Hence. The second law describes the direction of energy transfer in spontaneous processes

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1 Heat and Work

2 The first law of thermodynamics states that: Although energy has many forms, the total quantity of energy is constant. When energy disappears in one form, it appears simultaneously in other forms Hence U + E K + E P = Q W Heat and work are forms of energy that are transformed into other forms of energy. If friction is eliminated, work is efficiently transformed to potential, kinetic, electrical and heat energy by a conversion ratio of upto 100%. Heat on the other hand is readily lost to the surroundings and its conversion into work, mechanical or electrical energy does not exceed 40% efficiency because the flow of heat always takes place from the hotter to the cooler body and never in the reverse direction The second law describes the direction of energy transfer in spontaneous processes

3 Temperature rise was found to be C when 1 Joule of work was applied to 1 gram of adiabatically contained water at 14.5 C The quantitative relationship between work and heat was established as 1 J = calories The relationship between heat and work was established by Joule s experiments on water W = Q U = Q W Joule tried to convert mechanical energy of rotation of a fan into temperature rise of water. Other methods of work production used by Joule were turbulent motion, electric current, compression and friction. The same relationship between temperature rise and the amount of work applied to water was seen in all experiments:

4 An energy function that is only dependent on the internal condition of the system was necessary Internal energy U keeps a record of the energetic state of a system, excluding kinetic energy and potential energy Relativity concept of energies is used since absolute internal energy of a system cannot be measured at any state. Internal energy of system can only be defined or measured relative to a reference state U 2 (T 2,P 2 ) U = (U 2 U 1 ) U 1 (T 1,P 1 ) Thermodynamics deals with energy differences rather than absolute values of energies

5 Internal energy of the system is altered as a result of an energy exchange process. The system may absorb energies of different sorts from its surroundings during the process Example The work input to a tank by a paddle wheel is 5090 J. The heat transfer from the tank is 1500 J. Determine the change in the internal energy of the system. U = Q W = = 3590 J

6 Work is defined as a force F acting through a displacement x in the direction of the force: W = state 2 Fdx state 1 Work performed in a system is grouped into two: W = W expansion W other W other which includes electric, friction, turbulent work is often not present dw expansion = P external Adl, since area of a piston is constant = P external dv dl Work done by the system by expansion is considered positive and work done on a system by compression is considered negative Work is not a state function and is dependent on the path

7 During a process, the work done can be found by integrating between V 2 and V 1 only if the relationship between P ext and V is known W = V2 PdV V1 Expansion takes place instantaneously and irreversibly if the difference between P and P ext is great P ext >> P, W = P V Expansion will occur reversibly through a series of equilibrium states if P is only infinitesimally greater than P ext V2 P ext P, W = PdV V1 A reversible process can be defined as one that is performed in such a way that both the system and its local surroundings may be restored to their initial states without producing any permanent changes in the rest of the universe at the conclusion of process

8 Consider 3 processes with the same initial and final pressures P ext >> P, W = P V 1 atm P ext 3 atm P ex t P ex t Irreversible process gets less work out of the system than possible 1 atm 2 atm reversible 1 atm 3 atm 3 atm P ext P, W = V2 V PdV Reversible process spends more energy

9 Reversible processes are allowed to occur at infinitely slow rate to maintain a state of equilibrium of the system Reversible path represents the limiting case of maximum work output The path of a reversible expansion is shown by a curve on P-V coordinates since P 1 V Example 100 lt of an ideal gas at 300 K and 1 atm pressure expands reversibly to three times its volume by a) an isothermal process, b) an isobaric process. Calculate the work done in each case V2 a) W = PdV V1 b) W = P dv V2 V1

10 Question - Which of the following situations is considered to be a reversible process? a) A raw egg is thrown from a second story window to the ground below b) The pizza is put into the 200 C oven and baked for 15 minutes c) After the party, Ali damages his car by striking a traffic light d) The earthquake destroys an entire neighborhood e) None of these are reversible processes

11 Heat is defined as the form of energy that is transferred through the boundary of a system as a result of temperature difference Energy cannot be stored in the form of heat energy, it is transferred between bodies of different temperatures and the stored as the increase in internal energy of the heated body Heat is not a state function and is dependent on the path Computation of heat along different paths give different results When heat is transferred into the system, Q is considered positive When heat is transferred from the system, Q is considered negative An adiabatic path restricts transfer of energy in and out of the system as heat δq = 0, du = W The work is done by the system in this case at the expense of interal energy

12 Positive enthalpy for a process means it requires a heat input and is endothermic Negative enthalpy for a process means it releases heat and is exothermic A constant volume path restricts any expansion work dv = 0, δw = PdV = 0 so du = δq V, A constant pressure path results in the maximum amount of work obtained du = δq PdV δq P = du + PdV = d U + PV = dh Enthalpy, a state function, is the representation of heat at constant pressure process Most processes in practice occur at constant pressure E.g. Smelting, mixing, precipitation Enthalpies of all phase change and formation processes are tabulated

13 Enthalpies of formation of common compounds Compound ΔH f (kj/mol) Compound ΔH f (kj/mol) Compound ΔH f (kj/mol) Compound ΔH f (kj/mol) AgBr(s) H 2 O(l) C 2 H 2 (g) NH 4 Cl(s) AgCl(s) H 2 O 2 (l) C 2 H 4 (g) NH 4 NO 3 (s) AgI(s) H 2 S(g) C 2 H 6 (g) NO(g) Ag 2 O(s) H 2 SO 4 (l) C 3 H 8 (g) NO 2 (g) Ag 2 S(s) HgO(s) n-c 4 H 10 (g) NiO(s) Al 2 O 3 (s) HgS(s) n-c 5 H 12 (l) PbBr 2 (s) BaCl 2 (s) KBr(s) C 2 H 5 OH(l) PbCl 2 (s) BaCO 3 (s) KCl(s) CoO(s) PbO(s) BaO(s) KClO 3 (s) Cr 2 O 3 (s) PbO 2 (s) BaSO 4 (s) KF(s) CuO(s) Pb 3 O 4 (s) CaCl 2 (s) MgCl 2 (s) Cu 2 O(s) PCl 3 (g) CaCO MgCO 3 (s) CuS(s) PCl 5 (g) CaO(s) MgO(s) CuSO 4 (s) SiO 2 (s) Ca(OH) 2 (s) Mg(OH) 2 (s) Fe 2 O 3 (s) SnCl 2 (s) CaSO 4 (s) MgSO 4 (s) Fe 3 O 4 (s) SnCl 4 (l) CCl 4 (l) MnO(s) HBr(g) SnO(s) CH 4 (g) MnO 2 (s) HCl(g) SnO 2 (s) CHCl 3 (l) NaCl(s) HF(g) SO 2 (g) CH 3 OH(l) NaF(s) HI(g) So 3 (g) CO(g) NaOH(s) HNO 3 (l) ZnO(s) CO 2 (g) NH 3 (g) H 2 O(g) ZnS(s)

14 Zeroth law of thermodynamics states that If two bodies are in thermal equilibrium with a third body, they are in thermal equilibrium with each other and hence their temperatures are equal When a certain quantity of heat exchange takes place between the system and the surroundings, the temperature of the system changes Specific Heat: It is the amount of heat required to raise the temperature of a 1 kg mass 1 C or 1 K dq C dt dq CdT by integration Q C (T 2 Considering the mass m, Q mc (T - T ) T ) 1 KJ kg KJ

15 The knowledge of the final temperature of the system is not enough to determine the final state of the system since internal energy may also depend on volume and pressure The second variable should either be varied in a specified path or maintained constant during the change For a constant volume path recall that du = δq V Constant volume heat capacity C V = δq U = T 2 C V dt T 1 dt = du dt V, du = C V dt For a constant volume path recall that dh = δq P Constant pressure heat capacity C P = δq H = C P dt T 2 T 1 dt = dh dt V, dh = C P dt

16 Heat capacities are extensive properties with SI unit J/K However it is more convenient to use the heat capacity per unit quantity of the system, thus specific heat or molar heat capacity are used per mole of substance nc P =C P, and nc V =C V where c P and c V are the molar heat capacities Heat capacity is a function of temperature The C P and C V values for various gases are independent of temperature over a wide range of temperatures and are considered constant as an approximation to ideal gas law The values of molar heat capacities for various gases are C P = 5 2 and C V = 3 2 for monatomic gases (He, Ar, etc.) C P = 7 2 and C V = 5 2 for diatomic gases (H 2, O 2, CO, etc.) C P = 4 and C V = 3 for polyatomic gases (CO 2, CH 4, SO 2, etc.)

17 Since dh = du + PdV, C p is expected to be greater than C v If the temperature of an ideal gas is raised by dt at constant volume, all of the internal energy change equals the heat gained during the process However, if the temperature change happens at constant pressure, some energy will be needed to expand the system: PdV dt or P δv δt P Hence C p C v = P R P = R for ideal gases Ideal gas is defined to behave as a gas with infinite intermolecular distance, negligible intermolecular attraction and zero pressure. Hence, internal energy of an ideal gas is a function of temperature and independent of volume and pressure For real gases, C p C v = δv P + δu δt P δv T The derivation is as follows

18 Thus, du = C p = C p C v = δh, C δt v = δu P δt V H = U + PV δu δt P + P δu dv + δu δv T δt δu δt P = δv δt P δu δt V V dt, since U = f(v, T) δu δv + δv T δt P δu δt V C p C v = δu δv + δv T δt P δu δt V + P δv δt P δu δt V C p C v = In an attempt to evaluate δu δv δu δv + P δv T δt P δv δt P for gases, Joule performed an experiment T which involved filling a copper vessel with a gas at some pressure and connecting this vessel to a similar but evacuated vessel, allowing free expansion of the gas into evacuated vessel

19 Joule s Experiment Setup

20 Joule could not detect any change in the temperature of the system... δu U = 0, = 0 δv T As the system was adiabatic and no work was performed However Joule tried again with more precise instruments and detected a small temperature change which was undetectable by his first experimental setup due to the large heat capacity of the copper vessel which absorbed all the energy itself δv δu δt P δv represents the work done per degree rise in temperature in T expanding against the internal cohesive forces acting between the particles of the gas The term is small for gases and large for liquids and solids so C p C v gas > C p C v condensed For condensed substances C p C v

21 Q Experimental determination of heat capacities Calorimeters are used for determination of heat capacities. Amount of heat transferred from a sample to a calorimeter is calculated from the mass and specific heat of calorimeter T f Q P = H = C P dt, Q P = f(t) T 0 The process is repeated for several temperatures until a good fit of Q P versus T is obtained. Specific heat capacity of the sample is determined from temperature dependency of Q P Since Q P is the integral of C P vs T, the derivative of Q P vs T gives C P T

22 Empirical representation of heat capacities Variation of heat capacity with temperature for a substance is fitted to an expression of the form: C P = a + bt + c T 2 + dt3 + The analytical form of C P permits the calculation of enthalpy change between specific temperatures The heat required to change the temperature of the system is calculated as: H = H 2 H 1 = n T 2 T 2 C P dt = n a + bt + c T 1 T 1 T 2 dt Example Calculate the heat required when 2 moles of Cu is heated from 100 C to 800 C. C P = T J/mol K

23 54 grams of liquid silver at 1027 C and solid silver at 727 C are heated seperately. Calculate the heat required to raise their temperatures by 100 C. Molecular weight = 108 g Melting temperature = C

24 Heat capacity of many substances are unknown. Indirect calculation of the heat capacities of these materials is possible by the Neumann-Kopp method: xa(s) + yb(s) =A x B y (s) C p (A x B y ) = x (C p )A + y(c p )B Example Calculate the heat capacity of PbSO 4 PbO solid solution using the heat capacity of the solids PbSO 4 and PbO

25 Sensible Heat - The amount of heat that must be added when a substance undergoes a change in temperature from 298 K to an elevated temperature without a change in phase H = n T 298 C P dt = at + b 2 T2 c T T 298 = a + bt + c T 2 a b c 298 Heat of Transformation - The amount of heat that must be transferred when a substance completely undergoes a phase change without a change in temperature. Heat of Vaporization: The amount of heat added to vaporize a liquid or amount of heat removed to condense a vapor or gas Q ml v where: L latent heat of vaporization, KJ/kg m mass, kg, kg/sec Heat of Fusion: It is the amount of heat added to melt a solid or the amount of heat removed to freeze a liquid Q ml F where: L latent heat of fusion, KJ/kg m mass, kg, kg/sec

26 Example - Calculate the enthalpy change of 1 mole of copper when it is heated from room temperature to 1200 ºC ΔΗ m = 3100 cal/mol at 1083 ºC, C P (s)= T cal/mol.k, C P (l)= 7.50 cal/mol.k

27 Example - The normal freezing point of mercury is 38.9 ºC, and its molar enthalpy of fusion is ΔΗ fusion = 2.29 kj/mol. What is the enthalpy change of the system when the temperature of 50.0 g of Hg(l) is decreased from 25 ºC to -50 ºC? MW Hg =200.6 g, C P (s)= T /T cal/mol.k, C P (l)= T cal/mol.k