Entropy

Physics 1061 Entropy and the Second Law of Thermodynamics 1 Entropy We consider an ideal gas undergoing a cyclical process, as shown in Fig. 1. When evolving along the upper branch a → b the gas expands, thus doing positive work on the environment. When returning along the branch b → a to the initial state, the gas is compressed, so the environment is doing work on the gas. The area underneath the curve a → b is greater, than the area underneath the curve b → a , so the gas did more work than it was done on it. Thus the net work done by the gas, given by the area enclosed by the cycle, is a positive amount. According to the First Law of Thermodynamics the infinitesimal change of internal pressure volume a b area=net work done by gas Figure 1: A thermodynamic cycle proceeds clockwise. The enclosed area gives the net work done by the gas on the environment. energy of the gas is dE int = δW + δQ. Since the internal energy is a state function, the system has the same value of internal energy when it reaches state a again, after completing one full cycle. Thus integrating dE int over the complete cycle we get zero: I E int = 0 . We can say that the internal energy of the system is conserved over the full cycle. On the other hand, we have I δQ = - I δW = W net,output , namely the differential form δQ is not a full differential since the cyclical integral does not vanish. Integrating δQ over a cycle gives the total work done by the system on the environment. The only way the integral could vanish is if the process b → a exactly retraces the process a → b , but in reverse direction. In this case, the area enclosed by the cycle in pV-space is zero and the cyclic integrals vanish trivially. Nevertheless, we shall explore the idea of reversibility further. A quantity called entropy was introduced by Rudolf Clausius (1822-1888) in 1865, to describe quantita- tively the irreversibility in thermodynamic processes. By definition, the infinitesimal amount of entropy is defined as the full differential dS = δQ T , 1

where δQ is the amount absorbed by the system reversibly from its environment. Mathematically, the factor 1 /T serves as an integrating factor , which upon multiplication transforms the differential form δQ into a full differential dS . For any reversible cyclic process we have I dS = I δQ T = 0 . Thus, we say that entropy is conserved over a full cycle - the starting and the initial values of the entropy of the system are the same. The finite change of entropy when the system evolves reversibly between two states, is given by the integral Δ S = S f - S i = Z f i δQ T . For an isothermal reversible process the entropy change is Δ S = Q T . We see that entropy is measured in SI units of J/K , the same as the units for the Boltzmann constant k and the heat capacity C . Example: A block of 1 . 0 kg ice melts completely at 0 ◦ C . The latent heat of fusion of water is 333 kJ/kg . Thus the entropy change of the ice is Δ S = Q T = mL T = (1 . 0 kg )(333 kJ/kg ) (273 . 15 K ) ≈ 1 . 2 kJ/K. The reversible thermodynamic processes are an useful abstraction. In reality, all processes are irreversible. The entropy change for irreversible processes is calculated by using a reversible process that connects the same initial and final states. Example: Consider an ideal gas that is confined to the left half of a thermally insulated container, by a closed stopcock. When the stopcock is opened, the gas rushes to fill the entire container, doing no work in the process (no piston to push). The process is irreversible. The gas molecules will never return to the left half on their own. Since no heat was absorbed, and no work was done, according to the First Law of Thermodynamics, the internal energy, and respectively the temperature of the gas, will remain constant. To calculate the entropy for the adiabatic free expansion, we cannot simply use Δ S = S f - S i = Z f i δQ T . since this expression applies only to reversible processes connecting the initial and the final states. If we blindly plug in Q = 0, we will get that the entropy of the gas did not change, which will be incorrect. Instead, we use the fact that the gas temperature did not change, and use a reversible isothermal expansion that doubles the volume of the gas. For the reversible isothermal expansion we can write Δ S = Q T = - W T = nRT ln( V f /V i ) T = nR ln 2 = Nk ln 2 . We see that the change of entropy is positive. It is also proportional to the number of gas molecules, which means that the entropy is an extensive thermodynamic property, much like energy and volume. If a system is made of two subsystems A and B , then its entropy is the sum of the entropy of the subsystems: S = S A + S B . 2

Figure 3: (a) The cups hold equal amounts of water, but at different temperatures. (b) When the shutter is removed, the cups exchange heat and come to a final state, both with the same temperature. shutter. Once the shutter is removed, the cups begin to exchange thermal energy via thermal radiation, till they reach the same final temperature T f . This process is clearly irreversible. The two cups will never return on their own to their initial states, once they reach their final temperature. We can imagine that instead of undergoing the irreversible heat exchange process, just described, the cups instead were respectively heated and cooled, reversibly. This may be accomplished by placing the cup in contact with a heat reservoir. A heat reservoir is a body that has enormous heat capacity. Then adding heat to the reservoir would not change its temperature significantly. Earth’s oceans may be considered approximate heat reservoirs. We place one cup in contact with the reservoir and very, very slowly raise its temperature from T C to T f . Similarly, we place the hot cup in contact with the heat reservoir and this time very, very slowly cool it down to the final temperature. Figure 4: The cups can evolve from their initial to their final state in a reversible way, if we use a heat reservoir with controllable temperature. Proceeding reversibly, we can calculate the entropy change of each cup of water. For the hot cup we obtain Δ S H = Z T f T H mcdT T = mc ln T f T H . For the cold cup, we get Δ S C = Z T f T C mcdT T = mc ln T f T C . Assuming a perfect thermal balance, we have Q H + Q C = 0 , or mc ( T f - T C ) + mc ( T f - T H ) = 0 . Since the water amount in each cup is the same, we see that T H - T f = T f - T C . 4

Thus the equilibrium final temperature will be right in the middle between the hot and cold temperatures. T f = 1 2 ( T H + T C ) Then we can express the entropy change of the hot cup as Δ S H = mc ln T f T H = mc ln T H + T C 2 T H . For the cold cup we get Δ S C = mc ln T f T C = mc ln T H + T C 2 T C . The total entropy change is Δ S = Δ S H + Δ S C = mc ln T H + T C 2 T H + mc ln T H + T C 2 T C = mc ln ( T H + T C ) 2 4 T H T C . The argument of the logarithm is greater than 1. To show that, we test the inequality ( T H + T C ) 2 > 4 T H T C . Expanding the left-hand side gives T 2 H + 2 T H T C + T 2 C > 4 T H T C , which leads to ( T H - T C ) 2 > 0 , which is true. Thus we obtain that the entropy change of the system of two cups is a positive quantity. Notice that the thermal energy got distributed equally between the two cups and both reached the same final temperature. The hot cup lost heat, so its entropy decreased. The cold cup gained heat, so its entropy increased. But overall, the total entropy increased, regardless of the exact values for the temperatures. This and similar findings, eventually resulted in the formulation of the so called Entropy Principle : If an irreversible process occurs in a closed system, the entropy of that system always increases; it never decreases. 2 The ideal heat engine A heat engine is a device, operating on a cycle, that extracts heat from its environment and performs useful work. Let during the cycle the device absorbs heat Q input from sources in its environment, and delivers net amount of work W net,output to its environment. The thermal efficiency of the heat engine is defined as the ratio = Q input W net,output . The first scientific study on the efficiency of heat engines was conducted in 1824 by Nicolas L´ eonard Sadi Carnot (1796-1832), a French military engineer. In his work “Reflections on the Motive Power of Fire”, he set for himself the goal to answer two questions: 1. Is the work available from a heat source potentially unbounded? 2. Can heat engines, in principle, be improved by replacing the steam with some other working fluid? 5

3. Isothermal compression. Heat is released reversibly into the cold reservoir. 4. Isentropic compression. It is most helpful to represent the cycle on a temperature versus entropy diagram. The Carnot cycle takes a simple rectangular shape. The heat absorbed during the isothermal expansion is Q H = T H Δ S = T H ( S max - S min ) . The heat delivered to the cold reservoir during the isothermal compression is Q C = - T C Δ S = T C ( S min - S max ) . The efficiency of the Carnot engine is c = 1 - | Q C | Q H = 1 - T C Δ S T H Δ S = 1 - T C T H . We see that the efficiency of the ideal Carnot engine depends only on the temperature limits T H and T C and not on the working substance of the engine, its exact construction etc. We also see that the efficiency can be 100% only of the cold reservoir is at the absolute zero. Temperature Entropy T S S Δ S T H C max min Figure 6: The Carnot cycle takes rectangular shape when plotted on temperature vs. entropy diagram. The area enclosed by the cycle is equal to the net work done by the ideal Carnot engine. Any other engine that operates reversibly between the same temperature limits, will have lower effi- ciency than the Carnot engine. We consider such engine, operating between maximum temperature T H and minimum temperature T C . The engine cycle, drawn in temperature-entropy space is shown in Fig. 7. We assume that the curve describing the cycle is smooth, which is always true in practice, due to the inertia of the moving engine parts. The points a and b mark the states where the entropy reaches its minimum and maximum values. Along path a → b the entropy increases from S min to S max . The change of entropy is Δ S = Z b a δQ T = Q ab T ab , where Q ab is the total amount of heat added during the process a → b , and T ab is the average temperature during that process. Since Δ S > 0, the heat amount Q ab is positive. We identify the heat input with Q ab . 7

Similarly, along the reverse path b → a , we have - Δ S = Z a b δQ T = Q ba T ba , For this path Q ba < 0, so we identify it with the exhaust heat. The efficiency of this engine is = 1 - | Q exhaust | Q input = 1 - T ba Δ S T ab Δ S = 1 - T ba T ab Temperature Entropy T S S Δ S T H C max min a b Figure 7: A general type engine operates on a reversible cycle. The cycle is enclosed in a Carnot cycle. Furthermore, we notice that T H ≥ T ab , namely the average temperature for the path cannot be greater than the maximum temperature encountered along the path. Similarly, we conclude T C ≤ T ba . It follows then that T C T H ≤ T ba T ab and < c . Thus the Carnot efficiency provides a theoretical limit on the efficiency of heat engines operating in a given temperature range. This result is known as Carnot’s Theorem . For example, if the maximum temperature encountered in an engine cycle is T max = 600 K , and the minimum temperature is T min = 300 K , then the efficiency of this engine can never be greater than 50%, regardless of the particular design, materials, manufacturing technique etc. We can write the efficiency of the Carnot engine in the form c = 1 - T C T H = T H - T C T H = Δ T T H . We see that by decreasing the temperature T H , for a given temperature difference between “hot” and “cold”, we can increase the efficiency. 8

Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time. This statement is known as the Second Law of Thermodynamics. An alternative formulation was provided by Lord Kelvin, who in 1851 said: It is impossible for a self-acting machine, unaided by any external agency, to convey heat from one body to another at a higher temperature. It is impossible, by means of inanimate material agency, to derive mechanical effect from any portion of matter by cooling it below the temperature of the coldest of the surrounding objects. Max Planck provided yet another formulation in 1897: It is impossible to construct an engine which will work in a complete cycle, and produce no effect except the raising of a weight and cooling of a heat reservoir. The Second Law of Thermodynamics is unlike any other law of Physics. It is a prohibition. It states what is not possible. It is not possible to build a refrigerator that does not require work input. It is impossible for heat engine operating on a cycle, to convert its heat input completely into work, without any exhaust heat being dumped into the environment. A perpetual motion machine is a hypothetical device that can produce work indefinitely, without any external input of energy. In the history of mankind, many, many, such machines have been proposed. They violate either the First Law of Thermodynamics, and are then called perpetuum mobile of the first kind , or the Second Law of Thermodynamics, and are called then perpetuum mobile of the second kind . 4 Refrigerators and heat pumps Refrigerators are heat engines working in reverse. They take an external work as an input, remove heat Q removed from some enclosed space (the interior of the refrigerator), and dump exhaust heat | Q exhaust | in the external world. For refrigerators to work, the interior and exterior spaces must be thermally insulated from each other. That is, you cannot cool the room by opening the door of the refrigerator. Then whatever heat the fridge extract from the room, it will be dumped right back into it, leading to zero net effect. The same is true for air conditioners - the space that is supposed to be cooled (the room), must be thermally insulated from the space where the exhaust heat is dumped (the outside world). Figure 9: Principle of operation of a typical refrigerator. 10

The fluid circuit of a refrigerator contains a refrigerant coolant, which is the working substance, not air or ideal gas. Under normal conditions both sides, the cooling coils inside the refrigerator, and the condenser coils on the back of the refrigerator, contain liquid and vapor in phase equilibrium. The compressor, which is typically driven by an electric motor, compresses the fluid adiabatically and delivers it to the condenser coils at high pressure. The fluid temperature is higher than that of the air surrounding the condenser, so the refrigerant gives off heat | Q exhaust | and partially condenses to liquid. The fluid then expands adiabatically into an evaporator at a rate controlled by the expansion valve. As the fluid expands, it cools to the point where it is significantly colder than the air in the interior space. It absorbs then heat Q removed from that space, cooling it and partially vaporizing. Then the fluid enters the compressor and another cycle begins. To measure the efficiency of refrigerators, a quantity called coefficient of performance is used. It is defined as COP R = Q removed W . In the numerator we put the thing we want - heat removed. In the denominator we put the thing we pay for - the work done by the electric motor of the compressor. The maximum possible COP is given by that of an ideal Carnot refrigerator that operates between the same temperature limits: COP R ≤ T C T H - T C . Air conditioner operate on the same principle, except that the space from which we are removing heat is rather large, compared to the interior space of a fridge, and the thermal insulation between the inside and the outside “worlds” is not that good. To compensate for that, air conditioners typically employ a fan and a blower, in addition to a compressor. Figure 10: Air conditioner. A heat pump is similar to a refrigerator, except that it delivers heat | Q delivered | to an interior space (the house), and removes heat Q removed from the outside world. It requires work input W . The coefficient of performance of a heat pump is defined as COP HP = | Q delivered | W . The maximum possible coefficient of performance is that of a Carnot heat pump operating between the same temperature limits: COP HP ≤ T H T H - T C . A typical heat pump that is installed in residential buildings, would have a coefficient of performance around 3 . 0 or so. This means that we get 300 kW.h of thermal energy, but paying only for 100 kW.h of electrical 11

Evaluation We calculate COP HP = ($0 . 165 /kW.h )(30 kW.h/gallon ) $1 . 75 /gallon ≈ 2 . 83 Assessment The COP is a dimensionless quantity. All measurement units should cancel, which they do. The numerical value is of the right order of magnitude. The answer makes sense. 5 The Stirling and the Brayton thermodynamic cycles The Stirling and the Brayton cycles differ from the Carnot cycle only in half of their processes. The isentropic processes in Carnot’s cycle are replaced by constant volume processes in the Stirling cycle, while the isothermal stages are replaced by constant pressure processes in the Brayton cycle. d c a b min max a b c d min H C (a) (b) min max V V S S max Temperature Pressure Volume Entropy p p Brayton Cycle Stirling Cycle T T Figure 12: (a) Stirling cycle. (b) Brayton cycle. The working substance absorbs heat in processes colored in red, and releases heat in processes colored in blue. The Stirling engine is an external combustion engine, invented by Rev. Robert Stirling in 1816 to provide an alternative to steam engines, which were extremely dangerous devices at the time. Explosions due to malfunctioning steam valves were very common. Burns by hot steam are far more dangerous than burns caused by boiling water. Steam, due to its high latent heat of vaporization, can release large amounts of heat while condensing to liquid water. The engine uses typically air or Helium as its working substance. The working substance is sealed completely inside the body of the engine. The input heat is supplied externally. The source could be solar energy, heat generated in a radioactive decay, or even open flame. There is no combustion taking place inside the body of the engine. The engine operates much more quietly than any internal combustion engine. The idealized Stirling thermodynamic cycle is shown in Fig. 12. It consists of 1. isothermal expansion a → b 2. constant volume cooling b → c 3. isothermal compression c → d 4. constant volume heating d → a . 13

Let us assume that the temperature of the hot reservoir is T H = 600 K and of the cold reservoir is T C = 300 K . A Carnot engine will have an efficiency of 50% when operating between these temperature limits. The ratio r = V max V min is called the compression ratio of the engine. We shall calculate the efficiency of the engine by finding the ratio of the net work output to the heat input of the engine. The gas is clearly heated along b → c , where it cannot do any work. It is also absorbing heat when expanding isothermally, since then Δ E int = 0. According to the First Law of Thermodynamics Δ E int = 0 = Q ab + W ab , so we get Q ab = nRT H ln r. The heat absorbed at constant volume is Q da = nC V ( T H - T C ) . Thus our net heat input is Q input = Q ab + Q da = nRT H ln r + nC V ( T H - T C ) . Work is done only during the isothermal processes. The net work done by the gas is W net,output = W ab + W cd = nRT H ln r - nRT C ln r = nR ( T H - T C ) ln r. The efficiency is = W net,output Q input = nR ( T H - T C ) ln r nRT H ln r + nC V ( T H - T C ) = Δ T T H + ( C V /R )Δ T/ ln r . Notice that efficiency of the engine does not depend on the number of moles n ; on how much gas there is to do the work. This is always the case. The Carnot efficiency is c = Δ T T H . Thus we immediately see that < c For a compression ratio of 7, hot temperature 600 K , temperature difference of 300 K , and monoatomic gas (Helium), we get = 300 K 600 K + (300 K )(3 / 2) / ln 7 = 0 . 36 = 36% . In practice, thermal efficiencies of actual Stirling engines can be as high as 40%, compared to gasoline engines at 20% and diesel engines at about 30%. This is achieved by using a heat regenerator to capture some of the exhaust heat rejected during the constant volume cooling and re-inject it into the body of the engine. The main disadvantages of the Stirling engines is that they are rather bulky for a given power output, and their power cannot be easily controlled (no valves). The operation of gas turbines and jet engines is based on the Brayton cycle. They are internal combustion devices, where air is initially compressed and injected into a combustion chamber, together with the fuel. Upon combustion, the hot air-fuel mixture passes through a turbine (think of the analog of wind turbine) where it performs useful work by spinning the blades of the turbine. The exhaust gases are returned back to the atmosphere. The idealized Brayton cycle consists of the following stages: 1. combustion at constant pressure a → b 14