Chapter 7. Thermodynamics

Transcription

1 Chapter 7 Thermodynamics 7.0 Introduction Thermodynamics is a branch of physics, which deals with the energy and work of a system, was first studied in the 19 th century as scientists were first discovering how to build and operate steam engines. Thermodynamics deals with the large-scale response of a system in microscopic change, which can be observed and measured in experiments. Small-scale gas interactions are described by the kinetic theory of gasses, which is a compliment to thermodynamics. In this chapter, the laws of thermodynamic shall be discussed in great detail. 7.2 Review the Basis of Thermodynamic Let s review some basis of thermodynamic. They are covered in the following paragraphs. Heat is the mechanism by which energy is transferred between system and its environment because there is temperature difference between them. Internal energy U is the energy associated with the microscopic component of the system such as atom and molecule, when view from reference frame of rest with respect to the system. It includes kinetic energy and potential energy associated with the random translational, rotational, and vibrational motion of the atoms or molecules and intermolecular forces. For an ideal monoatomic gas, the internal energy is just the translational kinetic energy of the linear motion of the atom. For polyatomic gases, there is rotational and vibrational kinetic energy as well. In liquids and solids there is potential energy associated with the intermolecular attractive forces. A simplified visualization of the contributions to internal energy is shown in Fig Thermodynamics 101

2 Figure 7.1: Internal energy contribution of gas, liquid, and solid If the water is placed in a pot and heated by flame, the internal energy of water increases due to the heat transferred from the flame to the water. Comparison of internal energy for copper and water is shown in Fig Figure 7.2: Comparison of internal energy for copper and water To set-up a temperature scale, one has to pick a reproducible thermal phenomenon and arbitrary assigns a certain Kelvin temperature to its environment. This standard fixed-point temperature is select at the triple point of water. The triple point temperature T 3 of water is T 3 = K. The triple point temperature is the temperature where liquid water, solid ice, and water vapor can coexist in thermal equilibrium at a fixed set value of pressure and temperature. The triple point temperature can be measured using a constant volume gas thermometer shown in Fig. 7.3, filled with different type of gas such as hydrogen, helium or nitrogen. At constant volume, the pressure P of the in thermometer is dependent on the temperature of the liquid where it is immersed. Thermodynamics 102

3 Figure 7.3: Constant-volume gas thermometer Thus, T = Cp (7.1) where C is a constant and P the pressure is is equal to P = P o - hg. P o is the atmospheric pressure which is 1.01x10 5 pascal Pa (Newton/m 2 ) or 760 torr or 14.7 lb/in 2. At triple temperature the pressure shall be P 3, and then an equation relating it is P T K (7.2) P 3 However, it was found that at boiling point of water, different gas gives different result. Thus, it is necessary to measure the triple point temperature when the volume in the thermometer approaches zero, where is now independent of the gas type. Equation (7.2) shall then be changed to P T K lim. Figure 7.4 shows the result of the temperature-pressure plot V0 P 3 of constant-volume thermometer filled with helium, nitrogen, and hydrogen. Thermodynamics 103

4 Figure 7.4: Result of temperature-pressure plot of constant-volume thermometer filled with three-gas type There are two ways that the unit of heat is being defined. In British system, it is the British thermal unit BTU, in which one BTU is defined as the heat required to raise the temperature of 1.0 lb of water from 63 0 F to 64 0 F. Alternatively, one calorie is the energy required to raise the temperature of 1.0 g water from C and C. In 1948, scientist linked the heat like work defined as the energy transfer and defined one calorie c as equals to J, which is equal to 3.969x10-3 BTU. This shall mean 1.0 BTU = calorie = 1, J. The specific heat c m of a material is defined as the amount of energy necessary to raise the temperature of one kilogram of that material by one Q degree Celsius. In mathematical expression it is equal to c m, where m(t f T i ) m is the mass, Q is the heat or energy. Figure 7.5 shows the specific heats of some materials at room temperature. Substance Specific Heat cal/g-k J/kg-K Lead Tungsten Silver Copper Aluminum Brass Thermodynamics 104

5 Granite Glass Ice (-10 0 C) Mercury Ethyl alcohol Sea water Water Figure 7.5: specific heats of some materials at room temperature 7.3 Thermal Expansion When the temperature of material increases, its internal energy increases. This result its dimensions increase due to increase in inter-atomic or intermolecular distance that can be caused by vibration, rotational moment, and etc. If the temperature of metal rod of length L is raised by temperature T, its length is increased by an amount following equation (7.3). L = LT (7.3) where is the coefficient of linear expansion, which has unit per Kelvin or per degree Celsius. The volume of the material will also be increased with the increase of temperature, whereby the increase of temperature follows equation (7.4). V = VT (7.4) where is the coefficient of volume expansion. The relationship between coefficient of linear expansion and coefficient of volume expansion is = 3. Example 1 On a hot day in Las Vegas, an oil tanker loaded 37,000 liters of diesel fuel. The driver unloaded the entire load of diesel in Payson, Utah, where the temperature is 25.0 K lower than in Las Vegas. How many liters did he deliver? What is the percentage increase of the cost price per liter of diesel? Given that the coefficient of volume expansion of diesel is 9.54x10-4 / 0 C. Thermodynamics 105

6 Solution The change in volume of diesel in Payson is 37,000x9.54x10-4 x(-25) = liters. In Payson the driver delivers 37, = 36, liters of diesel. The increase of the cost of diesel is (37,000/36, )100 % = 2.44 % per liter. 7.4 Heat Transformation (Latent Heat) When a matter absorbs energy, it is not necessary used to raise the temperature of the material. The energy may be used to change from one phase or state to another. All matters can exist in three common phases. i.e. solid, liquid, and gas states. Take for an example, the temperature of solid ice and ice water is the same, and the temperature of boiling water and steam is the same. The energy absorbed by the solid ice is used to break the intermolecular force so that the molecules can become loosely bonded. This energy is termed as heat of fusion. It has unit J/kg. Boiling water absorbs energy to break the weak intermolecular bond into free molecular gas. The energy required is termed as heat of vaporization. Figure 7.6 shows the heat of fusion and heat of vaporization of the some substances. Melting Boiling Substance Melting Point K Heat of Fusion kj/kg Boling Point K Heat of Vaporization kj/kg Hydrogen Oxygen Water ,256 Mercury Lead , Copper 1, ,868 4,730 Silver 1, ,323 2,336 Figure 7.6: Heat of fusion and heat of vaporization of some materials From the table, it shows that the heat of vaporization is higher than heat of fusion, which is logically true because it requires less energy to break the solid bond into loose bond than to free the atom or molecule from inter-atomic or intermolecular bonds. Thermodynamics 106

7 Example 2 How much heat must be absorbed by ice of mass m = 720 g at 10 0 C to take it to liquid state at 15 0 C? Solution Three steps are to be considered which are shown in the diagram below. The energy required to raise the temperature of solid ice from 10 0 C to 0 0 C is 10x0.720x2,200J/kg-K = 15.84x10 3 J. The energy required to break the intermolecular bond to loose bond is the heat of fusion, which is equal to 0.72x333x10 3 = 239.8x10 3 J. The energy required to heat ice water from 0 0 C to 15 0 C is 0.72x15x4,190 = 45.25x10 3 J. The total energy required is x10 3 J. 7.5 Heat Transfer Mechanism Heat can transfer between a system and its environment via three mechanisms, which are conduction, convection, and radiation. Conduction Conduction is transfer of kinetic energy and vibrational energy of atoms to the atom of lower kinetic energy and vibrational energy. The rate of conduction dq P cond, which is also equal to follows equation (7.5). dt Thermodynamics 107

8 P cond TH TC k A (7.5) L where k is the thermal conductivity, L is the thickness of the transfer material, and A is the face contact area. Figure 7.7 illustrates the conduction. Figure 7.7: Thermal conduction between hot reservoir and cold reservoir via a slab Thermal resistance R is defined as R = k L (7.6) TH TC Equation (7.5) shall become Pcond A after substituting equation (7.6). If R there are two insulating layer of difference thermal conductivity k, it can be TH TC shown that the rate of conduction is Pcond A, where R 1 and R 2 are R1 R 2 respectively the thermal resistance of material 1 and material 2. If two materials are connected in parallel with the ends contact the hot and cold reservoirs then the rate of conductor P cond(total) shall be equal to sum of rate of conduction P cond(1) of material 1 and rate of conduction P cond(2) of material 2. Thus, equation (7.5), the rate of conduction P cond(total) is P cond(total) = A 1 T H T R 1 C + A 2 T H T R 2 C (7.7) Thermodynamics 108

9 Example 3 The wall of house is made of white pine, unknown materials, and a brick wall. The thickness L d of brick wall is two times the thickness of white pine L a. The thermal conductivity k d of the brick is five times the thermal conductivity k a of white pine. The temperature of indoor T 1 is = 25 0 C. The temperature T 2 = 20 0 C, T 5 = C. If steady state transfer of heat is attained, what is the temperature T 4 of the interface between brick and the unknown material? Solution Since steady state of transfer of heat is attained, the conduction rate of white T1 T2 T4 T5 pine is same as the brick wall, which is Pcond k a A and Pcond k d A. La Ld Thus, T1 T2 T4 T5 T4 T5 Pcond k a A k d A 5kaA. The equation is now contains one La Ld 2La unknown T 4 and T 4 is found to be 8 0 C. Convection Convection is occurred when liquid or gases come in contact with object whose temperature is higher than that of liquid or gas. The gas or liquid comes in contact hot object. Its temperature increases and becomes less dense. It buoyant force causes it to rise. The surround liquid or gas is then flowed in to take place of rising warm liquid or gas. The process continues is termed convection. An illustration is shown in Fig Figure 7.8: An illustration of convection of thermal heat transfer Thermodynamics 109

10 Radiation The transfer of heat between a system and its environment via electromagnetic waves is called thermal radiation. A very good example of thermal radiation is visible light. The rate of thermal radiation P rad at which an object emits energy via electromagnetic radiation depends on surface area A, temperature of that area and emissivity that has a value between 0 and 1, which surface dependent. A surface that has maximum emissivity of 1 is said to be a blackbody radiator. The rate of radiation P rad follows equation (7.8). 4 P rad = AT (7.8) where is the Stefan-Boltzmann constant that has value 5.67x10-8 J/s-m 2 -K 4. Indeed Stefan-Boltzmann constant is a universal constant that applies to all bodies, regardless of the nature of surface. The rate of absorption P abs of an object from its environment, which has uniform temperature T env, is 4 P abs = AT env (7.9) Owing to the fact that the object radiates energy to the environment and absorbs from the environment, the net rate of radiation P net shall be 4 4 P net = AT T (7.10) env Example 4 The supergiant star Betelgeuse in constellation Orion has a surface temperature about 2,900 K and emits a radiant power of approximately 4.0x10 30 W. The temperature is about half and the power is about 10,000 greater than that of the sun. Assuming that Betelgeuse and sun have perfect emissivity and spherical shape, calculate the radii of the supergiant and sun. Solution 4 Prad From Stenfan-Boltzmann equation, P rad = AT, the surface area A is A=. 4 T 30 Prad 4.0x10 The radius R of supergiant shall be R = = = T 4x5.67x10 x x10 11 m, which is larger than the orbit of mar, which is 2.28x10 11 m. Thermodynamics 110

11 Prad The radius of sun is R = 4 4T x10 4x5.67x10 x5800 = 8 4 = 7.04x10 8 m. Example 5 A wood-burning stove stands unused in a room where the temperature is 18 0 C. A fire is started inside the stove. Eventually the temperature of the stove surface reaches a constant temperature of C and the room warms to a constant temperature of 29 0 C. The stove has an emissivity of 0.9 and a surface area of 3.50 m 2. Determine the net radiant power generated by the stove when the stove (a) is unheated and has temperature equal to the room temperature (b) has temperature of C. Solution The power generated by unheated stove is P rad = 4 AT = 5.67x10 8 x0.9x3.5x291 4 = 1, W. - 4 The power absorbed from the surrounding by the stove is P rad = AT = 5.67x10-8 x0.9x3.5x291 4 = W. The net power generated by the stove is zero. The net power generated by the stove when it is heated is P net = A T T = 0.9x3.5x5.67x10 ( ) = W. env 7.6 Thermodynamic Processes A thermodynamic process is the way that a system changes from one state of thermal equilibrium to another state such as the gas may be in thermodynamic equilibrium for specified temperature, pressure, and volume. Heat transfer and work done will change this equilibrium state to another. Thermodynamic process can be divided into quasi-static (reversible) processes and irreversible processes. If a thermodynamic system undergoes a change from one state in thermal equilibrium to another state slowly enough such that in any instant of time the system is in thermal equilibrium is said to be reversible process. In reality this process does not exist. Any thermal process that is not a quasi-static process is an irreversible process. In nature, all thermodynamic processes are irreversible. There are four special types of quasi-static processes that are useful in equilibrium thermodynamics. The processes are characterized by the Thermodynamics 111

12 thermodynamic parameter that is kept constant during process while other parameters change. The types are: isothermal process where the temperature of the system is kept constant, isobaric process where the pressure is kept constant, isochoric process where the volume is kept constant and adiabatic process or isentropic process where no heat is transfer in or out of the system from or to the environment. 7.7 Laws of Thermodynamics There are three principal laws of thermodynamics, which are zeroth law the thermodynamic equilibrium law, first law - internal energy law, second law - entropy law. Each law leads to the definition of thermodynamic properties, which help to understand and predict the operation of a physical system. There is another law, which is third law state that it is impossible to reach zero Kelvin temperature in finite number steps. This law shall not be discussed. 7.8 Zeroth Law of Thermodynamics Zeroth law of thermodynamic involves some simple definitions of thermodynamic equilibrium and temperature. It is observed that some properties of an object, like the pressure in a volume of gas, the length of a metal rod, or the electrical conductivity of a wire, can change when the object is heated or cooled. Some of these phenomenons have been demonstrated in the earlier section. Zeroth law states that if two systems are at the same time in thermal equilibrium with a third system, they are in thermal equilibrium with each other. If the two objects are initially at different temperatures into physical contact, they will eventually achieve thermal equilibrium. During the process of reaching thermal equilibrium, heat is transferred between the objects and there is a change in the property of both objects such as their internal energy. Thermodynamic equilibrium leads to the large-scale definition of temperature. When two objects are in thermal equilibrium they have the same temperature. The details of the process of reaching thermal equilibrium are described in the first and second laws of thermodynamics. As an illustration in Fig. 7.5, object A and object B are in physical contact and in thermal equilibrium. Object B is also in thermal equilibrium with object C. There is initially no physical contact between object A and object C. But, if object A and object C are brought into contact, it is observed that they are in Thermodynamics 112

13 thermal equilibrium. This simple observation allows making of thermometers, in which it can be calibrated to measure the change in a thermal property such as the length of a column of mercury by putting the thermometer in thermal equilibrium with a known physical system. If the thermometer is brought into thermal equilibrium with any other system such as placing under tongue, the temperature of the other system is known by noting the change in the thermal property. Objects in thermodynamic equilibrium have the same temperature. Figure 7.5: Illustration of Zeroth law of thermodynamic 7.9 First law of thermodynamics First law of thermodynamics relates the various forms of energy, kinetic and potential, in a system to the work, which a system can perform and to the transfer of heat. It is used extensively in the discussion of heat engines. The first law of thermodynamics defines the internal energy E is equal to the difference of the heat transfer Q into a system and the work W done by the system. U = U 2 - U 1 = Q W (7.11) Heat Q is absorbed by the system uses to increase the internal energy U 2 U 1. The internal energy is used to perform work done W, resulting decrease of internal energy. This is the rationale how equation is formed. Heat removed from a system would be assigned a negative sign in the equation. Similarly work done on the system is assigned a negative sign. Figure 7.7 shows the interpretation of the law. Thermodynamics 113

14 In chemistry textbook, first law is written as U= Q + W. It is the same law, of course - the thermodynamic expression of the conservation of energy principle. It is just that W is defined as the work done on the system instead of work done by the system. In the context of physics, the common scenario is one adding heat to a volume of gas and using the expansion of the gas to do work, as in case pushing down of a piston in an internal combustion engine. In the context of chemical reactions and process, it may be more common to deal with situations where work is done on the system rather than by the system. Figure 7.7: Interpretation of first law of thermodynamic State 2 shows that there is decrease in volume, therefore work done is negative. v2 Work done W is also defined as W pdv. It is clearly shown that if V2 < V 1, the work done is negative. V1 Example 6 The temperature of three moles of a monoatomic ideal gas is reduced from temperature T i = 540 K to T f = 350 K by two different methods. In the first method 5,500 J of heat flows into the gas, while in the second method, 1,500 J of heat flows into it. In each case find the change of internal energy and (b) work done by the gas. Solution Since the gas is monoatomic type, the internal energy of the gas is only the translational kinetic energy, which follows equation 3 nrt. The change in 2 Thermodynamics 114

15 internal energy from temperature 540 K to 350 K shall be nr(t T ) 3 3x8.31( ) = -7,105 J. First method shall yield W = Q - U = 5,500 J +7,105 J = 12,605 J Second method shall yield 1,500 J + 7,105 J = 8,605 J. 3 2 f i = Let s understand the relationship of specific heat capacity and first law of thermodynamic and also look at some special process cases that applying first Law of Thermodynamics Specific Heat Capacities and First Law of Thermodynamics As mentioned in the earlier section the heat capacity is defined as Q = CmT. The heat capacity for a gas shall then be Q = CnT. Instead of mass, it is now replaced by number of mole. The internal energy U of an ideal monoatomic gas is U = 3 2 nrt. The change of internal energy U from temperature T i to T f shall be U = 3 nr(t f T i ). 2 For isobaric process, the work done is W = P(V f V i ). However, for ideal gas PV = nrt. This shall mean that the work done is also W = nr(t f T i ). For isochoric process, the work done is equal to zero since there is no volume change. Using the first law equation Q = U + W, for isochoric process Q = 3 3 nr(t f T i ). This implies that the specific heat C is equal to R 2 2 C V to denote specific heat for constant volume. 3 nr(t 2 5 R 2. We shall use Similarly the heat supplied for isobaric process is equal to Q = f T ) + nr(t f T i ). This shall mean that the specific heat capacity is i. We shall use C P to denote specific heat for constant pressure. Thermodynamics 115

16 7.9.2 Thermodynamic Process Let s look at several thermodynamic processes in detail here. Adiabatic process The Adiabatic process is a process that occurs so rapidly that there is no transfer of heat between the system and its environment. From first law of thermodynamics with Q = 0 shows that the change in internal energy is in the form of work done. Therefore, from equation (7.6) become U 2 - U 1 = W. Free expansion is an adiabatic process whereby there is no transfer of heat between the system and its environment, no work done on and by the system. Thus, from equation (7.6), Q = W = 0. For adiabatic condition PV = constant K, where is the ratio of C p /C v heat capacity at constant pressure and constant volume. This implies that P = K/V and the work done shall be diagram is shown in Fig.7.8. K(V W dv = v2 K V V 1 1 f V 1 1 i ). The pressure-volume Figure 7.8: The work done graph of an adiabatic process Note that for an ideal monatomic gas, value is equal to 5/3. Isochoric Process Isochoric process is a constant volume process that there is no change in volume. Thus, the work done is zero. Therefore, equation (7.6) becomes U 2 - U 1 = Q, Thermodynamics 116

17 which is equal to C V nt. From the ideal heat equation VP = nrt. This shall C V mean Q = V(Pf Pi ). The pressure-volume diagram is shown in Fig.7.9. R Figure 7.9: The work done graph of an isochoric process Cyclical process Cyclical process is a process after certain exchange of heat and work done, the system restores back to its initial state. Thus, there is no change in its internal energy. This implies that equation (7.6) becomes Q = W. Isothermal Process The isothermal process is a process whereby the temperature is kept constant that is no change of internal energy. This implies that equation (7.6) becomes Q = W. The work done is W pdv. Substituting P = v2 V1 nrt into the equation V v2 W nrt dv V = nrt V f ln. The pressure- volume diagram is shown in Fig. V 1 Vi Figure 7.10: The work done graph of an isothermal process Thermodynamics 117

18 Example 7 Two moles of the monatomic gas argon expand isothermally at temperature 298 K from an initial volume of V i = m 3 to a final volume of V f = m 3. Assuming that argon is an ideal gas, find (a) the work done by the gas, (b) the change in the internal energy of the gas and (c) the heat supplied to the gas. Solution Vf The work done by isothermal process is W = nrt ln = 2x8.31x298 ln Vi = 3,432.9 J. The internal energy of U = of kinetic energy is zero. 3 2 nrt. Since the temperature is constant, the change Using first law equation, the heat supplied Q is 3,432.9 J. Isobaric process Isobaric process is process where the pressure is kept constant. The work done W is W = PV = P(V f V i ). T = PV/(nR), thus the change in temperature T = P (V nr V ) f i. The heat absorbed shall be Q = C p (V f Vi ) shown in Fig P R. The PV diagram is Figure 7.11: The work done graph of an isobaric process Thermodynamics Potentials There are four thermodynamic potentials that are useful in the chemical thermodynamics of reactions and non-cyclic processes. They are internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy. Thermodynamics 118

19 Helmholtz free energy F states that F = U TS, where T is the absolute temperature and S is the final entropy. The system in an environment of temperature T, energy can be obtained by spontaneous heat transfer to and fro between environment and the system. Helmholtz free energy is a measure of the amount of energy that put in to create a system once the spontaneous energy transfer to the system from the environment is accounted. Enthalpy is H defined as H = U + PV. It is analogue to first law of thermodynamic Q = U + PV. It is a useful quantity for tracking chemical reactions. In an exothermic reaction, energy is released to a system. It has shown up in some measurable forms in terms of the state variables. An increase in the enthalpy H = U + PV may be associated with an increase in internal energy which can be measured by calorimeter, or with work done by the system, or a combination of the two. If the process changes the volume, as in a chemical reaction which produces gas, then work must be done to produce the change in volume. For a constant pressure process the work done to produce a volume change V is PV. Gibbs free energy G states that G = U TS + PV, where P is the absolute pressure and V is the final volume. As discussed in enthalpy, an additional amount of work PV must be done if the system is created from a very small volume into large volume system. As discussed in Helmholtz free energy, an environment at constant temperature T will contribute an amount TS to the system, reducing the overall energy necessary for creating the system. This net energy contribution for a system created in environment with temperature T from an initial very small volume is Gibbs free energy. The change in Gibbs free energy G, in a reaction is a very useful parameter. It can be treated as the maximum amount of work obtainable from a reaction. For example, in the oxidation of glucose, the change in Gibbs free energy is G = 686 kcal = 2,870 kj. This reaction is the main energy reaction in living cells. Example 8 Electrolysis of water in hydrogen and oxygen is a very good example for the application of the above mentioned thermodynamic potentials. This process is presumed is done at temperature 298K and one atmosphere pressure, and the relevant values are taken from a table of thermodynamic properties shown below. Thermodynamics 119

20 The process must provide the energy for the dissociation and the energy to expand the produced gases. Both of those are included in the change in enthalpy which is shown in the table. Quantity H 2 O H O 2 Change Enthalpy kj 0 0 H = kj Entropy J/K J/K 0.5 x J/K TS = 48.7 kj At temperature 298K and one atmosphere pressure, the system work is W = PV = (101.3 x 10 3 Pa)(1.5 moles)(22.4 x 10-3 m 3 /mol)(298k/273k) = 3,715.0 J Since the enthalpy H= U + PV, the change in internal energy U is then equal to U = H - PV = kj kj = kj. This change in internal energy must be accompanied by the expansion of the gases produced, so the change in enthalpy represents the necessary energy to accomplish the electrolysis. However, it is not necessary to put in this whole amount in the form of electrical energy. Since the entropy increases in the process of dissociation, the amount TS can be provided from the environment at temperature T. The amount must be supplied by the battery, which is actually the change in the Gibbs free energy. i.e. G = H - TS = kj kj = kj. Since the electrolysis process results in an increase in entropy, the environment contributes energy amounted to TS. Gibbs free energy tells the amount of energy in other forms must be supplied to get the process to proceed. Thermodynamics 120

21 7.10 Second Law of Thermodynamics Ice cream melts and a cold can of soda get warm-up when they are left in hot day. They never get colder when left in hot day. This spontaneous flow of heat is the focus of one of the most profound law of science, which is the second law of thermodynamics. They are several ways that second law of thermodynamic can be defined. Based on the above examples, the first statement of the second law of thermodynamics - heat flows spontaneously from a hot to a cold body. An explanation for this form of the second law can be obtained from Newton's laws and the microscopic description of how heat flow through conduction occurs when fast atoms collide with slow atoms, transferring some of their kinetic energy in the process. One might wonder why the fast atoms don't collide with the cool atoms and subsequently speed up, thereby gaining kinetic energy as the cool atoms lose kinetic energy - this would involve the spontaneous transfer of heat from a cool object to a hot object, in which it is the violation of the second law. From the laws of conservation of momentum and energy, in a collision between two objects, the faster object slows down and the slower object speeds up. It is possible to make a cool object in a warm place cooler like the refrigerator but this involves the input of some external energy. As such, the flow of heat is not spontaneous in this case. The second form of the second law of thermodynamic states heat cannot be completely converted into other forms of energy that places some practical restrictions on the efficiency like internal combustion and steam powered engines. The third statement of second law of thermodynamic states that there exists useful state variable called entropy, which is defined as the measure of number of state variable for a system at a given time. The change in entropy S is equal to the heat transfer Q divided by the temperature T. S = S f f S i = Q/T = i dq Q T TAvg (7.12) Thermodynamics 121

22 The entropy of the system and the environment will remain a constant if the process is reversible. If the initial and final states are denoted by S i and S f then S f = S i for reversible system. An example of a reversible process would be ideally forcing a flow through a constricted pipe. Ideal means no losses. As the flow moves through the constriction, the pressure, temperature, and velocity would change, but these variables would return to their original values downstream of the constriction. The state of the gas would return to its original conditions and the change of entropy of the system would be zero. The second law also states that if the physical process is irreversible, the entropy of the system and the environment MUST increase. The final entropy must be greater than the initial entropy. An example of an irreversible process is a hot object put in contact with a cold object. Eventually, they both attain the same equilibrium temperature. If objects are separated after attaining thermal equilibrium, they do not naturally return to their original different temperature states. The process of bringing them to the same temperature is irreversible Heat Transfer from Hot Object to Cold Object Let s look at how the second law describes why heat is transferred from the hot object with temperature T 1 to the cold object with temperature T 2. If heat is transferred from the hot object to the cold object, the amount of heat transferred is Q and the final equilibrium temperature for both objects is T f. The temperature of the hot object changes as the heat is transferred away from the object. The average temperature T h of the hot object during the process is the average of T 1 and T f. Similarly, for the cold object, the final temperature of the cold object is T f. The average temperature T c during the process is the average of T f and T 2. The entropy change for the hot object will be (-Q/T h ), with the minus sign applied because the heat is transferred away from the object. For the cold object, the entropy change is (Q/T c ), positive value because the heat is transferred into the object. So according to equation (7.12), the total entropy change for the whole system would be given by the equation. S f - S i = -Q/T h + Q/T c (7.13) where S i and S f being the final and initial values of the entropy. Temperature T h is greater than temperature T c, because T 1 is greater than T 2. The term Q/T c will always be greater than -Q/T h and therefore, S f will be greater than S i, as the second law predicts. Thermodynamics 122

23 If the heat was being transferred from the cold object to the hot object, then the final equation would be S f = S i + Q/T h -Q/T c. The signs on the terms would be changed because of the direction of the heat transfer. T h would still be greater than T c, and this would result in S f being less than S i. The entropy of the system would decrease, which would violate the second law of thermodynamics Heat Engines A heat engine is a device that uses heat to perform work. Essentially it has three features. It has a hot reservoir that supplies heat. Part of the heat is used to perform work by the working substance of the engine like gasoline- air mixture in the automobile engine. The remaining part of the heat is rejected at temperature lower than the input temperature called cold reservoir. The schematic of a heat engine is shown in Fig Figure 7.12: A schematic representation of a heat engine For an engine to be efficient, it must produce a relatively large amount of work from the input heat. The efficiency e of a heat engine is defined as the ratio of W the work done by the engine to the input heat Q H. i.e. e =. However, QH is also equal to Q H = Q C + W, thus, the efficiency is Q H Thermodynamics 123

24 e = Q C 1 (7.14) Q H Example 9 An automotive engine has an efficiency of 22.0% and produces 2,510 J of work. How much heat is rejected? Solution From equation (7.12), The heat rejected is Q C = Q H (1- e). But Q H is equal to Q C + W. Thus, Q C = (Q C + W)(1 - e). This implies that Q C = W/e - W = 2,510(1/0.22-1). = 8,899.0 J Carnot Engine According to a French engineer Sadi Carnot, in order to get maximum efficiency for an engine, the process within the engine must be reversible. This shall mean that the process will return to its initial states with no wasteful transfer of energy. Figure 7.13 shows the pressure-volume plot of the Carnot engine. Figure 7.13: Pressure-volume diagram of a Carnot engine During the process step a to b, which is an isothermal expansion, heat Q H is absorbed by the working substance from the hot reservoir. Since the process is an isothermal expansion process, the work done by the system is also equal to heat absorbed. During the process step c to d, the working substance is releasing heat Q C to the cold reservoir. Since this is an isothermal compression, work is Thermodynamics 124

25 done by the environment on the system. From concept of heat engine, there is no heat transfer between hot and cold reservoir directly. Thus, the process step bc and de must be the adiabatic process. The work done by the Carnot process shall be the orange color area of the graph shown in Fig According to first law of thermodynamic, U = Q W. Since the process is reversible, therefore the change of internal energy U is zero. Also for differential change dq = dw. The work done W shall be W = Q H Q C. From process step a to b there is a positive change of entropy S H = Q H /T H and during the process step c to d, there is a negative change of entropy S L = Q L /T L. There is no change of entropy for process step bc and da because there is not heat absorbed or released. The change of entropy shall be S = Q H /T H - Q L /T L. Since Carnot engine is a reversible engine, the change of entropy shall be zero. This implies that Q H /T H = Q L /T L. From the equation T H > T C, this shall mean Q H > Q C. This analysis confirms the second statement of second law of thermodynamic. The efficiency of the Carnot engine shall be e = W/Q H = (Q H Q C )/Q H = 1- T L /T H Refrigeration Refrigeration is a device that uses work to transfer energy from low temperature reservoir to high temperature reservoir. Figure 7.14 shows the schematic of a refrigerator. Figure 7.14: The schematic of a refrigerator Thermodynamics 125

26 A measure of the efficiency of a refrigerator is the coefficient of performance K, which is defined as the ratio of heat extracted from the cold reservoir to the work done. For an ideal Carnot refrigerator the coefficient of performance K QL Q L Q =. The change of entropy for ideal refrigerator shall be S = L. QH QL TL TH Notice that T H >T L, implying the S is a negative value, which is not allow for second law of thermodynamic. Therefore, there is no perfect refrigerator. Thermodynamics 126

27 Tutorials 1. A certain diet doctor encourages people to diet by drinking ice water. His theory is that the body must burn off enough fat to raise the temperature of water from C to the body temperature of C. How many liters of ice water would have to be consumed to burn-off 454 g of fat, assuming that this much fat required 3,500 Cal be transferred to ice water? Why this is not advisable to follow this diet. 2. A tank of water has been outdoor in cold weather and a slab of ice 4.0 cm thick has formed on its surface. The air above the ice is 10 0 C. Calculate the rate (centimeter per sec) of ice formation. Take the thermal conductivity of the ice to be cal/s-cm-k and density 0.92 g/cm 3. Assume the energy is transferred through the wall or bottom of the tank is negligible. 3. Find the change of entropy when a 2.3 kg block of ice melts slowly (reversible) at 273 K J of heat flowing spontaneously through a copper rod from a hot reservoir at 650 K to a cold reservoir at 350 K. Determine the amount the change in entropy for this irreversible process. The change in entropy shall be -1200/ /35 = 1.58 J/K. 5. A Carnot engine operates between temperature T H = 850 K and T L = 300 K. The engine performs 1200 J of work for each cycle that takes 0.2 s. Thermodynamics 127

28 (a). Find the efficiency of the engine. (b). How much heat Q H is extracted from high temperature reservoir for each cycle? And Q L delivered to cold reservoir? (c). What is the entropy change of the working substance for the energy transfer to it from the high temperature reservoir? From it to the low temperature reservoir? 6. If an inventor claims that he has invented an engine that has an efficiency of 75% when operated between boiling point and freezing point. Is it possible? Thermodynamics 128