THERMAL PHYSICS NOTES

Transcription

1 THERMAL PHYSICS NOTES PHYSICS B4B BAKERSFIELD COLLEGE Rick Darke (Instructor)

2 THERMODYNAMICS TERMS thermodynamics - that branch of physics which deals with heat and temperature (also called thermal physics) system - a definite quantity of matter enclosed by boundaries (real or imaginary) open system - a system, into or out of which mass may be transferred closed system - a system for which there is no transfer of mass across the boundaries temperature - an index of the average random translational kinetic energy of particles in a system; a relative measure of hotness or coolness heat (energy) - the energy exchanged between objects because of a difference of temperature; a measure of the random kinetic energy of molecules in a substance

3 THERMODYNAMICS TERMS thermal contact - a condition in which heat may be exchanged between two objects thermal equilibrium - the condition in which there is no net heat exchange between objects in thermal contact thermally isolated system - a system for which there is no transfer of heat energy across its boundaries completely isolated system - a system for which there is no transfer of mass or heat energy across its boundaries (thermally isolated and closed)

4 HOW TO MAKE A THERMOMETER STEP 1: Select a thermometric substance (a substance with some measurable property that is temperature dependent) with a temperature response you believe to be a linear function of temperature. example: liquid-in-glass thermometer details: Liquid mercury is the thermometric substance, and the volume of a fixed amount of liquid mercury is the measurable property that is known to be temperature dependent. In a capillary and reservoir of fixed diameters, the volume changes in the mercury with temperature will be observable as height changes in the mercury column in the capillary.

5 HOW TO MAKE A THERMOMETER STEP 2: Create a temperature scale by defining two fixed-point temperatures. example: Define the temperature of a water-ice equilibrium system at 1.0 atm to have a temperature of 0 C. Define the temperature of a watersteam equilibrium system at 1.0 atm to have a temperature of 100 C. Note that the above two fixed-point temperatures are defined, not measured. 100 C 0 C ice / water (1 atm) water / steam (1 atm)

6 HOW TO MAKE A THERMOMETER STEP 3: Create a graduated scale by dividing the interval between the fixed-point temperatures linearly into a number of divisions. This graduated scale can then be extrapolated in both directions from the two fixed-point temperatures.

7 P(atm) PHASE DIAGRAM OF H 2 O 218 vaporifusion zation curve liquid (water) 1.00 curve.006 solid (ice) sublimation curve triple point gas (steam) T( C)

8 DEFINITION OF THE KELVIN 1 kelvin (S.I. unit of temperature) = 1/ of the thermodynamic temperature of the triple-point of H 2 O (P = 0.61 kpa and T = 0.01 C) note: You must use kelvin temperatures in all expressions in which the (absolute) temperature T is involved, but you may use either kelvins or degrees celsius in expressions in which temperature difference ΔT is involved. examples: P = σeat 4 must use kelvins Q = mcδt may use kelvins or C

9 TEMPERATURE SCALES scale absolute? conversions Kelvin yes T K = T C Rankine yes T R = T F Celsius no T C = (T F - 32)(5/9) Fahrenheit no T F = 1.8T C

10 DON HERBERT (MR. WIZARD) Donald Jeffry Herbert (born Donald Herbert Kemske) [ ] was the creator and host of the shows "Watch Mr. Wizard" ( , ) and "Mr. Wizard's World" ( ), educational television programs for children devoted to science and technology. He also produced many short video programs about science and authored several popular books about science for children.

11 ABSOLUTE ZERO EXPERIMENT SETUP Four pressures of a constant volume of air were recorded for four known temperature environments into which the contained air was immersed psia 17.1 psia 9.6 psia 14.7 psia ice + water boiling water dry ice + alcohol room air T = 0.0 C T = C T = C T = 37.0 C

12 ABSOLUTE ZERO EXPERIMENT DATA bath T ( C) P (psia) dry ice + alcohol room air boiling water ice water

13 ABSOLUTE ZERO EXPERIMENT GRAPH P (psia) linear regression equation: P = psia + ( psia/ C)T r =.9965 regression-interpreted value of absolute zero: T(P = 0 psia) = -281 C T ( C)

14 ZEROTH LAW OF THERMODYNAMICS If two systems A and B are in thermal equilibrium with a third system C, then they will be in thermal equilibrium with each other if placed in thermal contact. Two objects in thermal equilibrium with each other are at the same temperature. system A system B system A no net heat flow and no net heat flow no net heat flow system C system C system B

15 LINEAR EXPANSION COEFFICIENTS material α at 20 C ( C -1 ) aluminum 24x10-6 brass 19x10-6 copper 17x10-6 concrete 12x10-6 steel 11x10-6 glass (ordinary) 9.0x10-6 glass (pyrex) 3.2x10-6 invar (Ni-Fe alloy) 0.9x10-6 VOLUME EXPANSION COEFFICIENTS material β at 20 C ( C -1 ) air 37x10-4 gasoline 9.6x10-4 mercury 1.8x10-4 ethanol 1.1x10-4

16 PHASE CHANGE TERMS freezing condensing ice water steam melting boiling sublimating resublimating

17 PRACTICE PROBLEM: THERMAL EXPANSION A solid copper sphere has a diameter of cm and is at room temperature (20 C). An aluminum plate has a circular cutout with a diameter of cm (also at room temperature). At what common temperature would the copper sphere just barely be able to pass through the hole in the aluminum plate? The coefficient of expansion of copper is 17x10-6 C -1, and the coefficient of expansion of aluminum is 24x10-6 C -1. Cu Al

18 PRACTICE PROBLEM: THERMAL EXPANSION A steel rod of circular cross-section and diameter 5.0 cm spans a 2.5-meter gap between two concrete fixtures. At 20 C the rod is not compressed and just touches each of the fixtures (exerting no force on them). What force is exerted on the fixtures when the temperature of the rod rises to 80 C, but the distance between the fixtures does not change? The coefficient of expansion of steel is 11x10-6 C - 1, and the Young's modulus of steel is 2.0x10 11 Pa. concrete steel concrete

19 PRACTICE PROBLEM: IDEAL GAS LAW How many moles of carbon dioxide would there be in a 3.5-cm 3 CO 2 cartridge at room temperature (20 C) if the gauge pressure of the gas in the cartridge is 500 psi? How many molecules of carbon dioxide would there be in the cartridge?

20 PRACTICE PROBLEM: IDEAL GAS LAW An aerosol can contains a gas whose gauge pressure is 2.0 atm at 22 C. Suppose that the can will rupture when the gauge pressure of the gas inside rises to 3.5 atm. If the can were tossed into a fire, at what temperature will the can rupture?

21 PRACTICE PROBLEM: IDEAL GAS LAW 66.0 ft 3 of air at atmospheric pressure and at 22 C is to be placed into a 10.0-liter scuba tank. Just after this transfer, it is found that the tank's gauge shows a pressure of 3000 psig. What temperature is the air immediately after the tank is filled?

22 PRACTICE PROBLEM: IDEAL GAS LAW A bubble is released from the bottom of a fresh-water lake, 15.0 meters below the surface. The gas in the bubble is 4 C when released and warms to 20 C by the time it reaches the surface. If the bubble had a volume of 2.0 cm 3 at release, what will be its volume when it reaches the surface?

23 HEAT ENERGY: UNITS & CONVERSIONS calorie: the amount of heat energy required to raise the temperature of 1 gram of water from 14.5 C to 15.5 C (also called the "15-degree calorie" or the "little calorie") Calorie: 1000 calories (also called the kilocalorie [kcal] or the "big calorie") Btu (British thermal unit): the amount of heat energy required to raise the temperature of 1 pound of water from 63 F to 64 F mechanical equivalent of heat (MEH): the conversion factor between mechanical energy and heat energy: 1 cal = J 1 Cal = 4186 J 1 Btu = 1055 J

24 JOULE'S EXPERIMENT In 1845 British physicist James Joule presented the paper "On the Mechanical Equivalent of Heat" to the British Association meeting in Cambridge. In this work he reported the results of his best-known experiment, in which he estimated the mechanical equivalent of heat to be 819 ft lbf/btu (4.41 J/cal). In 1850, Joule obtained a refined measurement of 773 ft lbf/ Btu (4.16 J/cal).

25 JOULE'S APPARATUS Joule's apparatus employed a falling weight, in which gravity does the mechanical work in spinning a paddle-wheel in an insulated barrel of water. The temperature of the water is increased through the viscous dissipation of mechanical energy which is converted into heat energy.

26 SPECIFIC HEATS material c (J/kg. C) c (cal/g. C) aluminum brass copper iron lead glass ice water steam ethanol

27 LATENT HEATS material M.P. ( C) L f (J/kg) L f (cal/g) H 2 O 0 333, aluminum , copper , lead , ethanol , helium , material B.P. ( C) L v (J/kg) L v (cal/g) H 2 O 100 2,260, aluminum ,400, copper ,060, lead , ethanol , helium ,

28 PRACTICE PROBLEM: HEAT ENERGY At Vernal Falls in Yosemite National Park, California, water in the Merced River plummets 97 meters from the rim of the falls to a pool below. What is the temperature increase of the water after dropping this distance? Hint: You need to consider two energy conversions.

29 PRACTICE PROBLEM: HEAT ENERGY Compute the amount of heat energy required to convert 10 grams of ice originally at -20 C to steam at 150 C?

30 PRACTICE PROBLEM: CALORIMETRY A styrofoam cup contains 100 grams of water at 50 C. A 40-gram piece of ice at -20 C is placed in the cup, and the system is allowed to come to thermal equilibrium. Assuming the cup does not take part in any heat sharing, describe thermal equilibrium reached by the system.

31 PRACTICE PROBLEM: CALORIMETRY A styrofoam cup contains 100 grams of water at 60 C. A 200-gram piece of ice at -40 C is placed in the cup, and the system is allowed to come to thermal equilibrium. Assuming the cup does not take part in any heat sharing, describe thermal equilibrium reached by the system.

32 PRACTICE PROBLEM: CALORIMETRY A styrofoam cup contains 100 grams of water at 20 C. A 200-gram piece of ice at -60 C is placed in the cup, and the system is allowed to come to thermal equilibrium. Assuming the cup does not take part in any heat sharing, describe thermal equilibrium reached by the system.

33 FIRST LAW OF THERMODYNAMICS The increase in internal energy (ΔE 12 ) of a system taken through a process from state 1 to state 2 is accounted for by the amount of heat energy (Q 12 ) transferred (added) to the system during the process and the amount of work (W 12 ) done on the system by the environment during the process. ΔE 12 = Q 12 + W 12 = Q 12 - S PdV The differential expression for the above is: W 12 ΔE 12 Q 12 de = dq + dw = dq - PdV

34 PRACTICE PROBLEM: FIRST LAW Calculate the increase in internal energy of 1.0 gram of H 2 O when it is taken from water at 100 C to steam at 100 C (assume that the process occurs at atmospheric pressure). H 2 O (g) H 2 O (l)

35 THERMODYNAMIC PROCESS A thermodynamic process is a continuous change in the state of a material. If the material is a gas, it is usual to describe the process graphically on a PV-diagram (where P is shown as a fucntion of V, and where the state variable temperature is suppressed). So graphically, a process is a path on the PV-diagram. An arrow is used to make the time-progression of the process obvious. P 1 P 2 P V 1 V 2 V

36 MOLAR SPECIFIC HEATS The molar specific heats (C V and C P ) of a gas are the constants used in the expressions Q = nc V ΔT and Q = nc P ΔT for computing the heat energy needed to raise the temperature of n moles of a gas ΔT degrees. C V = (ν/ 2)R is the molar specific heat at constant volume, and C P = (ν/2 + 1)R is the molar specific heat at constant pressure. The parameter ν is the effective number of degrees of freedom of internal energy sharing for the gas. gas ν C P (J/mol.K) C V (J/mol.K) γ = C P /C v He Ar H N O CO H

37 GENERAL PROCESS In any general processthat an ideal gas is taken through the three thermodynamic quantities Q, W, and ΔE can always be given by: Q 12 = (ν/2)δ(pv) + S PdV W 12 = -S PdV ΔE 12 = (ν/2)δ(pv) or, using nrt for PV: P 1 P 2 P Q 12 = n(ν/2)rδt + S PdV W 12 = -S PdV ΔE 12 = n(ν/2)rδt V 1 V 2 V where: Δ(PV) = P 2 V 2 - P 1 V 1 and ΔT = T 2 - T 1

38 ISOBARIC PROCESS An isobaric process is a process throughout which the system pressure remains constant. (P = constant or dp = 0 throughout the process). Q 12 = (ν/2 + 1)PΔV = n(ν/2 + 1)RΔT W 12 = - PΔV = - nrδt ΔE 12 = (ν/2)pδv = n(ν/2)rδt P P where: ΔV = V 2 - V 1 and ΔT = T 2 - T 1 V 1 V 2 V

39 PRACTICE PROBLEM: ISOBARIC PROCESS A weather balloon at sea level is filled with 600 m 3 of helium gas at 15.0 C and 1.00 atma pressure. The helium inside the balloon slowly absorbs heat energy from the warmer surrounding air. The balloon's volume increases, but the pressure of the helium inside remains constant. If the temperature of the helium increases to 25.0 C find: (1) W = the work done on the helium by the environment; (2) ΔE = the change in internal energy of the helium; and (3) Q = the heat energy absorbed by the helium. The effective ν for helium at these temperatures is 3.01.

40 ISOCHORIC PROCESS An isochoric process is a process throughout which the system volume remains constant. (V = constant or dv = 0 throughout the process). Q 12 = (ν/2)vδp = n(ν/2)rδt W 12 = 0 ΔE 12 = (ν/2)vδp = n(ν/2)rδt P 1 P where: ΔP = P 2 - P 1 and ΔT = T 2 - T 1 P 2 V V

41 PRACTICE PROBLEM: ISOCHORIC PROCESS A pressure tank has a volume of 30.0 m 3. This tank is filled with air to a gauge pressure of 5.00 atm early in the day when the temperature is 20.0 C. In the late afternoon it is found that the ambient temperature (and the air within the tank) has risen to 44.0 C. Use this information to find: (1) W = the work done on the air by the environment; (2) ΔE = the change in internal energy of the air; and (3) Q = the heat energy absorbed by the air. The effective ν for air at these temperatures is 5.03.

42 ISOTHERMAL PROCESS An isothermal process is a process throughout which the system temperature remains constant. (T = constant or dt = 0 throughout the process). Q 12 = PVln(P 2 / P 1 ) = nrtln(p 2 / P 1 ) W 12 = -PVln(P 2 / P 1 ) = -nrtln(p 2 / P 1 ) ΔE 12 = 0 P 1 P note: V 1 /V 2 can be subbed for P 2 /P 1 in the above expressions. PV in the above expressions is the product of any P-V combination along the isotherm. P 2 V 1 V 2 V

43 PRACTICE PROBLEM: ISOTHERMAL PROCESS During the upstroke of the handle of a bicycle pump, air at a temperature of 20 C and 1.00 atma pressure enters a 0.75-L volume of the cylinder. A slow downstroke compresses the air to a volume of 0.15 L. Assuming that the downstroke is slow enough that heat is conducted through the cylinder wall so that the temperature of the gas in the cylinder remains constant during the process, find: (1) W = the work done on the air by the environment; (2) ΔE = the change in internal energy of the air; and (3) Q = the heat energy absorbed by the air.

44 ADIABATIC PROCESS An adiabatic process is a process throughout which no heat energy is exchanged between the system and its surroundings. (PV γ = constant or dq = 0 throughout the process). γ = (ν + 2)/ν. Q 12 = 0 W 12 = (ν/2)δ(pv) = n(ν/2)rδt ΔE 12 = (ν/2)δ(pv) = n(ν/2)rδt P 1 P 2 P where: Δ(PV) = P 2 V 2 - P 1 V 1 and ΔT = T 2 - T 1 V 1 V 2 V

45 PRACTICE PROBLEM: ADIABATIC PROCESS During the upstroke of the handle of a bicycle pump, air at a temperature of 20 C and 1.00 atma pressure enters a 0.75-L volume of the cylinder. A quick downstroke compresses the air to a volume of 0.15 L. Assuming that the downstroke is rapid enough that there is no time for heat energy to be conducted through the cylinder wall, find: (1) W = the work done on the air by the environment; (2) ΔE = the change in internal energy of the air; and (3) Q = the heat energy absorbed by the air.

46 IDEAL GAS PROCESSES MATRIX process Q 12 W 12 ΔE 12 general (ν/2)δ(pv) + S PdV (ν/2)nrδt + S PdV -SPdV (ν/2)δ(pv) (ν/2)nrδt isobaric (ν/2 + 1)PΔV -PΔV (ν/2)pδv (constant P) (ν/2 + 1)nRΔT -nrδt (ν/2)nrδt isochoric (ν/2)vδp (ν/2)vδp 0 (constant P) (ν/2)nrδt (ν/2)nrδt isothermal PVln(P 2 / P 1 ) -PVln(P 2 / P 1 ) (constant P) nrtln(p 2 / P 1 ) -nrtln(p 2 / P 1 ) 0 adiabatic (ν/2)δ(pv) (ν/2)δ(pv) (constant PV γ 0 ) (ν/2)nrδt (ν/2)nrδt note: ΔP = P 2 - P 1, ΔV = V 2 - V 1, ΔT = T 2 - T 1, and Δ(PV) = P 2 V 2 - P 1 V 1

47 THERMAL CONDUCTIVITY The rate P at which heat is conducted through a slab of material (in W or Btu/h) with crosssectional area A and thickness L whose faces are maintained at temperatures T H and T C is given by: P = ka(t H - T C )/L L A where k is the thermal conductivity of the material of the slab (measured in W/m. C or Btu/h.ft. F). T H T C

48 THERMAL CONDUCTIVITIES substance k (W/m. C) substance k (W/m. C) silver 427 water 0.6 copper 397 rubber 0.2 gold 314 hydrogen 0.17 aluminum 238 helium 0.14 iron 80 asbestos 0.08 ice 2 wood concrete 0.8 oxygen glass 0.8 air 0.023

49 R-VALUES substance and thickness R-value (ft 2. F.h/Btu) drywall (0.5" thick) 0.45 glass pane (1/8" thick) 0.89 insulating glass (1/4") 1.54 concrete block (filled cores) 1.93 brick (4" thick) 4.00 styrofoam (1" thick) 5.00 fiber glass (3.5" thick) 10.9 fiber glass (6" thick) 18.8

50 PRACTICE PROBLEM: THERMAL CONDUCTIVITY A pane of window glass is 1/8" thick and has dimensions 1.5 m by 2.5 m. It separates two heat reservoirs whose temperatures are 10 C and 22 C. What is the rate of heat transport through the pane?

51 PRACTICE PROBLEM: THERMAL CONDUCTIVITY A copper and an aluminum cylinder are joined end-to-end as shown. Each has a diameter of 4.0 cm and a length of 25.0 cm. One end of the copper cylinder is maintained at 90 C and one end of the aluminum cylinder is maintained at 15 C. The thermal conductivity of copper is 397 W/m. C and the thermal conductivity of aluminum is 238 W/m. C. What is the steady state junction temperature? 90 C insulator Cu Al insulator 15 C

52 THERMAL RADIATION The rate P of radiant energy loss (watts) from the surface of an object with surface area A and surface temperature T is given by: P = σaet 4 A T where σ is the Stefan-Boltzmann constant (5.67x10-8 W/m 2. K 4 ), and e is the emissivity of the surface of the object (dimensionless parameter). This is called the Stefan-Boltzmann Law.

53 PRACTICE PROBLEM: THERMAL RADIATION The surface of a cube with sides of length 0.05 m has an emissivity of If the surface of the cube is maintained at a temperature of 550 C, estimate the radiant energy flux through the surface of the cube?

54 MOLAR SPECIFIC HEATS The molar specific heats (C V and C P ) of a gas are the constants used in the expressions Q = nc V ΔT and Q = nc P ΔT for computing the heat energy needed to raise the temperature of n moles of a gas ΔT degrees. C V = (ν/ 2)R is the molar specific heat at constant volume, and C P = (ν/2 + 1)R is the molar specific heat at constant pressure. The parameter ν is the effective number of degrees of freedom of internal energy sharing for the gas. gas ν C P (J/mol.K) C V (J/mol.K) γ = C P /C v He Ar H N O CO H

55 PRACTICE PROBLEM: MOLAR SPECIFIC HEAT A weather balloon is filled with 600 m 3 of helium gas at a temperature of 15.0 C. The gas inside the balloon slowly absorbs heat energy from the warmer surrounding air in such a way that the balloon's volume increases, but the pressure of the helium remains the same as that of the surrounding air (1.0 atm). How much heat energy would the helium need to absorb for its temperature to increase by 1.0 C? The molar specific heat at constant pressure of helium is 20.8 J/mol.K, and the molecular of weight of helium is 4.00 g/mol.

56 PRACTICE PROBLEM: MOLAR SPECIFIC HEAT A 600-m 3 pressure tank is filled with helium gas at a temperature of 15.0 C and pressure of 1.0 atm (absolute). The helium inside the tank slowly absorbs heat energy through the tank from the warmer surrounding air in such a way that the helium's pressure increases, but the volume of the tank (and helium) remains the same. How much heat energy would the helium need to absorb for its temperature to increase by 1.0 C? The molar specific heat at constant volume of helium is 12.5 J/mol.K.

57 SECOND LAW OF THERMODYNAMICS Kelvin-Planck formulation: It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the input of energy by heat from a reservoir and the performance of an equal amount of work. Clausius statement: It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another at a higher temperature without the input of energy by work. Lord Kelvin Max Planck Rudolf Clausius

58 ENTROPY Entropy (S) is a quantity describing the disorder of a system. It has dimensions of heat energy per temperature (cal/k or J/K). It is equal to the Boltzmann constant (k B ) times the natural log of the total number of ways (W) in which the system can arrange itself consistent with energy being conserved. S = k B lnw Entropy is an extensive physical quantity. The total entropy of a system is equal to the sum of the entropies of each of the system's components. n S total = Σ S i i=1

59 ENTROPY CHANGE Calculating the entropy change ΔS for a system is usually a more tractable problem than calculating the entropy itself. The change in entropy of a system taken through a process from state 1 to state 2 is most often derivable from: 2 ΔS 12 = S 2 - S 1 = S dq/t 1 example: The entropy change of m grams of a substance with specific heat c heated from T 1 to T 2 would be obtained as: 2 2 ΔS 12 = S dq/t = S mcdt/t = mcln(t 2 /T 1 ) 1 1

60 ENTROPY CHANGE: WARMING / COOLING In changing the temperature of a material from T 1 to T 2 without changing its phase, the change in entropy is given by: ΔS 12 = mcln(t 2 /T 1 ) or ncln(t 2 /T 1 ) example: Find the change in entropy of a gram gold krugerrand when it is heated from 20 C to 180 C. The specific heat of gold is.0308 cal/g. C.

61 ENTROPY CHANGE: PHASE CHANGE In changing the phase of a material without changing its temperature, the change in entropy is given by: ΔS = + ml/t example: Find the change in entropy of 2.5 kg of molten aluminum becoming a solid ingot at its melting point (660 C). The latent heat of fusion of aluminum is 397,000 J/kg.

62 ENTROPY CHANGE: FREE EXPANSION In allowing a gas to free-expand from initial volume V 1 to final volume V 2, the entropy change is given by: ΔS 12 = nrln(v 2 /V 1 ) example: Find the change in entropy when helium gas at 1.0 atm and 20 C, occupying a chamber whose volume is 0.5 m 3, is allowed to free-expand into an adjacent evacuated chamber whose (extra) volume is 2.5 m 3.

63 ENTROPY CHANGE: MIXING In mixing n A moles of gas A, initially occupying volume V A, with n B moles of gas B, initially occupying volume V B : ΔS 12 = n A Rln([V A + V B ]/V A ) + n B Rln([V A + V B ]/V B ) example: A chamber is divided into two compartments separated by a partition. Compartment A (H 2 gas): V A = 75 L, P A = 1.0 atm, and T A = 0 C. Compartment B (CO 2 gas): V B = 25 L, P B = 1.0 atm, and T B = 0 C. The partition is removed and the gases mix. Find the change in entropy of the system.

64 PRACTICE PROBLEM: ENTROPY CHANGE A 60-gram piece of ice at 0 C is placed in a container with 200 grams of water at 50 C. Find the temperature of this system after thermal equilibrium is attained. Use this result to find ΔS for: (1) the water; (2) the ice; and (3) the entire system for this process. Assume that the container does not transfer any heat energy.