E21-3 (a) We ll assume that the new temperature scale is related to the Celsius scale by a linear. T S = mt C + b, (0) = m( C) + b.

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1 E1-1 (a) We ll assume that the new temperature scale is related to the Celsius scale by a linear transformation; then T S = mt C + b, where m and b are constants to be determined, T S is the temperature measurement in the new scale, and T C is the temperature measurement in Celsius degrees One of our known points is absolute zero; T S = mt C + b, (0) = m( 7315 C) + b We have two other points, the melting and boiling points for water, (T S ) bp = m(100 C) + b, (T S ) mp = m(0 C) + b; we can subtract the top equation from the bottom equation to get (T S ) bp (T s )S mp = 100 C m We are told this is 180 S, so m = 18 S /C Put this into the first equation and then find b, b = 7315 Cm = S The conversion is then T S = (18 S /C )T C + (49167 S) (b) The melting point for water is S; the boiling point for water is 180 S above this, or S E1- T F = 9( 7315 deg C)/5 + 3 F = F E1-3 (a) We ll assume that the new temperature scale is related to the Celsius scale by a linear transformation; then T S = mt C + b, where m and b are constants to be determined, T S is the temperature measurement in the new scale, and T C is the temperature measurement in Celsius degrees One of our known points is absolute zero; T S = mt C + b, (0) = m( 7315 C) + b We have two other points, the melting and boiling points for water, (T S ) bp = m(100 C) + b, (T S ) mp = m(0 C) + b; we can subtract the top equation from the bottom equation to get (T S ) bp (T s )S mp = 100 C m We are told this is 100 Q, so m = 10 Q /C Put this into the first equation and then find b, b = 7315 C = 7315 Q The conversion is then T S = T C + (7315 S) (b) The melting point for water is 7315 Q; the boiling point for water is 100 Q above this, or Q (c) Kelvin Scale 65

2 E1-4 (a) T = (9/5)(6000 K 7315) + 3 = F (b) T = (5/9)(986 F 3) = 370 C (c) T = (5/9)( 70 F 3) = 57 C (d) T = (9/5)( 183 C) + 3 = 97 F (e) It depends on what you think is hot My mom thinks 79 F is too warm; that s T = (5/9)(79 F 3) = 6 C E1-5 T = (9/5)(310 K 7315) + 3 = 983 F, which is fine E1-6 (a) T = (5/9)(T 3), so T/10 = 3, or T = 30 F (b) T = (5/9)(T 3), so 13T/5 = 3, or T = 13 F E1-7 If the temperature (in Kelvin) is directly proportional to the resistance then T = kr, where k is a constant of proportionality We are given one point, T = 7316 K when R = 9035 Ω, but that is okay; we only have one unknown, k Then (7316 K) = k(9035 Ω) or k = 303 K/Ω If the resistance is measured to be R = 968 Ω, we have a temperature of T = kr = (303 K/Ω)(968 Ω) = 911 K E1-8 T = (510 C)/(008 V)V, so T = ( C/V)(0010 V) = 186 C E1-9 We must first find the equation which relates gain to temperature, and then find the gain at the specified temperature If we let G be the gain we can write this linear relationship as G = mt + b, where m and b are constants to be determined We have two known points: (300) = m(00 C) + b, (35) = m(550 C) + b If we subtract the top equation from the bottom we get 5 = m(350 C), or m = 149 C 1 Put this into either of the first two equations and which has a solution b = 70 Now to find the gain when T = 80 C: (300) = (0149 C 1 )(00 C) + b, G = mt + b = (0149 C 1 )(80 C) + (70) = 31 E1-10 p/p tr = (37315 K)/(7316 K) = 1366 E cm Hg is 1000 torr P He = (100 cm Hg)(373 K)/(7316 K) = cm Hg Nitrogen records a temperature which is 0 K higher, so P N = (100 cm Hg)(373 K)/(7316 K) = cm Hg The difference is 0073 cm Hg E1-1 L = ( /C )(33 m)(15c ) = m E1-13 L = ( /C )(00 in)(60c ) = in 66

3 E1-14 L = (75cm)[1 + ( /C )(18C )] = 733 cm E1-15 We want to focus on the temperature change, not the absolute temperature In this case, T = T f T i = (4 C) ( 50 C) = 47 C Then L = ( C 1 )(10 m)(47 C ) = m E1-16 A = αa T, so A = ( /C )(0 m)(30 m)(30c ) = m E1-17 (a) We ll apply Eq 1-10 The surface area of a cube is six times the area of one face, which is the edge length squared So A = 6(033 m) = 0661 m The temperature change is T = (750 C) (00 C) = 550 C Then the increase in surface area is A = αa T = ( C 1 )(0661 m )(550 C ) = m (b) We ll now apply Eq 1-11 The volume of the cube is the edge length cubed, so and then from Eq 1-11, V = (033 m) 3 = m 3 V = αv T = 3( C 1 )(00366 m 3 )(550 C ) = m 3, is the change in volume of the cube E1-18 V = V (1 + 3α T ), so V = (530 cm 3 )[1 + 3( /C )( 17 C )] = 5 cm 3 E1-19 (a) The slope is approximately /C (b) The slope is zero E1-0 r = (β/3)r T, so r = [( /K)/3]( m)(700 K) = m E1-1 We ll assume that the steel ruler measures length correctly at room temperature Then the 005 cm measurement of the rod is correct But both the rod and the ruler will expand in the oven, so the 011 cm measurement of the rod is not the actual length of the rod in the oven What is the actual length of the rod in the oven? We can only answer that after figuring out how the 011 cm mark on the ruler moves when the ruler expands Let L = 011 cm correspond to the ruler mark at room temperature Then L = α steel L T = ( C 1 )(011 cm)(50 C ) = cm is the shift in position of the mark as the ruler is raised to the higher temperature Then the change in length of the rod is not (011 cm) (005 cm) = 006 cm, because the 011 cm mark is shifted out We need to add 0055 cm to this; the rod changed length by 0115 cm The coefficient of thermal expansion for the rod is α = L L T = (0115 cm) (005 cm)(50 C ) = C 1 67

4 E1- A = ab, A = (a + a)(b + b) = ab + a b + b a + a b, so A = a b + b a + a b, = A( b/b + a/a + a b/ab), A(α T + α T ), = αa T E1-3 Solve this problem by assuming the solid is in the form of a cube If the length of one side of a cube is originally L 0, then the volume is originally V 0 = L 3 0 After heating, the volume of the cube will be V = L 3, where L = L 0 + L Then V = L 3, = (L 0 + L) 3, = (L 0 + αl 0 T ) 3, = L 3 0(1 + α T ) 3 As long as the quantity α T is much less than one we can expand the last line in a binomial expansion as V V 0 (1 + 3α T + ), so the change in volume is V 3αV 0 T E1-4 (a) A/A = (018%) = (036%) (b) L/L = 018% (c) V/V = 3(018%) = (054%) (d) Zero (e) α = (00018)/(100 C ) = /C E1-5 ρ ρ = m/v m/v = m/(v + V ) m/v m V/V Then ρ = (m/v )( V/V ) = ρβ T E1-6 Use the results of Exercise 1-5 (a) V/V = 3 L/L = 3(009%) = 076% The change in density is ρ/ρ = V/V = (076%) = 08 (b) α = β/3 = (08%)/3(40 C ) = /C Must be aluminum E1-7 The diameter of the rod as a function of temperature is d s = d s,0 (1 + α s T ), The diameter of the ring as a function of temperature is d b = d b,0 (1 + α b T ) 68

5 We are interested in the temperature when the diameters are equal, d s,0 (1 + α s T ) = d b,0 (1 + α b T ), α s d s,0 T α b d b,0 T = d b,0 d s,0, T = d b,0 d s,0, α s d s,0 α b d b,0 T = (99 cm) (3000 cm) ( /C )(3000 cm) ( /C )(99 cm), = 335 C The final temperature is then T f = (5 ) C = 360 E1-8 (a) L = L 1 + L = (L 1 α 1 + L α ) T The effective value for α is then (b) Since L = L L 1 we can write α 1 L 1 + α (L L 1 ) = αl, α = L L T = α 1L 1 + α L L L 1 = L α α α 1 α, The brass length is then 131 cm and the steel is 393 cm = (054 m) ( ) ( ) ( ) ( = 0131 m ) E1-9 At 100 C the glass and mercury each have a volume V 0 After cooling, the difference in volume changes is given by V = V 0 (3α g β m ) T Since m = ρv, the mass of mercury that needs to be added can be found by multiplying though by the density of mercury Then m = (0891 kg)[3( /C ) ( /C )]( 135C ) = kg This is the additional amount required, so the total is now 909 g E1-30 (a) The rotational inertia is given by I = r dm; changing the temperature requires r r = r + r = r(1 + α T ) Then I = (1 + α T ) r dm (1 + α T ) r dm, so I = αi T (b) Since L = Iω, then 0 = ω I + I ω Rearranging, ω/ω = I/I = α T Then ω = ( /C )(30 rev/s)(170 C ) = 15 rev/s 69

6 E1-31 This problem is related to objects which expand when heated, but we never actually need to calculate any temperature changes We will, however, be interested in the change in rotational inertia Rotational inertia is directly proportional to the square of the (appropriate) linear dimension, so I f /I i = (r f /r i ) (a) If the bearings are frictionless then there are no external torques, so the angular momentum is constant (b) If the angular momentum is constant, then L i = L f, I i ω i = I f ω f We are interested in the percent change in the angular velocity, which is ω f ω i ω i = ω f ω i 1 = I i I f 1 = ( ri r f ) 1 = ( ) 1 1 = 036% (c) The rotational kinetic energy is proportional to Iω = (Iω)ω = Lω, but L is constant, so K f K i K i = ω f ω i ω i = 036% E1-3 (a) The period of a physical pendulum is given by Eq 17-8 There are two variables in the equation that depend on length I, which is proportional to a length squared, and d, which is proportional to a length This means that the period have an overall dependence on length proportional to r Taking the derivative, P dp = 1 P r dr 1 P α T (b) P/P = ( C )(10C )/ = After 30 days the clock will be slow by E1-33 Refer to the Exercise 1-3 t = ( s)( ) = 907 s P = (3600 s)( C )( 0C )/ = 068 s E1-34 At C the aluminum cup and glycerin each have a volume V 0 After heating, the difference in volume changes is given by The amount that spills out is then V = V 0 (3α a β g ) T V = (110 cm 3 )[3( /C ) ( /C )](6C ) = 09 cm 3 E1-35 At 00 C the glass tube is filled with liquid to a volume V 0 After heating, the difference in volume changes is given by V = V 0 (3α g β l ) T The cross sectional area of the tube changes according to A = A 0 α g T 70

7 Consequently, the height of the liquid changes according to V = (h 0 + h)(a 0 + A) h 0 A, h 0 A + A 0 h, V/V 0 = A/A 0 + h/h 0 Then h = (18 m/)[( /C ) ( /C )](13 C ) = m E1-36 (a) β = (dv/dt )/V If pv = nrt, then p dv = nr dt, so (b) Kelvins (c) β 1/(300/K) = /K β = (nr/p)/v = nr/pv = 1/T E1-37 (a) V = (1 mol)(831 J/mol K)(73 K)/( Pa) = 5 10 m 3 (b) ( mol 1 )/( /cm 3 ) = E1-38 n/v = p/kt, so n/v = ( Pa)/( J/K)(95 K) = 5 part/cm 3 E1-39 (a) Using Eq 1-17, (b) Use the same expression again, n = pv RT = ( Pa)(47 m 3 ) = 113 mol (831 J/mol K)([1 + 73] K) V = nrt p = (113 mol)(831 J/mol K)([ ] K) ( Pa) = 0903 m 3 E1-40 (a) n = pv/rt = ( Pa)( m 3 )/(831 J/mol K)(315 K) = mol (b) T f = T i p f V f /p i V i, so T f = (315 K)( Pa)( m 3 ) ( Pa)( m 3 ) = 448 K E1-41 p i = ( ) lb/in = 389 lb/in p f = p i T f V i /T i V f, so p f = (389 lb/in )(99K)(988 in 3 ) (70 K)(100 in 3 ) The gauge pressure is then ( ) lb/in = 70 lb/in = 417 lb/in E1-4 Since p = F/A and F = mg, a reasonable estimate for the mass of the atmosphere is m = pa/g = ( Pa)4π( m) /(981 m/s ) = kg 71

8 E1-43 p = p 0 + ρgh, where h is the depth Then P f = Pa and V f = V i p i T f /p f T i, so p i = ( Pa) + (998 kg/m 3 )(981 m/s )(415 m) = Pa E1-44 The new pressure in the pipe is V f = (194 cm3 )( Pa)(96 K) ( Pa)(77 K) = 104 cm 3 p f = p i V i /V f = ( Pa)() = Pa The water pressure at some depth y is given by p = p 0 + ρgy, so y = (0 105 Pa) ( Pa) (998 kg/m 3 )(981 m/s ) = 103 m Then the water/air interface inside the tube is at a depth of 103 m; so h = (103 m) + (50 m)/ = 8 m P1-1 (a) The dimensions of A must be [time] 1, as can be seen with a quick inspection of the equation We would expect that A would depend on the surface area at the very least; however, that means that it must also depend on some other factor to fix the dimensionality of A (b) Rearrange and integrate, P1- First find A A = ln( T 0/ T ) t T T 0 d T T Then find time to new temperature difference t = ln( T 0/ T ) t This happens = 53 minutes later = t ln( T/ T 0 ) = At, 0 A dt, T = T 0 e At = ln[(9 C )/(5 C )] (45 min) = /min = ln[(9 C )/(1 C )] ( /min) = 978min P1-3 If we neglect the expansion of the tube then we can assume the cross sectional area of the tube is constant Since V = Ah, we can assume that V = A h Then since V = βv 0 T, we can write h = βh 0 T P1-4 For either container we can write p i V i = n i RT i We are told that V i and n i are constants Then p = AT 1 BT, where A and B are constants When T 1 = T p = 0, so A = B When T 1 = T tr and T = T b we have so A = 10 mm Hg/K Then T = (10 mm Hg) = A(373 K 7316 K), (90 mm Hg) + (10 mm Hg/K)(7316 K) (10 mm Hg/K) = 348 K Actually, we could have assumed A was negative, and then the answer would be 198 K 7

9 P1-5 Start with a differential form for Eq 1-8, dl/dt = αl 0, rearrange, and integrate: L L 0 dl = T T 0 αl 0 dt, T L L 0 = L 0 α dt, T 0 T L = L 0 (1 + T 0 α dt ) P1-6 L = αl T, so T t = 1 L αl t = ( m/s) ( /C )(18 10 m) = 03 C/s P1-7 (a) Consider the work that was done for Ex 1-7 The length of rod a is L a = L a,0 (1 + α a T ), while the length of rod b is The difference is L b = L b,0 (1 + α b T ) L a L b = L a,0 (1 + α a T ) L b,0 (1 + α b T ), = L a,0 L b,0 + (L a,0 α a L b,0 α b ) T, which will be a constant is L a,0 α a = L b,0 α b or L i,0 1/α i (b) We want L a,0 L b,0 = 030 m so k/α a k/α b = 030 m, where k is a constant of proportionality; k = (030 m)/ ( 1/( /C ) 1/( /C ) ) = m/c The two lengths are L a = ( m/c )/( /C ) = 0713 m for steel and for brass L b = ( m/c )/( /C ) = 0413 m P1-8 The fractional increase in length of the bar is L/L 0 = α T The right triangle on the left has base L 0 /, height x, and hypotenuse (L 0 + L)/ Then x = 1 (L 0 + L) L 0 = L 0 L L 0 With numbers, x = (377 m) ( /C )(3 C ) = m 73

10 P1-9 We want to evaluate V = V 0 (1 + β dt ); the integral is the area under the graph; the graph looks like a triangle, so the result is The density is then V = V 0 [1 + (16 C )(0000/C )/] = (10016)V 0 ρ = ρ 0 (V 0 /V ) = (1000 kg/m 3 )/(10016) = kg/m 3 P1-10 At 000 C the glass bulb is filled with mercury to a volume V 0 After heating, the difference in volume changes is given by V = V 0 (β 3α) T Since T 0 = 00 C, then T = T, if it is measured in C The amount of mercury in the capillary is V, and since the cross sectional area is fixed at A, then the length is L = V/A, or L = V (β 3α) T A P1-11 Let a, b, and c correspond to aluminum, steel, and invar, respectively Then cos C = a + b c ab We can replace a with a 0 (1 + α a T ), and write similar expressions for b and c Since a 0 = b 0 = c 0, this can be simplified to cos C = (1 + α a T ) + (1 + α b T ) (1 + α c T ) (1 + α a T )(1 + α b T ) Expand this as a Taylor series in terms of T, and we find Now solve: T = The final temperature is then 664 C cos C (α a + α b α c ) T cos(5995 ) 1 ( /C ) + ( /C ) ( /C ) = 464 C P1-1 The bottom of the iron bar moves downward according to L = αl T The center of mass of the iron bar is located in the center; it moves downward half the distance The mercury expands in the glass upwards; subtracting off the distance the iron moves we get h = βh T L = (βh αl) T The center of mass in the mercury is located in the center If the center of mass of the system is to remain constant we require m i L/ = m m ( h L)/; or, since ρ = mv = may, Solving for h, ρ i αl = ρ m (βh αl) h = ( /C )(100 m)[( kg/m 3 ) + ( kg/m 3 )] ( kg/m 3 )( /C ) = 017 m 74

11 P1-13 The volume of the block which is beneath the surface of the mercury displaces a mass of mercury equal to the mass of the block The mass of the block is independent of the temperature but the volume of the displaced mercury changes according to V m = V m,0 (1 + β m T ) This volume is equal to the depth which the block sinks times the cross sectional area of the block (which does change with temperature) Then h s h b = h s,0 h b,0 (1 + β m T ), where h s is the depth to which the block sinks and h b,0 = 0 cm is the length of the side of the block But h b = h b,0 (1 + α b T ), so 1 + β m T h s = h s,0 (1 + α b T ) Since the changes are small we can expand the right hand side using the binomial expansion; keeping terms only in T we get h s h s,0 (1 + (β m α b ) T ), which means the block will sink a distance h s h s,0 given by h s,0 (β m α b ) T = h s,0 [ ( /C ) ( /C ) ] (50 C ) = ( )h s,0 In order to finish we need to know how much of the block was submerged in the first place Since the fraction submerged is equal to the ratio of the densities, we have h s,0 /h b,0 = ρ b /ρ m = ( kg/m 3 )/( kg/m 3 ), so h s,0 = 397 cm, and the change in depth is 07 mm P1-14 The area of glass expands according to A g = α g A g T The are of Dumet wire expands according to A c + Ai = (α c A c + α i A i ) T We need these to be equal, so P1-15 P1-16 V = V 1 (p 1 /p )(T 1 /T ), so α g A g = α c A c + α i A i, α g r g = α c (r c r i ) + α i r i, α g (r c + r i ) = α c (r c r i ) + α i r i, r i r c = α c α g α c α i V = (347 m 3 )[(76 cm Hg)/(36 cm Hg)][(5 K)/(95 K)] = 559 m 3 75

12 P1-17 Call the containers one and two so that V 1 = 1 L and V = 318 L Then the initial number of moles in the two containers are The total is n 1,i = p iv 1 RT i and n,i = p iv RT i n = p i (V 1 + V )/(RT i ) Later the temperatures are changed and then the number of moles of gas in each container is The total is still n, so n 1,f = p fv 1 RT 1,f and n,f = p fv RT,f ( p f V1 + V ) = p i(v 1 + V ) R T 1,f T,f RT i We can solve this for the final pressure, so long as we remember to convert all temperatures to Kelvins, p f = p ( i(v 1 + V ) V1 + V ) 1, T i T 1,f T,f or p f = (144 atm)(1l L) (89 K) ( (1 L) (89 K) ) 1 (318 L) + = 174 atm (381 K) P1-18 Originally n A = p A V A /RT A and n B = p B V B /RT B ; V B = 4V A Label the final state of A as C and the final state of B as D After mixing, n C = p C V A /RT A and n D = p D V B /RT B, but P C = P D and n A + n B = n C + n D Then p A /T A + 4p B /T B = p C (1/T A + 4/T B ), or p C = (5 105 Pa)/(300 K) + 4( Pa)/(400 K) 1/(300 K) + 4/(400 K) = Pa P1-19 If the temperature is uniform then all that is necessary is to substitute p 0 = nrt/v and p = nrt/v ; cancel RT from both sides, and then equate n/v with n V P1-0 Use the results of Problem The initial pressure inside the bubble is p i = p 0 + 4γ/r i The final pressure inside the bell jar is zero, sop f = 4γ/r f The initial and final pressure inside the bubble are related by p i r i 3 = p f r f 3 = 4γr f Now for numbers: p i = ( Pa) + 4(5 10 N/m)/( m) = Pa and r f = ( Pa)( m) 3 4(5 10 N/m) = m P1-1 P1-76

Chapter 19 Solutions

Chapter 19 Solutions *19.1 (a) To convert from Fahrenheit to Celsius, we use T C = 5 9 (T F 32.0) = 5 (98.6 32.0) = 37.0 C 9 and the Kelvin temperature is found as T = T C + 273 = 310 K In a fashion identical