Chapter 18 Heat and the First Law of Thermodynamics

Transcription

1 Chapter 18 Heat and the First Law of Thermodynamics Heat is the transfer of energy due to the difference in temperature. The internal energy is the total energy of the object in its centerofmass reference frame. The amount of heat Q necessary to increase the temperature is proportional to the temperature difference and the mass of the object, = Δ = ΔT = mcδt where C is the heat capacity and c = C/m is called the specific heat. 1 calorie is the amount of heat needed to increase the temperature of 1 g of water by 1 0 C, 1 cal = J. In the US heat is often measured in Btu (British thermal unit amount of heat needed to increase the temperature of 1 pound of water by 1 0 F, The specific heat of water, 1 Btu = 252 cal = kj 1

2 =1 =4.184 The heat capacity per mole is called the molar heat capacity, = / = mc/n Example 1. The specific heat of bismuth is kj/kg K. How much heat is needed to raise the temperature of 2kg of bismuth from 20 0 C to C? = mcδt = 2kg kj/kg K 480 K = kj Calorimetry is measuring specific heat using calorimeters (isolated thermal baths of known heat capacity). Example 2. One puts 1.2kg of lead initially at C into aluminum calorimeter (mass 400g) with 1kg of water initially at 18 0 C. What is the specific heat of lead if the final temperature of the mixture is C and the specific heat of the container is 0.9kJ/kg K? The heat received by the calorimeter Q = m al c al ΔT + m wat c wat ΔT =0.4kg(0.9kJ/kg K)2.7K+1kg(4.18kJ/kg K)2.7K=( )kJ=12.269kJ 2

3 According to the energy conservation law, the lead sample lost the same amount of heat, Q = m lead c lead ΔT = 1.2kg 80K c lead = kJ. Therefore, c lead =12.269kJ/(1.2kg 80K) = kj/kg K. Phase Transformations and Latent Heat Common examples of phase changes are melting (solid to liquid), fusion or crystallization (liquid to solid), vaporization and boiling (liquid to gas), condensation (gas or vapor to liquid), etc. A change in phase at a given pressure occurs at a particular temperature and continues until the transformation is complete without a change in temperature. The heat necessary to complete the phase transformation Q = ml, where L is called the latent heat. 3

4 Example 3. How much heat is needed to transform 2kg of ice at 1 atm and 10 0 C into steam? First, the ice should be warmed to 0 0 C: Q 1 = mcδt = 2kg 2.05kJ/K kg 10K = 41kJ Second, the ice should be melted (use the latent heat L from the Table): Q 2 = ml f = 2kg 333kJ/kg = 666kJ The heat is also needed to bring the temperature from 0 0 C to C: Q 3 = mcδt = 2kg 4.18kJ/K kg 100K = 836kJ Vaporization of all this water requires Q 4 = ml v = 2kg 2.26MJ/kg = 4.52MJ The overall amount of heat is then Q = Q 1 + Q 2 +Q 3 + Q 4 = 41kJ + 666kJ +836kJ kJ = 6.063MJ 4

5 The First Law of Thermodynamics The temperature of a system can be raised not only by bringing in some heat, but also by using mechanical energy (the mechanical equivalent of heat), such as, for example, by rotating a paddle inside water (Joule s experiment). The energy conservation law can be written as ΔE int = Q in + W on where ΔE int is the change in the internal energy of the system, Q in stands for the heat brought into the system and W on for the work done on the system. In the differential form this equation represents the first law of thermodynamics, de int = dq in + dw on. If all internal energy of the gas is just the translational kinetic energy, then E int = nrt. The equation becomes more complicated for gases consisting of more complex molecule with internal vibrational and rotational modes. 5

6 Work and PV diagram for a gas!" # =$ % %=&'%=&(, * # =!" # = &(. * # =, &( Work can be calculated using the PV diagram for a gas. For an isobaric compression,.. * # = / &( = &/ ( = &(( 1 ( 2 )= &Δ( 6

7 Initial and final point on a PV diagram can be connected by different paths: Each path corresponds to a different amount of work, /. &( Amount of work on the gas is different for each process! Isometric heating (at constant volume) Followed by an isoba Ric compression Isobaric compression followed by an isometric heating Isothermal (T = const) compression In the case of isothermal compression ( slow compression), * # =, &(=, = 45/ 6 = 45. ( (. =45 7

8 Most of engines and refrigerators are based on cyclic processes. Example 4. Consider a cyclic process A B C D A below. Let us find a total work W done on the gas during the cycle and the total amount of heat Q transferred into it. In all cyclic processes the total change in internal energy ΔE int = Q + W = 0. Segments B C and D A are isometric meaning that there is no work, W BC = W DA = 0. Process A Bis isobaric expansion and < ':=, & 2 (= & 2 (( : ( ' ) Process C D is isobaric compression = and 8 7=, & 1 (= & 1 (( 7 ( ) 9 The total work W = ': + : ' = (& 2 & 1 )(( : ( ' ) 8

9 and the total heat is Q =ΔE int W = (& 2 & 1 )(( : ( ' ). Heat Capacities of Gases Heat capacities C = Q/ΔT depend on the processes during which they are measured. C V is the value of heat capacity measured at constant volume V, and C P at constant pressure P. At constant volume W = 0, Q V = ΔE int, and = /5. At constant pressure dw = PdV, dq V = de int + PdV and? = 6@ ABC 6D +&(/5. According to the gas law, V = nrt/p (/5 = nr/p and? = ( +4 For simple gases, in which the whole internal energy is the kinetic energy of translational motion, E int = (3/2)nRT and C V = (3/2)nR, C P = (5/2)nR. 9

10 Heat Capacities of Diatomic Gases A diatomic molecule has two rotation axes, y and x. Its kinetic energy contains usual terms describing translational motion and two rotational terms, = E FG H 2 + E FG I 2 + E FG J 2 + E K Hω x 2 + E K Iω y 2 Such a molecule has 5 quadratic degrees of freedom and, according to the equipartition theorem, The heat capacity of the gas E int =5 E nrt = L nrt = 6@ ABC 6D = L nr,? = ( +4 = (7/2)nR 10

11 Example mol of oxygen O 2 are heated from 20 0 C to C. a) What amount of heat is needed if the volume is kept constant? b) What amount of heat is needed if the pressure is kept constant? c) How much work does the gas do? a) Q = C V ΔT = (5/2)nRΔT = (5/2) 3mol 8.314J/mol K 120K = 7.483kJ b) Q = C P ΔT = (7/2)nRΔT = (7/2) 3mol 8.314J/mol K 120K = kJ c) In a) the volume does not change and W a = 0. In b) the volume changes, and the work, according to the first law, W on = ΔE int Q = (5/2)nRΔT (7/2)nRΔT = nrδt = 2.993kJ In addition to rotational modes, large molecules also have vibrational modes which contribute to heat capacity 11

12 Heat Capacities of Solids A solid can be modeled as a regular array of atoms or molecules located near fixed equilibrium positions and connected to each other by springs. The energy of a molecule consists of translational and vibrational parts: = E FG H 2 + E FG I 2 + E FG J 2 + E Hx 2 + E I" 2 + E JM 2 According to the equipartition theorem, all these six modes contribute equally to the internal energy and heat capacity (DulongPetit law): E int = 6 E nrt = 3nRT, C = 3nR, c = 3R The equipartition theorem breaks down when the quantum effects in rotation become important. 12

13 The QuasiStatic Adiabatic Compression of Gases The adiabatic processes are the processes in which no heat is transferred in or out of the systems. The system is either well insulated or the process takes place very quickly. Since Q = 0, the first law becomes C V dt = 0 PdV. Combining this with the gas equation, we get C V dt = nrtdv/v or dt/t + (nr/c V )dv/v = 0. After integration lnt + (nr/c V )lnv = const or TV nr/c V = const Since C P C V = nr, nr/c V = C P /C V 1 = γ 1 where γ = C P /C V is called the adiabatic constant. Therefore during the adiabatic process, TV γ1 = const and PV γ = const. Since Q = 0, the work done on the gas in an adiabatic compression is ΔW on = ΔE int = C V ΔT, Or, using the gas law, ΔW on = 9 N (P fv f P i V i ) = (P f V f P i V i )/(γ1) 13

14 For monatomic gases γ = C P /C V = 5/3, for diatomic gases γ = 7/5. Example 6. a) How much work is needed to pump the air adiabatically into the 1L tire to raise its pressure from 1atm to 583kPa b) What would be the pressure in the tire after the pump is removed and the temperature returns to 20 0 C? a) Since Q = 0, W = C V ΔT. The temperature can be found from the alternative form of the law of adiabatic compression, T γ /P γ1 = const: W = C V T i [(P f /P i ) 11/γ 1] The final temperature is T f = 293K[(583kPa/101.3kPa) 15/7 1] = 483K and the work W = C V (483K 293K) = (5/2)(P f V f /T f )(483K 203K) = 634J b) Since the air in the tire cools at constant volume, P 3 = T 3 P 2 /T 2 = (293K/483K)2.64atm = 1.6atm 14

15 Speed of sound waves The speed of sound wave is usually given by equation G= : 6OP /Q (wait until Chapter 15!) where ρ is the mass density, Q= R ( = R = R& 45 45, & and : 2!S2 is the adiabatic bulk modulus : 6OP = ( 6? = γp. 6 In the end, the sound velocity in a gas is G= T45/R. 15

16 Review of Chapter 18 Heat Q energy transferred due to a temperature difference ΔT Heat is often measured in calories; 1cal = 4.184J Heat capacity: C = Q/ΔT; specific heat: c = C/m; molar specific heat: c =C/n Heat capacity at constant volume: C V ; at constant pressure: C P = C V + nr > C V The internal energy of monatomic gas E int = (3/2)nRT; diatomic gas E int = (5/2)nRT The internal energy determines the heat capacity: C V = de int /dt Latent heat is the amount of heat needed for the phase transformation per unit mass, L = Q/m The first law of thermodynamics: ΔE int = Q in + W on 16

17 Isometric process: V = const W = 0 Isobaric process: P = const W on = PΔV Isothermal process: T = const W on = nrtln(v f /V i ) Adiabatic process: Q = 0 W on = C V ΔT Adiabatic process in ideal gas: T γ /P γ1 = const; TV γ1 = const; PV γ = const γ = C P /C V adiabatic constant Work W on = PdV Equipartition theorem: for each quadratic degree of freedom E int gains E 45 DulongPetit law: the molar specific heat of most solids is c =3R Speed of sound in gases: G= T45/R 17