of an ideal monatomic gas is UU= (= (33//22))nRT nRT , it follows that , it follows that ∆∆U = U = ((33//22))nR(TnR(Tff – T – Tii)) , ,

where where TTii and and TTff are the initial and final Kelvin are the initial and final Kelvin

temperatures. With this substitution, the relation temperatures. With this substitution, the relation ∆∆UU = -= -WW becomes becomes


When an ideal gas expands adiabatically, it does When an ideal gas expands adiabatically, it does positive work, so positive work, so WW is positive. is positive.

Therefore, the term Therefore, the term TTii - - TTff is also positive, and the is also positive, and the

final temperature of the gas must be less than the final temperature of the gas must be less than the initial temperature. initial temperature.

The internal energy of the gas is reduced to The internal energy of the gas is reduced to provide the necessary energy to do the work, and provide the necessary energy to do the work, and because the internal energy is proportional to the because the internal energy is proportional to the Kelvin temperature, the temperature decreases. Kelvin temperature, the temperature decreases.


Figure Figure bb shows a plot of pressure versus shows a plot of pressure versus volume for the adiabatic process. The volume for the adiabatic process. The adiabatic curve (red) intersects the adiabatic curve (red) intersects the isotherms (blue) at the higher initial isotherms (blue) at the higher initial temperature [temperature [TT ii = = PPiiVVii/(/(nRnR)] and the lower )] and the lower

final temperature [final temperature [TTff = = PPffVVff/(/(nRnR)]. The )]. The

colored area under the adiabatic curve colored area under the adiabatic curve represents the work done.represents the work done.


The equation that gives the adiabatic curve The 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 derived can be derived using integral calculus. The result is using integral calculus. The result is

where the exponent where the exponent γγ is the ratio of the is the ratio of the specific heat capacities at constant pressure specific heat capacities at constant pressure and constant volume, and constant volume, γγ = = ccPP//ccVV. .


THE SECOND LAW OF THE SECOND LAW OF THERMODYNAMICSTHERMODYNAMICS

THE SECOND LAW OF THE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOW THERMODYNAMICS: THE HEAT FLOW STATEMENTSTATEMENT

Heat flows spontaneously from a substance at a Heat flows spontaneously from a substance at a higher temperature to a substance at a lower higher temperature to a substance at a lower temperature and does not flow spontaneously in temperature and does not flow spontaneously in the reverse directionthe reverse direction

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

A A heat engineheat engine is any device that uses heat to perform is 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 high Heat is supplied to the engine at a relatively high temperature from a place called the temperature from a place called the hot reservoir.hot reservoir.

2.

Part of the input heat is used to perform work by the Part of the input heat is used to perform work by the working substanceworking substance of the engine, which is the material of the engine, which is the material within the engine that actually does the work (e.g., the within 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 a The remainder of the input heat is rejected at a temperature lower than the input temperature to a place temperature lower than the input temperature to a place called the called the cold reservoircold reservoir

This schematic representation This schematic representation of a heat engine shows the input of a heat engine shows the input heat (magnitude = heat (magnitude = QQHH) that ) that

originates from the hot originates from the hot reservoir, the work (magnitude reservoir, 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 cold the engine rejects to the cold reservoir. reservoir.

To be highly efficient, a heat engine must produce To be highly efficient, a heat engine must produce a relatively large amount of work from as little a relatively large amount of work from as little input heat as possible. Thus, the input heat as possible. Thus, the efficiency eefficiency e of a of a heat engine is defined as the ratio of the work heat engine is defined as the ratio of the work WW done by the engine to the input heat done by the engine to the input heat QQHH: :

If the input heat were converted entirely into If the input heat were converted entirely into work, the engine would have an efficiency of work, the engine would have an efficiency of 1.00, since 1.00, since W = QW = Q

HH; such an engine would be ; such an engine would be

100% efficient. 100% efficient.

What is it that allows a heat engine to What 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 maximum proposed that a heat engine has maximum efficiency when the processes within the engine efficiency when the processes within the engine are reversible. are reversible. A reversible process is one in A reversible process is one in which both the system and its environment which both the system and its environment can be returned to exactly the states they were can 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 and the system and its environment can be returned to their initial its environment can be returned to their initial states. states.

A process that involves an energy-dissipating A process that involves an energy-dissipating mechanism, such as friction, cannot be mechanism, such as friction, cannot be reversible because the energy wasted due to reversible because the energy wasted due to friction would alter the system or the friction would alter the system or the environment or both. environment or both.

CARNOT'S PRINCIPLE: AN CARNOT'S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE ALTERNATIVE STATEMENT OF THE SECOND LAW OF THERMODYNAMICSSECOND LAW OF THERMODYNAMICS

No irreversible engine operating between two No irreversible engine operating between two reservoirs at constant temperatures can have a greater reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between efficiency than a reversible engine operating between the same temperatures. Furthermore, all reversible the same temperatures. Furthermore, all reversible engines operating between the same temperatures engines operating between the same temperatures have the same efficiency.have the same efficiency.

Natural Limits on the Efficiency of a Natural Limits on the Efficiency of a Heat EngineHeat Engine

Consider a hypothetical engine that receives 1000 J Consider a hypothetical engine that receives 1000 J of heat as input from a hot reservoir and delivers of heat as input from a hot reservoir and delivers 1000 J of work, rejecting no heat to a cold reservoir 1000 J of work, rejecting no heat to a cold reservoir whose temperature is above 0 K. Decide whether whose temperature is above 0 K. Decide whether this engine violates the first or the second law of this engine violates the first or the second law of thermodynamics, or both.thermodynamics, or both.

The first law of thermodynamics is an expression The first law of thermodynamics is an expression of energy conservation. From the point of view of of energy conservation. From the point of view of energy conservation, nothing is wrong with an energy conservation, nothing is wrong with an engine that converts 1000 J of heat into 1000 J of engine that converts 1000 J of heat into 1000 J of work. work.

Energy has been neither created nor destroyed; it Energy has been neither created nor destroyed; it has only been transformed from one form (heat) to has only been transformed from one form (heat) to another (work).another (work).

This engine does, however, violate the second This engine does, however, violate the second law of thermodynamics. Since all of the input law of thermodynamics. Since all of the input heat is converted into work, the efficiency of the heat is converted into work, the efficiency of the engine is 1, or 100%.engine is 1, or 100%.

Since we know that Since we know that TTCC is above 0 K, it is clear is above 0 K, it is clear

that the ratio that the ratio TTCC//TTHH is greater than zero, so the is greater than zero, so the

maximum possible efficiency is less than 1, or maximum 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 to An ideal or Carnot heat pump is used to heat a house to a temperature of heat a house to a temperature of TTHH = 294 = 294

K (21 °C). How much work must be done K (21 °C). How much work must be done by the pump to deliver by the pump to deliver QQHH = 3350 J of heat = 3350 J of heat

into the house when the outdoor into the house when the outdoor temperature temperature 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 be applies to the heat pump. Thus, the work can be determined from determined from W = QW = Q

HH - - QQCC, provided we can , provided we can

obtain a value for obtain a value for QQCC, the heat taken by the pump , the heat taken by the pump

from the outside. To determine from the outside. To determine QQCC, we use the fact , we use the fact

that the pump is a Carnot heat pump and operates that the pump is a Carnot heat pump and operates reversibly. Therefore, the relation reversibly. Therefore, the relation QQCC//QQHH = = TTCC//TTHH

applies. Solving it for applies. Solving it for QQCC, we obtain , we obtain QQCC = = QQHH((TTCC//TTHH). ).

Using this result, we find that Using this result, we find that

ReasoningReasoning

SolutionSolution

(a)(a) At an indoor temperature of At an indoor temperature of TTHH = 294 K and an = 294 K and an

outdoor temperature of outdoor temperature of TTCC = 273 K, the work needed = 273 K, the work needed

is is

(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, so except that it is now cooler outside, so TTCC = 252 K. = 252 K.

The necessary work is , which is more than The necessary work is , which is more than in part (a).in part (a).

To introduce the idea of entropy we recall the To introduce the idea of entropy we recall the relation relation QQCC//QQHH = = TTCC//TTHH that applies to a Carnot that applies to a Carnot

engine. This equation can be rearranged as engine. This equation can be rearranged as QQCC//TTCC = =

QQHH//TTHH, which focuses attention on the heat , which focuses attention on the heat QQ divided divided

by the Kelvin temperature by the Kelvin temperature TT. The quantity . The quantity QQ//TT is is called the change in the entropy called the change in the entropy ∆∆SS: :

ENTROPYENTROPY

the SI unit for entropy is a joule per kelvin (J/K).

THE SECOND LAW OF THE SECOND LAW OF THERMODYNAMICS STATED IN THERMODYNAMICS STATED IN TERMS OF ENTROPYTERMS OF ENTROPY

The total entropy of the universe does not The total entropy of the universe does not change when a reversible process occurs change 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 of This Figure shows 1200 J of heat flowing spontaneously heat flowing spontaneously through a copper rod from a hot through a copper rod from a hot reservoir at 650 K to a cold reservoir at 650 K to a cold reservoir at 350 K. Determine reservoir at 350 K. Determine the amount by which this the amount by which this irreversible process changes the irreversible process changes the entropy of the universe, entropy of the universe, assuming that no other changes assuming that no other changes occur.occur.

ReasoningReasoning

The hot-to-cold heat flow is irreversible, so The hot-to-cold heat flow is irreversible, so the relation the relation ∆∆SS = ( = (QQ//TT))RR is applied to a is applied to a

hypothetical process whereby the 1200 J of hypothetical process whereby the 1200 J of heat is taken reversibly from the hot heat is taken reversibly from the hot reservoir and added reversibly to the cold reservoir and added reversibly to the cold reservoir.reservoir.

SolutionSolution

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

Order to DisorderOrder to Disorder

Find the change in entropy that results when a Find 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 constant Since the phase change occurs reversibly at a constant temperature, the change in entropy can be found by temperature, the change in entropy can be found by using, using, ∆∆SS = ( = (QQ//TT))RR, where , where QQ is the heat absorbed by is the heat absorbed by

the melting ice. This heat can be determined by using the melting ice. This heat can be determined by using the relation the relation Q = mLQ = mL

ff, where , where mm is the mass and is 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 any It is not possible to lower the temperature of any system to absolute zero in a finite number of system to absolute zero in a finite number of steps.steps.

termodinamika

Science

system gains heat

heat q

system changes

net flow of heat

given heat values

surroundings thermodynamics

j of heat flows

work w