19 Temperature CHAPTER OUTLINE 19.1 Temperature and the Zeroth Law of Thermodynamics 19.2 Thermometers and the Celsius Temperature Scale 19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale 19.4 Thermal Expansion of Solids and Liquids 19.5 Macroscopic Description of an Ideal Gas * An asterisk indicates a question or problem new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ19.1 Answer (b) The markings are now farther apart than intended, so measurements made with the heated steel tape will be too short—but only by a factor of × 10−5 of the measured length OQ19.2 Answer (d) Remember that one must use absolute temperatures and pressures in the ideal gas law Thus, the original temperature is TK = TC + 273.15 = 25 + 273.15 = 298 K, and with the mass of the gas constant, the ideal gas law gives ⎛ P2 ⎞ ⎛ V2 ⎞ ⎛ 1.07 × 106 Pa ⎞ T2 = ⎜ ⎟ ⎜ ⎟ T1 = ⎜ ( 3.00) ( 298 K ) = 191 K ⎝ 5.00 × 106 Pa ⎟⎠ ⎝ P1 ⎠ ⎝ V1 ⎠ OQ19.3 Answer (d) From the ideal gas law, with the mass of the gas constant, P2V2/T2 = P1V2/T1 Thus, ⎛V ⎞⎛T ⎞ ⎛ 1⎞ P2 = ⎜ ⎟ ⎜ ⎟ P1 = ⎜ ⎟ ( ) P1 = 2P1 ⎝ 2⎠ ⎝ V2 ⎠ ⎝ T1 ⎠ OQ19.4 Answer (a) As the temperature increases, the brass expands This would effectively increase the distance d from the pivot point to the 997 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 998 Temperature center of mass of the pendulum, and also increase the moment of inertia of the pendulum Since the moment of inertia is proportional I to d2, and the period of a physical pendulum is T = 2π , the mgd period would increase, and the clock would run slow OQ19.5 Answer (c) TC = 5 (TF − 32 ) = (162 − 32 ) = 72.2°C, then, 9 TK = TC + 273.15 = 72.2 + 273.15 = 345 K OQ19.6 Answer (c) From the ideal gas law, with the mass of the gas constant, P2V2/T2 = P1V2/T1 Thus, ⎛ P ⎞⎛T ⎞ V2 = ⎜ ⎟ ⎜ ⎟ V1 = ( ) ( 1) ( 0.50 m ) = 2.0 m ⎝ P2 ⎠ ⎝ T1 ⎠ OQ19.7 Answer (d) If glass were to expand more than the liquid, the liquid level would fall relative to the tube wall as the thermometer is warmed If the liquid and the tube material were to expand by equal amounts, the thermometer could not be used because the liquid level would not change with temperature OQ19.8 The ranking is (a) = (b) = (d) > (e) > (c) We think about nRT/V in each case Since R is constant, we need only think about nT/V, and units of mmol⋅K/cm3 are as convenient as any: (a) 2⋅3/1 = 6, (b) 6, (c) 4, (d) 6, (e) OQ19.9 Answer (d) Cylinder A must be at lower pressure If the gas is thin, PV = nRT applies to both with the same value of nRT for both Then A will be at one-third the absolute pressure of B OQ19.10 (i) Answer (a) Call the process isobaric cooling or isobaric contraction The rubber wall is easy to stretch The air inside is nearly at atmospheric pressure originally and stays at atmospheric pressure as the wall moves in, just maintaining equality of pressure outside and inside The air is nearly an ideal gas to start with, and stays fairly ideal—fairly far from liquefaction—even at 100 K The water vapor liquefies and then freezes, and the carbon dioxide turns to dry ice, but these are minor constituents of the air Thus, as the absolute temperature drops to 1/3 of its original value and the volume will drop to 1/3 of what it was (ii) Answer (c) As noted above, the pressure stays nearly constant at atm OQ19.11 Answer (c) For a quick approximation, multiply 93 m and 17 and © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 999 1/(1 000 000 °C) and say 5°C for the temperature increase To simplify, multiply 100 and 100 and 1/1 000 000 for an answer in meters: it is on the order of cm OQ19.12 Answer (b) Around atmospheric pressure, 0°C is the only temperature at which liquid water and solid water can both exist OQ19.13 Answer (b) When a solid, containing a cavity, is heated, the cavity expands in the same way as it would if filled with the material making up the rest of the object OQ19.14 Answer (e) 9 TF = TC + 32 = ( −25° ) + 32° = −13° F 5 ANSWERS TO CONCEPTUAL QUESTIONS CQ19.1 The coefficient of linear expansion must be greater for mercury than for glass, otherwise the interior of a glass thermomter would expand more and the mercury level would drop See OQ19.7 CQ19.2 (a) The copper’s temperature drops and the water temperature rises until both temperatures are the same (b) The water and copper are in thermal equilibrium when their temperatures are the same CQ19.3 (a) PV = nRT predicts V going to zero as T goes to zero (b) The ideal-gas model does not apply when the material gets close to liquefaction and then turns into a liquid or solid The molecules start to interact all the time, not just in brief collisions The molecules start to take up a significant portion of the volume of the container CQ19.4 Air pressure decreases with altitude while the pressure inside the bags stays the same; thus, that inside pressure is greater than the outside pressure CQ19.5 (a) No The thermometer will only measure the temperature of whatever is in contact with the thermometer The thermometer would need to be brought to the surface in order to measure its temperature, since there is no atmosphere on the Moon to maintain a relatively consistent ambient temperature above the surface (b) It would read the temperature of the glove, since it is in contact with the glove CQ19.6 The coefficient of expansion of metal is larger than that of glass When hot water is run over the jar, both the glass and the lid expand, but at different rates Since all dimensions expand, the inner © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1000 Temperature diameter of the lid expands more than the top of the jar, and the lid will be easier to remove CQ19.7 CQ19.8 (a) As the water rises in temperature, it expands or rises in pressure or both The excess volume would spill out of the cooling system, or else the pressure would rise very high indeed (b) Modern cooling systems have an overflow reservoir to accept the excess volume when the coolant heats up and expands (a) The sphere expands when heated, so that it no longer fits through the ring With the sphere still hot, you can separate the sphere and ring by heating the ring This more surprising result occurs because the thermal expansion of the ring is not like the inflation of a blood-pressure cuff Rather, it is like a photographic enlargement; every linear dimension, including the hole diameter, increases by the same factor The reason for this is that the atoms everywhere, including those around the inner circumference, push away from each other The only way that the atoms can accommodate the greater distances is for the circumference—and corresponding diameter—to grow This property was once used to fit metal rims to wooden wagon wheels If the ring is heated and the sphere left at room temperature, the sphere would pass through the ring with more space to spare ANS FIG CQ19.8 (b) Heating the ring increases its diameter, the sphere can pass through it easily The hole in the ring expands as if it were filled with the material of the ring CQ19.9 Two objects in thermal equilibrium need not be in contact Consider the two objects that are in thermal equilibrium in Figure 16.1(c) The act of separating them by a small distance does not affect how the molecules are moving inside either object, so they will still be in thermal equilibrium CQ19.10 (a) One mole of H2 has a mass of 2.016 g (b) One mole of He has a mass of 4.002 g (c) One mole of CO has a mass of 28.010 g © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 1001 SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 19.2 Thermometers and the Celsius Temperature Scale   Section 19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale   P19.1 (a) By Equation 19.2, 9 TF = TC + 32 = ( 41.5°C ) + 32 = ( 74.7 + 32 ) °F = 107°F 5 (b) P19.2 (a) Yes The normal body temperature is 98.6°F, so the patient has a high fever and needs immediate attention Consider the freezing and boiling points of water in each scale: 0°C and 100°C; 32°F and 212°F We see that there are 100 Celsius units for every 180 Fahrenheit units: ΔTC 100°C = ΔTF 180°F (b) → ΔTC = 5 ΔTF ) = ( 57.0 ) °C = 31.7°C ( 9 The Kelvin unit is the same size as the Celsius unit: T = TC + 273.15 → ΔT = ΔTC ⎛ 1K⎞ ⎛ 1K⎞ ΔT = ΔTC ⎜ o ⎟ = 31.7 K ⎜ o ⎟ = 31.7 K ⎝ C⎠ ⎝ C⎠ P19.3 (a) By Equation 19.2, 9 TF = TC + 32 = ( −78.5 ) + 32 = −109°F 5 And, from Equation 19.1, T = TC + 273.15 = ( −78.5 + 273.15 ) K = 195 K (b) Again, 9 TF = TC + 32 = ( 37.0 ) + 32 = 98.6°F 5 T = TC + 273.15 = ( 37.0 + 273.15 ) K = 310 K P19.4 (a) The relationship between the Kelvin and Celsius scales is given by Equation 19.1: T = TC + 273.15 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1002 Temperature Thus 20.3 K converts to TC = T − 273.15 = 20.3 K − 273.15 K = −253°C (b) The relationship between the Celsius and Fahrenheit scales is, from Equation 19.2, TF = TC + 32°F Thus –253°C converts to 9 5 TF = TC + 32°F = ( −253°C ) + 32°F = −423°F P19.5 (a) By Equation 19.2, 9 TF = TC + 32.0°F = ( −195.81°C ) + 32.0 = −320°F 5 (b) Applying Equation 19.1, T = TC + 273.15 = −195.81°C + 273.15 = 77.3 K *P19.6 (a) To convert from Fahrenheit to Celsius, we use TC = (TF − 32.0) The temperature at Furnace Creek Ranch in Death Valley is TC = 5 (TF − 32.0) = ( 134°F − 32.0) = 56.7°C 9 and the temperature at Prospect Creek Camp in Alaska is TC = (b) 5 (TF − 32.0) = ( −79.8°F − 32.0) = –62.1°C 9 We find the Kelvin temperature from Equation 19.1, T = TC + 273.15 The record temperature on the Kelvin scale at Furnace Creek Ranch in Death Valley is T = TC + 273.15 = 56.7°C + 273.15 = 330 K and the temperature at Prospect Creek Camp in Alaska is T = TC + 273.15 = −62.11°C + 273.15 = 211 K P19.7 Since we have a linear graph, we know that the pressure is related to the temperature as P = A + BTC , where A and B are constants To find A and B, we use the given data: 0.900 atm = A + B ( −78.5°C ) [1] © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 1003 and 1.635 atm = A + B ( 78.0°C ) [2] Solving Equations [1] and [2] simultaneously, we find: A = 1.27 atm and B = 4.70 × 10−3 atm °C Therefore, P = 1.27 atm + ( 4.70 × 10−3 atm °C ) TC (a) At absolute zero the gas exerts zero pressure (P = ), so TC = (b) −1.27 atm = − 270°C 4.70 × 10−3 atm °C At the freezing point of water, TC = and P = 1.27 atm + = 1.27 atm At the boiling point of water, TC = 100°C, so P = 1.27 atm + ( 4.70 × 10−3 atm °C )( 100°C ) = 1.74 atm   Section 19.4 P19.8 Thermal Expansion of Solids and Liquids Each section can expand into the joint space to the north of it We need think of only one section expanding Using Equation 19.4, −1 ΔL = Liα ΔT = ( 25.0 m ) ⎡⎣12.0 × 10−6 ( °C ) ⎤⎦ ( 40.0°C ) = 1.20 cm : (a) By Equation 19.4, −1 ΔL = α Li ΔT = ⎡⎣ 9.00 × 10−6 ( °C ) ⎤⎦ ( 30.0 cm )( 65.0°C ) = 0.176 mm (b) The diameter is a linear dimension, so Equation 19.4 still applies: −1 ΔL = α Li ΔT = ⎡⎣ 9.00 × 10−6 ( °C ) ⎤⎦ ( 1.50 cm )( 65.0°C ) = 8.78 × 10−4 cm = 8.78 µm (c) Using the volumetric coefficient of expansion β, and Vi = π d L / 4, ΔV = β Vi ΔT ≈ 3α VΔT © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1004 Temperature ΔV = β Vi ΔT ≈ 3α Vi ΔT ⎞ −1 ⎛ 30.0 (π )( 1.50 ) = ⎡⎣ 9.00 × 10−6 ( °C ) ⎤⎦ ⎜ cm ⎟ ( 65.0°C ) ⎝ ⎠ = 0.093 cm P19.10 The horizontal section expands according to ΔL = α Li ΔT −1 Δx = ⎡⎣17 × 10−6 ( °C ) ⎤⎦ ( 28.0 cm )( 46.5°C − 18.0°C ) = 1.36 × 10−2 cm ANS FIG P19.10 The vertical section expands similarly by −1 Δy = ⎡⎣17 × 10−6 ( °C ) ⎤⎦ ( 134 cm )( 28.5°C ) = 6.49 × 10−2 cm The vector displacement of the pipe elbow has magnitude Δr = Δx + Δy = ( 0.136 mm )2 + ( 0.649 mm )2 = 0.663 mm and is directed to the right below the horizontal at angle ⎛ Δy ⎞ ⎛ 0.649 mm ⎞ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ = 78.2° ⎝ Δx ⎠ ⎝ 0.136 mm ⎟⎠ Δr = 0.663 mm to the right at 78.2° below the horizontal P19.11 The wire is 35.0 m long when TC = −20.0°C ΔL = Liα (T − Ti ) Since α = α ( 20.0°C ) = 1.70 × 10−5 ( °C ) −1 for Cu, −1 ΔL = ( 35.0 m ) ⎡⎣1.70 × 10−5 ( °C ) ⎤⎦ [ 35.0°C − ( −20.0°C )] = +3.27 cm *P19.12 For the dimensions to increase, ΔL = α Li ΔT: 1.00 × 10−2 cm = [ 1.30 × 10−4 ( °C )−1 ]( 2.20 cm ) (T − 20.0°C ) T = 55.0°C © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 *P19.13 1005 By Equation 19.4, ΔL = α Li ΔT = [ 11 × 10−6 ( °C )−1 ]( 300 km )[ 35°C − ( −73°C )] = 1.54 km The expansion can be compensated for by mounting the pipeline on rollers and placing Ω -shaped loops between straight sections They bend as the steel changes length *P19.14 By Equation 19.4, ΔL = α Li ΔT = [ 22 × 10−6 ( °C )−1 ]( 2.40 cm ) ( 30.0°C ) = 1.58 × 10−3 cm *P19.15 (a) Following the logic in the textbook for obtaining Equation 19.6 from Equation 19.4, we can express an expansion in area as ΔA = 2α Ai ΔT = [ 17.0 × 10−6 ( °C )−1 ]( 0.080 m ) ( 50.0°C ) = 1.09 × 10−5 m = 0.109 cm (b) The length of each side of the hole has increased Thus, this represents an increase in the area of the hole *P19.16 By Equation 19.6, ΔV = ( β − 3α ) Vi ΔT = [ 5.81 × 10−4 ( °C )−1 − ( 11.0 × 10−6 ( °C )−1 )] × ( 50.0 gal ) ( 20.0°C ) = 0.548 gal *P19.17 (a) By Equation 19.4, L = Li ( + αΔT ) , and ( 5.050 cm = 5.000 cm ⎡1 + 24.0 × 10−6 ( °C ) ⎣ −1 )(T − 20.0°C)⎤⎦ which gives T = 437°C (b) We must get LAl = LBrass for some ΔT, or Li, Al ( + α Al ΔT ) = Li, Brass ( + α Brass ΔT ) ( 5.000 cm ⎡1 + 24.0 × 10−6 ( °C ) ⎣ ( −1 ) ΔT ⎤⎦ = 5.050 cm ⎡1 + 19.0 × 10−6 ( °C ) ⎣ −1 ) ΔT ⎤⎦ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1006 Temperature Solving for ΔT, ΔT = 080°C T = × 103°C so (c) P19.18 No Aluminum melts at 660°C (Table 17.2) Also, although it is not in Table 17.2, internet research shows that brass (an alloy of copper and zinc) melts at about 900°C We solve for the temperature T at which the brass ring would fit over the aluminum cylinder LAl ( + α Al ΔT ) = LBrass ( + α Brass ΔT ) ΔT = T − Ti = ΔT = LAl − LBrass LBrassα Brass − LAlα Al 10.02 cm − 10.00 cm (10.00 cm ) (19.0 × 10−6 (°C)−1 ) − (10.02 cm ) ( 24.0 × 10−6 (°C)−1 ) ΔT = −396 = T − 20.0 → T = −376°C The situation is impossible because the required T = –376°C is below absolute zero P19.19 (a) The original volume of the acetone we take as precisely 100 mL After it is finally cooled to 20.0°C, its volume is { } −1 Vf = Vi ( + βΔT ) = ( 100 ml ) + ⎡⎣1.50 × 10−4 ( °C ) ⎤⎦ ( −15.0°C ) = 99.8 mL (b) P19.20 (a) Initially, the volume of the acetone reaches the 100-mL mark on the flask, but the acetone cools and the flask warms to a temperature of 32.0 °C Thus, the volume of the acetone decreases and the volume of the flask increases This means the acetone will be below the 100-mL mark on the flask The material would expand by ΔL = α Li ΔT, or ΔL = αΔT, but Li instead feels stress F YΔL = A Li −1 = YαΔT = ( 7.00 × 109 N m ) ⎡⎣12.0 × 10−6 ( C° ) ⎤⎦ ( 30.0°C ) = 2.52 × 106 N m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1028 Temperature ω f = ω i [ − α ΔT ] = −2 25.0 rad s ( ⎡1 − 17 × 10 ( °C ) ⎣ −6 −1 )( 830°C)⎤⎦ = 25.0 rad s 0.972 = 25.7 rad s P19.71 Visualize the molecules of various species all moving randomly The net force on any section of wall is the sum of the forces of all of the molecules pounding on it For each gas alone, P1 = N kT N kT N kT and P2 = and P3 = , etc V V V For all gases, P1V1 + P2V2 + P3V3 … ( N + N + N …) kT and ( N1 + N + N …) kT = PV Also, V1 = V2 = V3 =…= V ; therefore, P = P1 + P2 + P3 …   Challenge Problems P19.72 (a) At 20.0°C, the unstretched lengths of the steel and copper wires are Ls ( 20.0°C ) = ( 2.000 m ) ⎡⎣1 + ( 11.0 × 10−6  °C−1 )( −20.0°C )⎤⎦ = 1.999 56 m Lc ( 20.0°C ) = ( 2.000 m ) ⎡⎣1 + ( 17.0 × 10−6  °C−1 )( −20.0°C )⎤⎦ = 1.999 32 m Under a tension F, the length of the steel and copper wires are F ⎤ Ls′ = Ls ⎡⎢1 + ⎣ YA ⎥⎦ s and F ⎤ Lc′ = Lc ⎡⎢1 + ⎣ YA ⎥⎦ c where Ls′ + Lc′ = 4.000 m Since the tension F must be the same in each wire, we solve for F: F= ( Ls′ + Lc′ ) − ( Ls + Lc ) Ls Ys As + Lc Yc Ac © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 1029 When the wires are stretched, their areas become As = π ( 1.000 × 10−3 m ) ⎡⎣1 + ( 11.0 × 10−6  °C –1 )( −20.0 )⎤⎦ = 3.140 × 10−6 m 2 Ac = π ( 1.000 × 10−3 m ) ⎡⎣1 + ( 17.0 × 10−6  °C –1 )( −20.0 )⎤⎦ = 3.139 × 10−6 m 2 2 Recall Ys = 20.0 × 1010 Pa and Yc = 11.0 × 1010 Pa Substituting into the equation for F, we obtain F = ⎡⎣ 4.000 m − ( 1.999 56 m + 1.999 32 m ) ⎤⎦ × 1.999 56 m 1.999 32 m + 10 −6 10 ( 20.0 × 10 Pa)( 3.140 × 10 ) m (11.0 × 10 Pa)( 3.139 × 10−6 ) m2 F = 125 N (b) To find the x coordinate of the junction, ⎡ ⎤ 125 N Ls′ = ( 1.999 56 m ) ⎢1 + 10 −6 ⎥ ⎢⎣ ( 20.0 × 10 N m ) ( 3.140 × 10 m ) ⎥⎦ = 1.999958 m Thus the x coordinate is −2.000 + 1.999958 = −4.20 × 10−5 m P19.73 (a) We find the linear density from the volume density as the massper-volume multiplied by the volume-per-length, which is the cross-sectional area µ= 1 ρ (π d ) = π (1.00 × 10−3 m)2 ( 7.86 × 103 kg/m ) 4 = 6.17 × 10−3 kg/m (b) Since f1 = v T T and v = , then f1 = , and we have for the 2L µ 2L µ tension F = T = µ ( 2Lf1 ) = ( 6.17 × 10−3 kg/m ) ⎡⎣ ( 0.800 m )( 200 s −1 ) ⎤⎦ 2 = 632 N (c) At 0°C, the length of the guitar string will be ( Lactual = L0 C + ) F = 0.800 m AY © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1030 Temperature Where L0°C is the unstressed length at the low temperature We know the string’s cross-sectional area A= ( π4 )(1.00 × 10 and modulus −3 m)2 = 7.85 × 10−7 m Y = 20.0 × 1010 N/m2 Therefore, F 632 N = = 4.02 × 10−3 AY ( 7.85 × 10 –7 m ) ( 20.0 × 1010 N m ) and L0 C = 0.800 m = 0.796 m + 4.02 × 10 –3 Then at 30°C, the unstressed length is L30°C = (0.796 m) ⎡⎣1 + (30.0°C) ( 11.0 × 10−6 °C−1 ) ⎤⎦ = 0.797 m With the same clamping arrangement, F′ ⎤ ⎡ 0.800 m = ( 0.797 m ) ⎢1 + ⎣ A′Y ⎥⎦ where F ′ and A′ are the new tension and the new (expanded) cross-sectional area Then F′ 0.800 = – = 3.693 × 10 –3 A′Y 0.797 and F ′ = A′Y(3.693 × 10−3 ) F ′ = (7.85 × 10 m )(20.0 × 10 N/m )(3.693 × 10 ) (1 + αΔT)2 –7 10 –3 −4 F ′ = (580 N)(1 + 3.30 × 10 ) = 580 N f′ (d) Also the new frequency f1′ is given by = f1 so P19.74 (a) f1′ = (200 Hz) F′ , F 580 N = 192 Hz 632 N P0V P′V ′ because the amount of gas remains constant = T T′ The volume increases by Ah when the piston rises: V ′ = V + Ah © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 1031 When the piston compresses the piston by h, so the spring force increases by F = kx = kh, increasing the external pressure on the piston by kh/A: V ′ = V + Ah ANS FIG P19.74 Using the particle in equilibrium model applied to the piston, kh ⎞ ⎛ ⎛ T′ ⎞ ⎜⎝ P0 + ⎟⎠ (V + Ah ) = P0V ⎜⎝ ⎟⎠ A T (1.013 × 10 ( N m + 2.00 × 105 N m h ) × 5.00 × 10−3 m + ( 0.010 0 m ) h ) ⎛ 523 K ⎞ = ( 1.013 × 105 N m ) ( 5.00 × 10−3 m ) ⎜ ⎝ 293 K ⎟⎠ 2 000h2 + 2 013h − 397 = Taking the positive root, h = (b) −2 013 + 2 689 = 0.169 m 4 000 2.00 × 103 N m ) ( 0.169 m ) ( kh P′ = P + = 1.013 × 10 Pa + A 0.010 0 m P′ = 1.35 × 105 Pa P19.75 Each half of the spherical container is a particle in equilibrium Therefore, using the result of Problem 14.58, ∑ F  = 0   →   Ffrom gas  = Fholding hemispheres together →   P (π r ) =  F Pr A = σ ( 2π rt )   →   t =    A 2σ where σ is the yield strength of the steel Find the mass of the steel sphere: Pr ⎡ ⎛ Pr ⎞ ⎤ mSt  =  ρStV  =  ρSt ( 4π r t ) =  ρSt ⎢ 4π r ⎜  = 2πρSt ⎝ 2σ ⎟⎠ ⎥⎦ σ ⎣ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1032 Temperature Find the pressure of the helium in the tank: ⎛ nRT mHe ⎜ PV  = nRT     →    P =   =  V MHe ⎜ ⎜⎝ ⎞ RT ⎟ 3⎟ πr ⎟ ⎠ Substitute into the previous equation: ⎡ ⎛ r ⎢ mHe ⎜ mSt  = 2πρSt   ⎢ σ ⎢ MHe ⎜ ⎜⎝ ⎣ ⎞⎤ RT ⎟ ⎥ ρSt mHe  =  RT ⎟ ⎥ σ MHe πr ⎟ ⎥ ⎠⎦ Find the buoyant force on the balloon: B =  ρair gVballoon  =  ρair g PV nRT m RT  =  ρair g  =  ρair g He P0 P0 MHe P0 where we use the pressure of helium in the balloon P = P0 = atmospheric pressure Find the net force on the balloon and tank: ∑ F  = B − mHe g − mSt g  =  ρair g mHe RT ρSt mHe g  − mHe g −  RT MHe P0 σ MHe ⎛ RT ρSt RT ⎞         = mHe g ⎜ ρair − 1 −  P0 MHe σ MHe ⎟⎠ ⎝ ⎡ RT ⎛ ρair ρSt ⎞ ⎤         = mHe g ⎢ −  − 1 ⎥ ⎜ ⎟ ⎣ MHe ⎝ P0 σ ⎠ ⎦ Evaluate the brackets: ⎡ RT ⎛ ρair ⎤ ρSt ⎞  −   − 1⎥   ⎢ ⎜ ⎟ σ ⎠ ⎣ MHe ⎝ P0 ⎦ ⎧⎪ ⎡ ( 8.314 J/mol ⋅ K )( 293 K ) =  ⎨ ⎢ 4  ×  10 –3  kg/mol ⎪⎩ ⎣ { ⎛ 1.20 kg/m 3 ⎛ 7 860 kg/m ⎞ ⎞ ⎤ ⎫⎪  −  ⎜⎝ 1.013 × 105  Pa ⎜⎝ 5 × 108  N/m ⎟⎠ ⎟⎠ ⎥ − 1⎬ ⎦ ⎪⎭ = ⎡⎣( 6.09 × 105  m /s ) } × ( 1.184 6 × 10−5  s /m  − 2.358 × 10−5  s /m ) ⎤⎦ − = −8.146 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 1033 Because the net force is negative, the balloon cannot lift the tank If we can vary the strength of the steel, let’s find out how strong the steel must be by evaluating σ to make the net force positive We want the following to be true: RT ⎛ ρair ρSt ⎞  −   − 1 > 0 MHe ⎜⎝ P0 σ ⎟⎠ Manipulating this inequality gives, RT ⎛ ρair ρSt ⎞  −   > 1 MHe ⎜⎝ P0 σ ⎟⎠ ρair ρSt M ρSt M ρ  −   >  He     →    −   >  He  −  air P0 σ RT σ RT P0 ρSt M ρ −MHe P0  +  ρair RT     →         ρSt −MHe P0  +  ρair RT ρSt P0 RT     →    σ  >     −MHe P0  +  ρair RT     →    ⎤ 860 kg/m ) ( 1.013 × 105  Pa ) ( 8.314 J/mol ⋅ K )( 293 K ) ( 3⎡ = ⎢ ⎥ –3 ⎢⎣ – ( 4  ×  10  kg/mol ) ( 1.013 × 10  Pa ) + ( 1.20 kg/m ) ( 8.314 J/mol ⋅ K )( 293 K ) ⎥⎦ σ = 11.6 × 108  N/m  = 2.3σ actual No, the steel would need to be 2.3 times stronger P19.76 With piston alone, T = constant, so PV = P0V0 or P ( Ahi ) = P0 ( Ah0 ) ⎛h ⎞ With A = constant, P = P0 ⎜ ⎟ ⎝ hi ⎠ But, P = P0 + mp g A , where mp is the mass of the piston Thus P0 + h0 ⎛h ⎞ = P0 ⎜ ⎟ , which reduces to hi = + mp g / P0 A A ⎝ hi ⎠ mp g © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1034 Temperature With the dog of mass M on the piston, a very similar calculation (replacing mp by mp + M) gives h′ = h0 ( ) + mp + M g / P0 A Thus, when the dog steps on the piston, it moves downward by Δh = hi − h′ = 50.0 cm + ( 20.0 kg ) ( 9.80 m s ) ⎡⎣( 1.013 × 105 Pa ) π ( 0.400 m ) ⎤⎦ − 50.0 cm + ( 45.0 kg ) ( 9.80 m s ) ⎡⎣( 1.013 × 105 Pa ) π ( 0.400 m ) ⎤⎦ Δh = 2.38 mm (b) P = const, so V V′ = T Ti Ahi Ah′ = , T Ti or giving ( ) + mp + M g / P0 A ⎛h ⎞ T = Ti ⎜ i ⎟ = Ti ⎝ h′ ⎠ + mp g / P0 A + ( 45.0 kg ) ( 9.80 m s ) ⎡⎣( 1.013 × 105 Pa ) π ( 0.400 m ) ⎤⎦ = 293 K + ( 20.0 kg ) ( 9.80 m s ) ⎡⎣( 1.013 × 105 Pa ) π ( 0.400 m ) ⎤⎦ T = 294.4 K = 21.4 o C P19.77 (a) (b) dL = α dT: L ⎛ Lf ⎞ dL ⇒ ln ⎜ ⎟ = αΔT ⇒ L f = Li eαΔT ⎝ Li ⎠ Li L Ti Li ∫ α dT = ∫ Ti ( )( ) ⎡ 2.00×10−5 °C−1 100°C ⎤ ⎦ L f = ( 1.00 m ) e ⎣ ( = 1.002 002 m ) L′f = 1.00 m ⎡⎣1 + 2.00 × 10−5 °C−1 ( 100°C ) ⎤⎦ = 1.002 000 : L f − L′f Lf (c) = 2.00 × 10−6 = 2.00 × 10−4% ( )( ) ⎡ 2.00×10−2 °C−1 100°C ⎤ ⎦ L f = ( 1.00 m ) e ⎣ = 7.389 m L′f = 1.00 m ⎡⎣1 + ( 0.020 0°C−1 )( 100°C )⎤⎦ = 3.000 m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 L f − L′f Lf 1035 = 59.4% (d) P19.21 redone: We start with dV = βVdT dV = β dT V → → Vf = Vi e βΔT where we assume β (and α) remains constant over the temperature range ΔT Thus, for the turpentine, Vt, final = Vt e βt ΔT and for the aluminum cylinder, VAl, final = VAl e 3α Al ΔT , where we assume α remains constant over the temperature range ΔT new (a): ΔV = Vt e βt ΔT − VAl e 3α Al ΔT ( ΔV = Vi e βt ΔT − e 3α Al ΔT ) 9.00×10 = ( 000 cm ) ⎡⎢ e( ⎣ −4 ) C−1 ( 60.0°C ) 24.0×10−6°C−1 )( 60.0°C ) ⎤ −e ( ⎥⎦ ΔV = 102 mL of turpentine spills, new (b): The volume of the turpentine remaining in the cylinder at 80.0°C is the same as the volume of the aluminum cylinder at 80.0°C: Vt,remaining = VAl, final = VAl e 3α Al ΔT = ( 000 cm ) e ( ) 24.0×10−6 ( °C )−1 ( 60.0°C ) = 009 cm 2.01 L remains in the cylinder at 80.0 °C new (c): The volume of turpentine at 80.0°C we found in part new (b), Vt, remaining = VAl e 3α Al ΔT , shrinks when the temperature changes by –ΔT: Vt, final = Vt, remaining e − βt ΔT = VAl e 3α Al ΔT e − βt ΔT = VAl e( 3α Al − βt )ΔT © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1036 Temperature Vt, final = VAl e 3α Al ΔT e − βt ΔT = VAl e( = ( 000 cm ) e ⎣ ( 3α Al − βt )ΔT )( ) ⎡ 24.0×10−6°C−1 − 9.00×10−4 C−1 ⎤( 60.0°C ) ⎦ Vt, final  903 cm Find the percentage of the cylinder that is empty at 20.0°C: VAl − VAl e( 3α Al − βt ) ΔT = − e( 3α Al − βt ) ΔT VAl e Find the empty height of the cylinder above the turpentine: ( − e( 3α Al − βt ) ΔT )( 20.0 cm) = 0.969 cm and the turpentine level at 20.0°C is 0.969 cm below the cylinder’s rim P19.78 (a) Let xL represent the distance of the stationary line below the top edge of the plate The normal force on the lower part of the plate is mg ( − x ) cos θ and the force of kinetic friction on it is µ k mg ( − x ) cos θ up the roof Again, µ k mgx cos θ acts down the roof on the upper part of the plate The near-equilibrium of the plate requires ∑ Fx = − µ k mgx cos θ + µ k mg ( − x ) cos θ − mg sin θ = −2 µ k mgx cos θ = mg sin θ − µ k mg cos θ µ k x = µ k − tan θ tan θ x= − 2 µk ANS FIG P19.78(a) and the stationary line is indeed below the top edge by xL = L⎛ tan θ ⎞ 1− k â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 (b) 1037 With the temperature falling, the plate contracts faster than the roof The upper part slides down and feels an upward frictional force µ k mg ( − x ) cosθ The lower part slides up and feels downward frictional force µ k mgx cosθ The equation ∑ Fx = is then the same as in part (a) and the stationary line is above the L⎛ tan θ ⎞ bottom edge by xL = ⎜ − 2⎝ µ ⎟⎠ k ANS FIG P19.78(b) (c) Start thinking about the plate at dawn, as the temperature starts to rise As in part (a), a line at distance xL below the top edge of the plate stays stationary relative to the roof as long as the temperature rises The point P on the plate at distance xL above the bottom edge is destined to become the fixed point when the temperature starts falling As the temperature rises, point P on the plate slides down the roof relative to the upper fixed line from (L – xL – xL) to ( L − xL − xL ) ( + α ΔT ) , a change of ΔLplate = ( L − xL − xL )α ΔT The point on the roof originally under point P at the beginning of the expansion moves down not quite as much from (L – xL – xL) to ( L − xL − xL ) ( + α ΔT ) relative to the upper fixed line; a change of ΔLroof = L ( − x − x )α 1ΔT When the temperature drops, point P remains stationary on the roof while the roof contracts, pulling point P back by approximately ΔLroof Therefore, relative to the upper fixed line, point P has moved down the roof ΔLplate − ΔLroof Its displacement for the day is ΔL = ΔLplate − ΔLroof = (α − α ) ( L − xL − xL ) ΔT ⎡ L⎛ tan θ ⎞ ⎤ = (α − α ) ⎢ L − ⎜ − ⎥ (Th − Tc ) 2⎝ µ k ⎟⎠ ⎦ ⎣ ⎛ L tan θ ⎞ = (α − α ) ⎜ (Th Tc ) k â 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1038 Temperature ANS FIG P19.78(c) At dawn the next day the point P is farther down the roof by the distance ΔL It represents the displacement of every other point on the plate (d) ⎛ L tan θ ⎞ (Th − Tc ) ⎝ µ k ⎟⎠ (α − α ) ⎜ = ( 24 × 10−6  °C−1 − 15 × 10−6  °C−1 ) ⎡ (1.20 m)tan 18.5° ⎤ ×⎢ ⎥⎦ ( 32.0°C ) 0.42 ⎣ = 0.275 mm (e) If α < α , the forces of friction reverse direction relative to parts (a) and (b) because the roof expands more than the plate as the temperature rises and less as the temperature falls The diagram in part (a) then applies to temperature falling and the diagram in part (b) applies to temperature rising A point on the plate xL from the top of the plate (which becomes the upper fixed line later when the plate contracts) moves upward from the lower fixed line by ΔLplate , and when the temperature drops, the upper fixed line of the plate is carried down the roof by ΔLroof , so the net change in the plate’s position is ΔLroof − ΔLplate , same as before (up to a sign because now ΔLroof > ΔLplate ) The plate creeps down the roof each day by an amount given by the same expression (with α and α interchanged) P19.79 See ANS FIG P19.79 Let 2θ represent the angle the curved rail subtends We have Li + ΔL = 2θ R = Li ( + αΔT ) and sin θ = Thus, θ= Li L = i R 2R Li ( + αΔT ) = ( + αΔT ) sin θ 2R © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 1039 From Table 19.1, α = 11 × 10−6 ( o C ) , and ΔT = 25.0°C – 20.0°C = 5.00°C We must solve the transcendental equation −1 θ = ( + αΔT ) sin θ = ( 1.000 005 ) sin θ If your calculator is designed to solve such an equation, it may find the zero solution Homing in on the nonzero solution gives, to five digits, θ = 0.018 165 rad = 1.040 8° Now, h = R − R cosθ = Li ( − cosθ ) sin θ This yields h = 4.54 m , a remarkably large value compared to ΔL = 5.50 cm ANS FIG P19.79     © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1040 Temperature ANSWERS TO EVEN-NUMBERED PROBLEMS P19.2 (a) 31.7° C; (b) 31.7 K P19.4 (a) −253° C; (b) −423° F P19.6 (a) 56.7°C and –62.1°C; (b) 330 K and 211 K P19.8 1.20 cm P19.10 Δr = 0.663 mm to the right at 78.2° below the horizontal P19.12 55.0°C P19.14 1.58 × 10–3 cm P19.16 0.548 gal P19.18 Required T = −376° C is below absolute zero P19.20 (a) 2.52 × 106 N/m2; (b) the concrete will not fracture P19.22 (a) 396 N; (b) −101° C; (c) The original length divides out, so the answers would not change P19.24 (a) P19.26 1.20 mol P19.28 In each pump-up-and-discharge cycle, the volume of air in the tank doubles Thus 1.00 L of water is driven out by the air injected at the first pumping, 2.00 L by the second, and only the remaining 1.00 L by the third Each person could more efficiently use his device by starting with the tank half full of water, instead of 80% full P19.30 (a) 1.17 × 10−3 kg; (b) 11.5 mN; (c) 1.01 kN; (d) molecules must be moving very fast P19.32 4.39 kg P19.34 6.64 × 10–27 kg P19.36 2.42 × 1011 molecules P19.38 7.13 m P19.40 m1 − m2 = P19.42 ~102 kg P19.44 ⎛ Pgf + P0 ⎞ m f = mi ⎜ ⎟ ⎝ Pgi + P0 ⎠ ρ0 ; (b) m + βΔT P0VM ⎛ 1 ⎞ − R ⎜⎝ T1 T2 ⎟⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 P19.46 35.016 m P19.48 2.86 atm P19.50 95.0°; T falls below 100°C, so steam condenses, and the expensive apparatus falls (assuming that the boiling point does not change significantly with the change in pressure) 1041 VΔT A P19.52 Δh = ( β − 3α ) P19.54 ⎛ + α AlTC ⎞ (a) θ = sin −1 ⎜ ⎟⎠ ; (b) Yes; (c) Yes, ⎝ ⎛ + α AlTC ⎞ (d) θ = sin −1 ⎜ ⎟ ; (e) 61.0°; (f) 59.6° ⎝ ( + α invarTC ) ⎠ P19.56 0.523 kg P19.58 (a) β = P19.60 θ= ; (b) 3.66 × 10−3 K−1; (c) β He = 3.665 × 10−3 K −1 , this agrees T within 0.06% of the tabulated value; (d) β He = 3.67 × 10−3 K −1 , this agrees within 0.2% of the tabulated value (α − α ) Li ΔT ; (b) In the expression from part (a), θ is directly Δr proportional to ΔT and also to (α − α ) Therefore, θ is zero when either of these quantities becomes zero; (c) the bimetallic strip bends the other way P19.62 See P19.62 for the full solution P19.64 (a) Particle in equilibrium model; (b) On the piston, ∑ F = Fgas − Fg − Fair = : ∑ F = PA − mg − P0 A = ; (c) h = nRT mg + P0 A PM ; (b) 1.33 kg/m3 RT P19.66 (a) P19.68 y ≈ L αΔT/2 P19.70 (a) No torque acts on the disk so its angular momentum is constant Yes: it increases As the disk cools, its radius, and hence, its moment of inertia decrease Conservation of angular momentum then requires that its angular speed increase; (b) 25.7 rad/s P19.72 (a) 125 N; (b) −4.20 × 10−5 m P19.74 (a) 0.169 m; (b) 1.35 × 10 Pa P19.76 (a) 2.38 mm; (b) 294.4 K = 21.4° C © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1042 P19.78 Temperature (a) ∑ Fx = ; (b) With the temperature falling, the plate contracts faster than the roof The upper part slides down and feels an upward frictional force µ k mg ( − x ) cosθ The lower part slides up and feels downward frictional force µ k mgx cosθ The equation ∑ Fx = is then the same as in part (a), and the stationary line is above the bottom edge L⎛ tan θ ⎞ by xL = ⎜ − ; (c) See P19.78(c) for the full explanation; (d) 2⎝ µ ⎟⎠ k 0.275 mm; (e) The plate creeps down the roof each day by an amount given by the same expression (with α2 and α1 interchanged) © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ... is the exterior pressure Likewise, the final absolute pressure in the cylinder is Pf,abs = Pgf + P0, where Pgf is the final gauge pressure The initial and final masses of gas in the cylinder are... P19.46 We must first convert both the initial and final temperatures to Celsius: TC = Thus, (TF − 32 ) Tinitial = T final = (T F , initial (T F , final ) − 32 = ) − 32 = 9 (15.000 − 32.000) = −9.444°C... K = 288.2 K and the final absolute temperature is T f = T f ,C + 273.15 = ( 45 + 273.15 ) K = 318.2 K The ideal gas law, with volume and quantity of gas constant, gives the final absolute pressure