Heat Engines, Entropy and the Second Law of Thermodynamics Chapter 22 Heat Engines, Entropy and the Second Law of Thermodynamics

First Law of Thermodynamics – Review The first law states that a change in internal energy in a system can occur as a result of energy transfer by heat, by work, or by both. The law makes no distinction between the results of heat and the results of work.

First Law of Thermodynamics – Review In according with first law of thermodynamics the energy is always conserved What does it mean to conserve energy if the total amount of energy in the universe does not change regardless of what we do? The first law of thermodynamic does not tell the whole story: Energy is always conserved, but some forms of energy are more useful than others.

First Law – Missing Pieces There is an important distinction between heat and work that is not evident from the first law The first law makes no distinction between processes that occur spontaneously and those that do not An example is that it is impossible to design a device that takes in energy and converts it all to work

The Second Law of Thermodynamics The possibility or impossibility of putting energy to use is the subject of the Second Law of Thermodynamics. For example, it is easy to convert work into thermal energy, but it is impossible to remove energy as heat from a single reservoir and convert it internally into work with no other changes. This experimental fact is one statement of the second law of thermodynamics.

The Second Law of Thermodynamics Kelvin’s statement: No system can take energy as heat from a single reservoir and convert it entirely into work without additional net changes in the system or its surroundings. Clausus statement: A process whose only net result is to transfer energy as heat from a cooler object to a hotter one is impossible.

The Second Law of Thermodynamics A common example of conversion of work into heat is movement with friction. Suppose you spend two minutes pushing a block along a tabletop in a closed path, leaving the block in its initial position. Also, suppose that the block-table system is initially in thermal equilibrium with the surroundings.

The Second Law of Thermodynamics The work you do on the system is converted into internal energy of the system and the block-table system becomes warmer → the system is no longer in thermal equilibrium with its surroundings → the system will transfer energy as heat to its surroundings until it returns to the thermal equilibrium

The Second Law of Thermodynamics Because the final and initial states of the system are the same, the 1-st Law of TD dictates that the energy transferred to the environment as heat equals to the work done on the system. The reverse process never occur – a block and table that are warm will never spontaneously cool by converting their internal energy into the work that causes the block to push your hand around the table!

The Second Law of Thermodynamics Such an amazing occurrence would not violate the first law of thermodynamics or any other physical law, it does, however, violate the second law of thermodynamics. There is a lack of symmetry in the roles played by heat and work that is not evident from the first law. This lack of symmetry is related to the fact that some processes are irreversible.

The Second Law of Thermodynamics Establishes which processes do and which do not occur Some processes can occur in either direction according to the first law They are observed to occur only in one direction according to the second law.

Irreversible Processes An irreversible process is one that occurs naturally in one direction only No irreversible process has been observed to run backwards An important engineering implication is the limited efficiency of heat engines

Heat Engine A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work A heat engine carries some working substance through a cyclical process

Heat Engine. The working substance absorbs energy by heat from a high temperature energy reservoir (Qh) Work is done by the engine (Weng) Energy is expelled as heat to a lower temperature reservoir (Qc)

Heat Engine. Since it is a cyclical process, ΔEint = 0 Its initial and final internal energies are the same Therefore, Qnet = Weng The work done by the engine equals the net energy absorbed by the engine The work is equal to the area enclosed by the curve of the PV diagram The working substance is a gas

Thermal Efficiency of a Heat Engine Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature We can think of the efficiency as the ratio of what you gain to what you give

More About Efficiency In practice, all heat engines expel only a fraction of the input energy by mechanical work Therefore, their efficiency is always less than 100% To have e = 100%, energy expelled as heat to a lower temperature reservoir, QC, must be 0

Perfect Heat Engine No energy is expelled to the cold reservoir It takes in some amount of energy and does an equal amount of work e = 100% It is an impossible engine

During each cycle a heat engine absorbs 200 J of heat from a hot reservoir, does work, and exhausts 160 J to a cold reservoir. What is the efficiency of the engine?

During each cycle a heat engine absorbs 200 J of heat from a hot reservoir, does work, and exhausts 160 J to a cold reservoir. What is the efficiency of the engine? Qh= 200J W = Qh – Qc = 200J – 160J = 40J

Heat Pumps and Refrigerators Heat engines can run in reverse This is not a natural direction of energy transfer Must put some energy into a device to do this Devices that do this are called heat pumps or refrigerators Examples A refrigerator is a common type of heat pump An air conditioner is another example of a heat pump

Heat Pump Process Energy is extracted from the cold reservoir, QC Energy is transferred to the hot reservoir, Qh Work must be done on the engine, W

Coefficient of Performance The effectiveness of a heat pump is described by a number called the coefficient of performance (COP) In heating mode, the COP is the ratio of the heat transferred in to the work required

COP, Heating Mode COP is similar to efficiency Qh is typically higher than W Values of COP are generally greater than 1 It is possible for them to be less than 1 We would like the COP to be as high as possible

COP, Cooling Mode In cooling mode, you “gain” energy from a cold temperature reservoir A good refrigerator should have a high COP Typical values are 5 or 6

A refrigerator has a coefficient of performance equal to 5. 00 A refrigerator has a coefficient of performance equal to 5.00. The refrigerator takes in 120 J of energy from a cold reservoir in each cycle. Find (a) the work required in each cycle and (b) the energy expelled to the hot reservoir.

A refrigerator has a coefficient of performance equal to 5. 00 A refrigerator has a coefficient of performance equal to 5.00. The refrigerator takes in 120 J of energy from a cold reservoir in each cycle. Find (a) the work required in each cycle and (b) the energy expelled to the hot reservoir. (a) (b) Heat expelled = Heat removed + Work done.

Carnot Engine A theoretical engine developed by Sadi Carnot A heat engine operating in an ideal, reversible cycle (now called a Carnot cycle) between two reservoirs is the most efficient engine possible This sets an upper limit on the efficiencies of all other engines

Carnot’s Theorem No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs All real engines are less efficient than a Carnot engine because they do not operate through a reversible cycle The efficiency of a real engine is further reduced by friction, energy losses through conduction, etc.

Carnot Cycle Overview of the processes in a Carnot cycle

Carnot Cycle, A to B A → B is an isothermal expansion The gas is placed in contact with the high temperature reservoir, Th The gas absorbs heat |Qh| The gas does work WAB in raising the piston

Carnot Cycle, B to C B → C is an adiabatic expansion The base of the cylinder is replaced by a thermally insulated wall No heat enters or leaves the system The temperature falls from Th to Tc . The gas does work WBC

Carnot Cycle, C to D The gas is placed in contact with the cold temperature reservoir C → D is an isothermal compression The gas expels energy Qc Work WCD is done on the gas

Carnot Cycle, D to A D → A is an adiabatic compression The gas is again placed against a thermally nonconducting wall So no heat is exchanged with the surroundings The temperature of the gas increases from Tc to Th The work done on the gas is WDA

Carnot Cycle, PV Diagram The work done by the engine is shown by the area enclosed by the curve, Weng The net work is equal to |Qh| – |Qc| DEint = 0 for the entire cycle

Efficiency of a Carnot Engine Carnot showed that the efficiency of the engine depends on the temperatures of the reservoirs Temperatures must be in Kelvins All Carnot engines operating between the same two temperatures will have the same efficiency

Notes About Carnot Efficiency Efficiency is 0 if Th = Tc Efficiency is 100% only if Tc = 0 K Such reservoirs are not available Efficiency is always less than 100% The efficiency increases as Tc is lowered and as Th is raised In most practical cases, Tc is near room temperature, 300 K So generally Th is raised to increase efficiency

Carnot Heat Pump COPs In heating mode: In cooling mode:

A Carnot heat engine uses a steam boiler at 100C as the high-temperature reservoir. The low-temperature reservoir is the outside environment at 20.0C. Energy is exhausted to the low-temperature reservoir at the rate of 15.4 W. (a) Determine the useful power output of the heat engine. (b) How much steam will it cause to condense in the high-temperature reservoir in 1.00 h?

A Carnot heat engine uses a steam boiler at 100C as the high-temperature reservoir. The low-temperature reservoir is the outside environment at 20.0C. Energy is exhausted to the low-temperature reservoir at the rate of 15.4 W. (a) Determine the useful power output of the heat engine. (b) How much steam will it cause to condense in the high-temperature reservoir in 1.00 h? The efficiency is:

A Carnot heat engine uses a steam boiler at 100C as the high-temperature reservoir. The low-temperature reservoir is the outside environment at 20.0C. Energy is exhausted to the low-temperature reservoir at the rate of 15.4 W. (a) Determine the useful power output of the heat engine. (b) How much steam will it cause to condense in the high-temperature reservoir in 1.00 h? (a) The useful power output is: (b)

Gasoline Engine In a gasoline engine, six processes occur during each cycle For a given cycle, the piston moves up and down twice This represents a four-stroke cycle The processes in the cycle can be approximated by the Otto cycle

Otto Cycle The PV diagram of an Otto cycle is shown at right The Otto cycle approximates the processes occurring in an internal combustion engine

The Conventional Gasoline Engine

Gasoline Engine – Intake Stroke During the intake stroke, the piston moves downward A gaseous mixture of air and fuel is drawn into the cylinder Energy enters the system as potential energy in the fuel O → A in the Otto cycle

Gasoline Engine – Compression Stroke The piston moves upward The air-fuel mixture is compressed adiabatically The temperature increases The work done on the gas is positive and equal to the negative area under the curve A → B in the Otto cycle

Gasoline Engine – Spark Combustion occurs when the spark plug fires This is not one of the strokes of the engine It occurs very quickly while the piston is at its highest position Conversion from potential energy of the fuel to internal energy B → C in the Otto cycle

Gasoline Engine – Power Stroke In the power stroke, the gas expands adiabatically This causes a temperature drop Work is done by the gas The work is equal to the area under the curve C → D in the Otto cycle

Gasoline Engine – Valve Opens This is process D → A in the Otto cycle An exhaust valve opens as the piston reaches its bottom position The pressure drops suddenly The volume is approximately constant So no work is done Energy begins to be expelled from the interior of the cylinder

Gasoline Engine – Exhaust Stroke In the exhaust stroke, the piston moves upward while the exhaust valve remains open Residual gases are expelled to the atmosphere The volume decreases A → O in the Otto cycle

Otto Cycle Efficiency If the air-fuel mixture is assumed to be an ideal gas, then the efficiency of the Otto cycle is g is the ratio of the molar specific heats V1/V2 is called the compression ratio

Otto Cycle Efficiency Typical values: Compression ratio of 8 g = 1.4 e = 56% Efficiencies of real engines are 15% to 20% Mainly due to friction, energy transfer by conduction, incomplete combustion of the air-fuel mixture

Diesel Engines Operate on a cycle similar to the Otto cycle without a spark plug The compression ratio is much greater and so the cylinder temperature at the end of the compression stroke is much higher Fuel is injected and the temperature is high enough for the mixture to ignite without the spark plug Diesel engines are more efficient than gasoline engines

A 1. 60-L gasoline engine with a compression ratio of 6 A 1.60-L gasoline engine with a compression ratio of 6.20 has a useful power output of 102 hp. Assuming the engine operates in an idealized Otto cycle, find the energy taken in and the energy exhausted each second. Assume the fuel-air mixture behaves like an ideal gas with γ= 1.40.

A 1. 60-L gasoline engine with a compression ratio of 6 A 1.60-L gasoline engine with a compression ratio of 6.20 has a useful power output of 102 hp. Assuming the engine operates in an idealized Otto cycle, find the energy taken in and the energy exhausted each second. Assume the fuel-air mixture behaves like an ideal gas with γ= 1.40. We have assumed the fuel-air mixture to behave like a diatomic gas. Now

A 1. 60-L gasoline engine with a compression ratio of 6 A 1.60-L gasoline engine with a compression ratio of 6.20 has a useful power output of 102 hp. Assuming the engine operates in an idealized Otto cycle, find the energy taken in and the energy exhausted each second. Assume the fuel-air mixture behaves like an ideal gas with γ= 1.40.

Irreversibility and Disorder There are many irreversible processes that can not be described by the heat-engine or refrigerator statements of the second law, such as a glass falling to the floor and breaking or a balloon popping. However all irreversible processes have one thing in common – the system and its surroundings moves towards a less ordered state.

Suppose a box containing a gas of mass M at temperature T is moving along a frictionless table with a velocity vcm. The total kinetic energy of the gas has two components: that associated with the movement of the center of the mass ½Mvcm2, and the energy of the motion of its molecules relative to its center of the mass

The center of the mass energy ½Mv2cm is ordered mechanical energy that could be converted entirely into work. For example, if a weight were attached to a moving box by a string passing over a pulley, this energy could be used to lift the weight. The relative energy of the molecules is the internal thermal energy of the gas, which is related to its temperature T. It is random, non-ordered energy that can not be converted entirely into work.

Now, suppose that the block hits a fixed wall and stops Now, suppose that the block hits a fixed wall and stops. This inelastic collision is clearly an irreversible process. The ordered mechanical energy of the gas is converted into random internal energy and the temperature of the gas rises. The gas still has the same total energy, but now all of the energy is associated with the random motion of the molecules about of center of the mass of the gas, which is now at rest.

Entropy Thus, the gas become less ordered (more disordered), and gas lost some of its ability to do work. There is a thermodynamic function called entropy S that is a measure of the disorder of the system. Entropy S, like pressure P, volume V, temperature T, and internal energy Eint, is a function of the state of the system. The change in entropy dS of the system as it goes from one state to another is defined as where dQrev is the energy that must be transferred to the system as heat in the reversible process that bring the system from the initial state to the final state.

Entropy of an Ideal Gas Consider an arbitrary reversible quasi-static process in which a system consisting of an ideal gas adsorbs an amount of heat dQrev. According to the first law, dQrev is related to dEint and W by or If we will divide each term by T:

Entropy of an Ideal Gas We will assume Cv to be constant and by integrating we will find the entropy change of an ideal gas that undergoes a reversible expansion from an initial state of volume V1 and temperature T1 to a final state of volume V2 and temperature T2

Entropy and the Second Law Entropy is a measure of disorder The entropy of the Universe increases in all real processes This is another statement of the second law of thermodynamics

Entropy and Heat The original formulation of entropy deals with the transfer of energy by heat in a reversible process If dQrev is the amount of energy transferred by heat when a system follows a reversible path then the change in entropy, DS is

Entropy and Heat The change in entropy depends only on the endpoints and is independent of the path followed The entropy change for an irreversible process can be determined by calculating the change in entropy for a reversible process that connects the same initial and final points

More About Change in Entropy dQrev is measured along a reversible path, even if the system may have followed an irreversible path The meaningful quantity is the change in entropy and not the entropy itself For a finite process,

Change in Entropy The change in entropy of a system going from one state to another has the same value for all paths connecting the two states The finite change in entropy depends only on the properties of the initial and final equilibrium states

DS for a Reversible Cycle S = 0 for any reversible cycle In general, This integral symbol indicates the integral is over a closed path

Entropy Changes in Irreversible Processes To calculate the change in entropy in a real system, remember that entropy depends only on the state of the system Do not use Q, the actual energy transfer in the process Distinguish this from Qrev , the amount of energy that would have been transferred by heat along a reversible path Qrev is the correct value to use for DS

Entropy Changes in Irreversible Processes In general, the total entropy and therefore the total disorder always increases in an irreversible process The total entropy of an isolated system undergoes a change that cannot decrease This is another statement of the second law of thermodynamics

Entropy Changes in Irreversible Processes If the process is irreversible, then the total entropy of an isolated system always increases In a reversible process, the total entropy of an isolated system remains constant The change in entropy of the Universe must be greater than zero for an irreversible process and equal to zero for a reversible process

Heat Death of the Universe Ultimately, the entropy of the Universe should reach a maximum value At this value, the Universe will be in a state of uniform temperature and density All physical, chemical, and biological processes will cease The state of perfect disorder implies that no energy is available for doing work This state is called the heat death of the Universe

DS in Thermal Conduction The cold reservoir absorbs Q and its entropy changes by Q/Tc At the same time, the hot reservoir loses Q and its entropy changes by -Q/Th Since Th > Tc , the increase in entropy in the cold reservoir is greater than the decrease in entropy in the hot reservoir Therefore, DSU > 0 For the system and the Universe

DS in a Free Expansion Consider an adiabatic free expansion Q = 0 and cannot be used since that is for an irreversible process

DS in Free Expansion For an isothermal process, this becomes Since Vf > Vi , DS is positive This indicates that both the entropy and the disorder of the gas increase as a result of the irreversible adiabatic expansion

DS in Calorimetric Processes The process is irreversible because the system goes through a series of nonequilibrium states Assuming the specific heats remain constant and no mixing takes place: If mixing takes place, this result applies only to identical substances DS will be positive and the entropy of the Universe increases

Entropy on a Microscopic Scale We can treat entropy from a microscopic viewpoint through statistical analysis of molecular motions A connection between entropy and the number of microstates (W) for a given macrostate is S = kB ln W The more microstates that correspond to a given macrostate, the greater the entropy of that macrostate This shows that entropy is a measure of disorder

Entropy, Molecule Example One molecule in a two-sided container has a 1-in-2 chance of being on the left side Two molecules have a 1-in-4 chance of being on the left side at the same time Three molecules have a 1-in-8 chance of being on the left side at the same time

Entropy, Molecule Example Extended Consider 100 molecules in the container The probability of separating 50 fast molecules on one side and 50 slow molecules on the other side is (½)100 If we have one mole of gas, this is found to be extremely improbable

Entropy, Marble Example Suppose you have a bag with 50 red marbles and 50 green marbles You draw a marble, record its color, return it to the bag, and draw another Continue until four marbles have been drawn What are possible macrostates and what are their probabilities?

Entropy, Marble Example, Results The most ordered are the least likely The most disorder is the most likely

What change in entropy occurs when a 27 What change in entropy occurs when a 27.9-g ice cube at –12C is transformed into steam at 115C?

What change in entropy occurs when a 27 What change in entropy occurs when a 27.9-g ice cube at –12C is transformed into steam at 115C? As the ice melts its entropy change is

What change in entropy occurs when a 27 What change in entropy occurs when a 27.9-g ice cube at –12C is transformed into steam at 115C? As liquid water warms from 273 K to 373 K, As the water boils and the steam warms, The total entropy change is