Physics exercises_solution: Chapter 17

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Physics exercises_solution: Chapter 17

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 17

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Nội dung Text: Physics exercises_solution: Chapter 17

  1. 17.1: From Eq. 17.1, a) 9 5 62.8  32  81.0F. b) 9 556.7   32  134.1F. c) 9 531.1  32  88.0F. 17.2: From Eq. 17.2, a) 5 941.0  32  5.0C. b) 5 9107  32  41.7C. c) 5 / 9 18  32  27.8C. 17.3: 1 C  9 F, so 40.0  72.0 F 5 T2  T1  70.0 F  140.2F 17.4: a) (5 9) (45.0  (4.0))  27.2 C. b) (5 9) (56.0  44)  55.6 C. 17.5: a) From Eq. (17.1), 9 540.2  32  104.4F, which is cause for worry. b) 9 512  32  53.6F, or 54F to two figures. 17.6: 9 511.8  21.2 F 17.7: 1 K  1 C  9 F , so a temperature increase of 10 K corresponds to an increase 5 of 18 F . Beaker B has the higher temperature. 17.8: For (b), TC  TK  10.0 C. Then for (a), TF  9 TC  5 9 5  10.0 C  18.0 F. 17.9: Combining Eq. (17.2) and Eq. (17.3), 5 TK  TF  32  273.15, 9 and substitution of the given Fahrenheit temperatures gives a) 216.5 K, b) 325.9 K, c) 205.4 K. 17.10: (In these calculations, extra figures were kept in the intermediate calculations to arrive at the numerical results.) a) TC  400  273.15  127C, TF  (9 / 5)(126.85)  32  260F. b) TC  95  273.15  178C, TF  (9 / 5)(178.15)  32  289F. c) TC  1.55  107  273.15  1.55  107C, TF  (9 / 5)(1.55  107 )  32  2.79  107F.
  2. 17.11: From Eq. (17.3), TK  (245.92C)  273.15  27.23 K. 17.12: From Eq. (17.4), (7.476)(273.16 K)  2042.14 K  273.15  1769C. 17.13: From Eq. (17.4), (325.0 mm) 273.16 K   444 mm. 373.15 K 17.14: On the Kelvin scale, the triple point is 273.16 K, so R  (9/5)273.15 K  491.69R. One could also look at Figure 17.7 and note that the Fahrenheit scale extends from  460F to  32F and conclude that the triple point is about 492 R. 17.15: From the point-slope formula for a straight line (or linear regression, which, while perhaps not appropriate, may be convenient for some calculators), 4.80  10 4 Pa (0.01C)  (100.0C)  282.33C, 6.50  10 4 Pa  4.80  10 4 Pa which is  282C to three figures. b) Equation (17.4) was not obeyed precisely. If it were, the pressure at the triple   point would be P  (273.16) 6.501015Pa  4.76 10 4 Pa. 373. 4   17.16: T  L  αL0   25  102 m  2.4  10 5 C 62.1 m  168 C, 1 so the temperature is 183 C . 17.17: L0 T  (1.2  105 (C) 1 )(1410 m)(18.0 C  (5.0)C)  0.39 m. 17.18: d  d  d (1  αT )  (0.4500 cm)(1  (2.4  105 (C) 1 )(23.0 C  (78.0C)))  0.4511 cm  4.511 mm. 17.19: a) αD0 T  (2.6  105 (C) 1 )(1.90 cm) (28.0C)  1.4  103 cm, so the diameter is 1.9014 cm. b) αD0 T  3.6  10 3 cm, so the diameter is 1.8964 cm.
  3. 17.20: αT  (2.0  105 (C) 1 )(5.00 C  19.5 C)  2.9  104. 17.21:  α  (L) ( L0 T )  2.3  10 4 m  40.125  10 2  m 25.0 C  2.3  105 C . 1 V V0 1.5010 3 17.22: From Eq. (17.8), T  β  5.110 5 K 1  29.4C, so T  49.4C. 17.23  β V0 T  75  10 5 C 1 1700 L 9.0C  11 L, so there is 11 L of air. 17.24: The temperature change is T  18.0 C  32.0 C  14.0 C. The volume of ethanol contracts more than the volume of the steel tank does, so the additional amount of ethanol that can be put into the tank is Vsteel  Vethanol   βsteel  βethanol V0 T     3.6  10 5 C  75  10 5 (C) 1 2.80 m3  14.0 C  0.0280 m3 1 17.25: The amount of mercury that overflows is the difference between the volume change of the mercury and that of the glass; βglass  18.0  10 5 K 1  8.95 cm3    1.7  10 5 C . 1  1000 cm 55.0C  3  17.26: a) A  L2 , A  2 LL  2 LL L2  2 LL A0 . But L L  αT , and so A  2αTA0  2α A0 T . A  2α AT  2 (2.4  105 (C) 1 ) (π  .275 m  ) 12.5C   1.4  10 4 m 2 . 2 b) πD 2 π  1.350 cm   1.431 cm 2 . 2 17.27: a) A0  4 4 b) A  A0 1  2α T   1.431 cm 2 1  2 1.20  105C 150C   1.437 cm 2 .
  4. 17.28: (a ) No, the brass expands more than the steel. (b) call D the inside diameter of the steel cylinder at 20C At 150C : DST  DBR D  DST  25.000 cm  D BR D  αST D T  25 cm  αBR (25 cm)T 25 cm(1  αBR T ) D  1  αST T   (25 cm) 1  (2.0  10 5 (C) 1 )(130C)  1  (1.2  10 5 (C) 1 )(130 C)  25.026 cm 17.29: The aluminum ruler expands to a new length of L  L0 (1  αT )  (20.0 cm)[1  (2.4  10 5 (C) 1 )(100 C)]  20.048 cm The brass ruler expands to a new length of L  L0 (1  αT )  (20.0 cm)[1  (2.0  10 5 (C) 1 )(100 C)]  20.040 cm The section of the aluminum ruler will be longer by 0.008 cm 17.30: From Eq. (17.12), F  YαTA  (0.9  1011 Pa)(2.0  10 5 (C) 1 )(110C)(2.01 10 4 m 2 )  4.0  104 N. 17.31: a) α  (L L0 T )  (1.9  10 2 m) (1.50 m)(400 C)   3.2  105 (C) 1. b) YαT  Y L L0  (2.0  1011 Pa)(1.9  10 2 m) (1.50 m)  2.5  109 Pa. 17.32: a) L  αTL  (1.2  105 K 1 )(35.0 K)(12.0 m)  5.0  103 m. b) Using absolute values in Eq. (17.12), F  YαT  (2.0  1011 Pa)(1.2  105 K 1 )(35.0 K)  8.4  107 Pa. A
  5. 17.33: a) (37C  (20C))(0.50 L)(1.3  103 kg L) (1020 J kg  K )  38 J b) There will be 1200 breaths per hour, so the heat lost is (1200)(38 J)  4.6  10 4 J. Q mcT (70 kg)(3480 J kg  K )(7 C) 17.34: t    1.4  103 s, about 24 min. P P (1200 W ) 17.35: Using Q=mgh in Eq. (17.13) and solving for Τ gives gh (9.80 m s 2 )(225 m) T    0.53 C. c (4190 J kg.K ) 17.36: a) The work done by friction is the loss of mechanical energy, 1  1  mgh  m(v12  v2 )  (35.0 kg ) (9.80 m s 2 )(8.00 m) sin 36.9  (2.50 m s) 2  2 2  2   1.54  103 J. b) Using the result of part (a) for Q in Eq. (17.13) gives  T  1.54  103 J  35.0 kg 3650 J kg  K   1.21  10 2 C. 17.37: 210 C  20C1.60 kg 910 J kg  K   0.30 kg 470 J kg  K   3.03  105 J. 17.38: Assuming Q  0.60  10  K , T  0.60   10  K 6 1 MV 2  6 1 1.80 kg 7.80 m s  2 2 2  45.1 C. mc mc   8.00  10 3 kg 910 J kg  K  17.39: 85.0 C  20.0 C1.50 kg 910 J kg  K   1.80 kg 4190 J kg  K   5.79  10 J. 5
  6. 17.40: a) Q  mcT  0.320 kg 4190 J kg  K 60.0 K   8.05  10 4 J. b) t  Q P  8.0510 4 J 200 W  402 s. 17.41: a) c  Q  120 s65.0 W   2.51  103 J kg  K. mT 0.780 kg 22.54 C  18.55 C b) An overstimate; the heat Q is in reality less than the power times the time interval. 17.42: The temperature change is T  18.0 K, so c Q  gQ    9.80 m s 2 1.25  104 J   240 J kg  K. mT w T 28.4 N 18.0 K  17.43: a) Q  mcT , c  470 J kg  K We need to find the mass of 3.00 mol:   m  nM  3.00 mol 55.845  10 3 kg mol  0.1675 kg T  Q mc  8950 J  0.1675 kg 470 J kg  K   114 K  114 C b) For m  3.00 kg, T  Q mc  6.35 C c) The result of part (a) is much larger; 3.00 kg is more material than 3.00 mol. Qmelt 10,000 J min 1.5 min  17.44: (a) LF    30,000 J kg m 0.50 kg Q (b) Liquid : Q  mcT  c  mT c 10,000 J min 1.5 min   1,000 J kg  C 0.50 kg 30C Solid : c  Q  10,000 J min 1.0 min   1300 J kg  C mT 0.50 kg 15 C
  7. 17.45: a) Qwater  Qmetal  0 m water c water Twater  mmetal c metal Tmetal  0 1.00kg 4190 J kg  K 2.0 C  0.500 kg cmetal  78.0 C  0 cmetal  215 J/kg  K b) Water has a larger specific heat capacity so stores more heat per degree of temperature change. c) If some heat went into the styrofoam then Qmetal should actually be larger than in part (a), so the true c metal is larger than we calculated; the value we calculated would be smaller than the true value. 17.46: a) Let the man be designated by the subscript m and the “‘water” by w, and T is the final equilibrium temperature.  mm C m Tm  mw C w Tw  mmCm T  Tm   mw Cw T  Tw  mm C m Tm  T   mw C w T  Tw  m m C m Tm  m w C w Tw Or solving for T, T  mm C m  m w C w . Inserting numbers, and realizing we can change K to C , and the mass of water is .355 kg, we get (70.0 kg) (3480 J kg  K ) (37.0C )  (0.355 kg) (4190 J kg  C ) (12.0C ) T (70.0 kg)(3480 J kg. C )  (0.355 kg) (4190 J kg  C ) Thus, T  36.85C. b) It is possible a sensitive digital thermometer could measure this change since they can read to .1C. It is best to refrain from drinking cold fluids prior to orally measuring a body temperature due to cooling of the mouth. 17.47: The rate of heat loss is Q t.   Q t mCT t , or t  mCt Q ( t ) . Interesting numbers, ( 70.355 kg)(3480 J kg. C)(0.15 C) t  7 10 6 J day  0.005 d, or t  7.6 minutes. This may acount for mothers taking the temperature of a sick child several minutes after the child has something to drink. 17.48: Q  m(cT  Lf )   (0.350 kg) (4190 J kg  K )(18.0 K)  334  103 J kg   1.43  10 J  34.2 kcal  136 Btu. 5
  8. 17.49: Q  m(cice Tice  Lf  cwater Twater  LV )  (2100 J kg  K)(10.0 C)  334  103 J kg   (12.0  10 kg) 3    (100 C)(4190 J kg  K)  2256  103 J kg     3.64  10 J  8.69 kcal  34.5 Btu. 4 Q mcT (0.550 kg)(2100 J kg  K)(15.0 K) 17.50: a) t     21.7 min . P P (800 J min) mLf ( 0.550 kg)(33410 3 J kg ) b) P  (800 J min )  230 min, so the time until the ice has melted is 21.7 min  230 min  252 min. 17.51: (4000 lb) 2.205 lb kg) (334  103 J kg)  7.01 kW  2.40  104 Btu hr. (86,400 s) 17.52: a) m(cT  Lv )  (25.0  10 3 kg)(4190 J kg  K)(66.0 K)  2256  103 J kg   6.33  104 J. b) mcT  (25.0  103 kg)(4190 J kg  K)(66.0 K)  6.91  103 J. c) Steam burns are far more severe than hot-water burns. 17.53: With Q  m(cT  Lf ) and K  (1 / 2)mv 2 , setting Q  K and solving for v gives v  2(130 J Kg  K)(302.3 C  24.5  103 J kg)  357 m s .
  9. McT (70.0 kg)(3480 J kg  K)(1.00 K) 17.54: a) msweat    101 g. Lv (2.42  106 J kg) b) This much water has a volume of 101 cm 3 , about a third of a can of soda. 17.55: The mass of water that the camel saves is McT (400 kg)(3480 J kg  K)(6.0 K)   3.45 kg, Lv (2.42  106 J kg) which is a volume of 3.45 L. 17.56: For this case, the algebra reduces to  ((200)(3.00  103 kg ))(390 J kg  K )(100.0 C)      (0.240 kg )(4190 J kg )(20.0 C)  T   35.1C.  ((200)(3.00  10 3 kg )(390 J kg  K )      (0.240 kg )(4190 J kg  .K )    17.57: The algebra reduces to  ((0.500 kg)(390 J kg  K )  (0.170 kg)(4190 J kg  K ))(20.0 C       (0.250 kg)(470 J kg  K )(85.0C) T   27.5C  ((0.500 kg)(390 J kg  K )  (0.170 kg)(4190 J kg  K ))        (0.250 kg)(470 J kg  K )  17.58: The heat lost by the sample is the heat gained by the calorimeter and water, and the heat capacity of the sample is Q ((0.200 kg)(4190 J kg  K )  (0.150 kg)(390 J kg  K ))(7.1 C) c  mT (0.0850 kg)(73.9 C) =1010 J kg  K, or 1000 J kg  K to the two figures to which the temperature change is known.
  10. 17.59: The heat lost by the original water is  Q  (0.250 kg)(4190 J kg  K )(45.0 C)  4.714  104 J, and the mass of the ice needed is Q mice  cice Tice  Lf  c water Twater (4.714  104 ) J  (2100 J kg  K) (20.0 C)  (334  103 J kg)  (4190 J kg  K )(30.0 C)  9.40  10 2 kg  94.0 g. 17.60: The heat lost by the sample (and vial) melts a mass m, where Q ((16.0 g)(2250 J kg  K )  (6.0 g)(2800 J kg  K ))(19.5K) m   3.08 g. Lf (334  103 J kg ) Since this is less than the mass of ice, not all of the ice melts, and the sample is indeed cooled to 0C. Note that conversion from grams to kilograms was not necessary. (4.00 kg)(234 J kg  K)(750 C) 17.61:  2.10 kg. (334  103 J kg) 17.62: Equating the heat lost by the lead to the heat gained by the calorimeter (including the water-ice mixtue), mP b c Pb (200C  T )  (mw  mice )c wT  mcu ccu T  mice Lf . Solving for the final temperature T and using numerical values,  (0.750 kg)(130 J kg  K)(255 C)      (0.018 kg)(334  103 J kg)  T   21.4C.  (0.750 kg)(130  J kgK)      (0.178 kg)(4190 J kg  K)    (0.100 kg)(390 J kg  K)    (The fact that a positive Celsius temperature was obtained indicates that all of the ice does indeed melt.)
  11. 17.63: The steam both condenses and cools, and the ice melts and heats up along with the original water; the mass of steam needed is (0.450 kg)(334  103 J kg)  (2.85 kg)(4190 J kg  K)(28.0 C) msteam  2256  103 J kg  (4190 J kg)(72.0 C)  0.190 kg. 17.64: The SI units of H and dQ are both watts, the units of area are m 2 , temperature dt difference is in K, length in meters, so the SI units for thermal conductivity are [ W][m] W  . [m ][K] m  K 2 17.65: a) 0.450Km  222 K m. b)(385 W m  K)(1.25  10-4 m 2 )(400 K m)  10.7 W. 100 c)100.0C  (222 K m)(12.00  102 m)  73.3C. 17.66: Using the chain rule, H  dQ  Lf dm and solving Eq. (17.21) for k, dt dt dm L k  Lf dt AT (8.50  10 3 kg) (60.0  10 2 s)  (334  103 J kg ) (600 s) (1.250  10 4 m 2 )(100 K)  227 W m  K.
  12. 17.67: (Although it may be easier for some to solve for the heat flow per unit area, part (b), first the method presented here follows the order in the text.) a) See Example 17.13; as in that example, the area may be divided out, and solving for temperature T at the boundary, k foam Lfoam Tin  k wood Lwood Tout T N k foam Lfoam   k wood Lwood  0.010 W m  K  2.2 cm19.0 C  0.080 W m  K  3.0 cm 10.0 C 0.010 W m  K  2.2 cm  0.080 W m  K  3.0 cm  5.8C. Note that the conversion of the thickness to meters was not necessary. b) Keeping extra figures for the result of part, (a), and using that result in the temperature difference across either the wood or the foam gives H foam H wood   0.010 W m  K  19.0 C   5.767 C A A 2.2  10 2 m  0.080 W m  K   5.767C   10.0C 3.0  10 2 m  11 W m 2 . 17.68: a) From Eq. 17.21,  H  0.040 W m  K  1040 m 2   140 K    196 W, 2 4.0  10 m or 200 W to two figures. b) The result of part (a) is the needed power input. 17.69: From Eq. 17.23, the energy that flows in time t is Ht  AT t    125 ft 2 34F 5.0 h   708 Btu  7.5  105 J. R  30 ft 2  F  h Btu 
  13. 17.70: a) The heat current will be the same in both metals; since the length of the copper rod is known,  H  385.0 W m  K  400  10 4 m 2  1.00 m   5.39 W. 35.0 K b) The length of the steel rod may be found by using the above value of H in Eq. 17.21 and solving for L2 , or, since H and A are the same for the rods, L2  L k2 T2  1.00 m  50.2 W m  K 65.0 K   0.242 m. k T 385.0 W m  K 35.0 K  17.71: Using H  Lv dm (see Problem 17.66) in Eq. (17.21), dt dm L T  Lv dt kA (0.390 kg) (0.85  102 m)  (2256  10 J kg) 3  5.5 C, (180 s) (50.2 W m  K )(0.150 m 2 ) and the temperature of the bottom of the pot is 100 C  6 C  106 C. 17.72: Q T  kA t L  W   300 K  150 J s   50.2  A   m. K   0.500 m  A  4.98  10 3 m 2 2 D A  πR  π   2 2 D  4A π  4(4.98  10 3 m 2 ) π  8.0  10 2 m  8.0 cm
  14. 17.73: H a  H b (a  aluminum, b  brass) A(150.0C  T ) A(T  0 C) H a  ka , H b  kb La Lb (It has been assumed that the two sections have the same cross-sectional area.) A(150.0C  T ) A(T  0C) ka  kb La Lb (2050 W m  K)(150.0 C  T ) (109.0 W m  K)(T  0C)  0.800 m 0.500 m Solving for T gives T  90.2C 17.74: From Eq. (17.25), with e  1, a) (5.67  108 W m 2  K 4 )(273 K) 4  315 W m 2 . b) A factor of ten increase in temperature results in a factor of 10 4 increase in the output; 3.15  10 6 W m 2 . 17.75: Repeating the calculation with Ts  273 K  5.0 C  278 K gives H  167 W. 17.76: The power input will be equal to H net as given in Eq. (17.26); P  Aeσ (T 4  Ts4 )  (4π (1.50  10 2 m)2 )(0.35)(5.67  10 8 W m 2  K 4 )((3000 K) 4  (290 K) 4 )  4.54  103 W. H 150W 17.77: A   2.10 cm 2 eσ T 4  8 2 4  0.35 5.67  10 W m  K 2450K 4
  15. 17.78: The radius is found from R A   H σT 2   H 1 . 4π 4π 4πσ T 2 Using the numerical values, the radius for parts (a ) and (b ) are Ra  2.7  10 W 32 1  1.61  1011 m  4 5.67  10 W m  K 11,000 K  8 2 4 2  Rb  2.10  10 W 23 1  5.43  106 m  4 5.67  10 W m  K 10,000 K  82 4 2  c) The radius of Procyon B is comparable to that of the earth, and the radius of Rigel is comparable to the earth-sun distance. 17.79: a) normal melting point of mercury:  30 C  0.0 normal boiling point of mercury: 357 C  100.0 100   396 C so 1   3.96 C Zero on the M scale is  39 on the C scale, so to obtain TC multiple T by 3.96 and then subtract 39 : TC  3.96TM  39 1 Solving for TM gives TM  TC  39 3.96 1 The normal boiling point of water is 100C; TM  100  39  35.1 3.96 b)10.0   39.6 C 17.80: All linear dimensions of the hoop are increased by the same factor of α  T , so the increase in the radius of the hoop would be    Rα  T  6.38  106 m 1.2  105 K 1 0.5 K   38 m.
  16. 17.81: The tube is initially at temperature T0 , has sides of length L0 volume V0 , density ρ0 , and coefficient of volume expansion β. a) When the temperature increase to T0  T , the volume changes by an amount m m V , where V  β V0 T . Then, ρ  , or eliminating V , ρ  . V0  V V0  βV0 T Divide the top and bottom by V0 and substitute ρ0  m V0 . Then m V0 ρ0 ρ or ρ  . This can be rewritten as ρ  ρ0 1  β T  1. Then V0 V0  βV0  T V0 1  βT using the expression 1  x   1  nx, where n  1, ρ  ρ0 1  β T . n b) The copper cube has sides of length 1.25 cm  .0125 m, and T  70.0 C  20.0 C  50.0 C.   V  βV0 T  5.1  10 5  C .0125 m  50.0 C   5  10 9 m3 . 3 Similarly, ρ  8.9  103 kg m 3 (1  (5.1  105  C)(50.0 C)), or ρ  8.877  103 kg m 3 ; extra significant figures have been keep. So ρ  23 kg m3 .
  17. 17.82: (a) We can use differentials to find the frequency change because all length changes are small percents . Let m be the mass of the wire v  F μ  F (m L)  FL m v f  and   2 L(fundamental)  FL m 1 F f v   2L 2 mL F f  L (only L changes due to heating) L f 1 2 ( 1 )( F mL) 1 2 ( mL2 )L 2 F 1 L   f 1 2 F mL 2 L 1 1 f  (αT ) f  (1.7  10 5 (C) 1 )(40C)(440 C)(440 Hz)  0.15 Hz 2 2 The frequency decreases since the length increases (b) v  F μ  FL m v 1 ( FL m) 1 2 ( F m)L L αT  2   v FL m 2L 2 1  (1.7  10 5 (C) 1 )(40C)  3.4  10  4  0.034% 2 (c)   2 L    2L    22LL  LL  αT   (1.7  10 5C 1 )(40C)  6.8  10 4  0.068% :  it increases 17.83: Both the volume of the cup and the volume of the olive oil increase when the temperature increases, but β is larger for the oil so it expands more. When the oil starts to overflow, Voil  Vglass  (1.00  10 3 m) A, where A is the cross-sectional area of the cup. Voil  V0, oil βoil T  (9.9 cm) Aβoil T Vglass  V0, glass β glass T  (10.0 cm) Aβglass T (9.9 cm) Aβoil T  (10.0 cm) Aβglass T  (1.00  103 m) A The area A divides out. Solving for T gives T  15.5 C T2  T1  T  37.5C
  18. 17.84: Volume expansion: dV  βV dT dV dT Slope of graph β  V V Construct the tangent to the graph at 2C and 8C and measure the slope of this line. 3 At 22C : Slope   0.10Ccm and V  1000 cm3 3  0.10 cm3 3C β 3  3  10 5 (C) 1 1000 cm The slope in negative, as the water contracts or it is heated. At 8 C : slope  0.24Ccm and V  1000 cm 3 3 4  0.24 cm3 4C  β 3  6  10 5 (C) 1 1000 cm The water now expands when heated. 17.85: La  Ls  0.40 cm (a  aluminum, s  steel) L  L0α T , so (24.8 cm)(2.4  105 (C ) 1 ))T  (34.8 cm)(1.2  105 (C ) 1 ))T  0.40 cm T  395 C T2  T1  T  415C 17.86: a) The change in height will be the difference between the changes in volume of the liquid and the glass, divided by the area. The liquid is free to expand along the column, but not across the diameter of the tube, so the increase in volume is reflected in the change in the length of the columns of liquid in the stem.  V liquid   V glass V b)  h   ( β liquid  β glass )  T A A 6 (100  10 m3 )  (8.00  10 4 K 1  2.00  105 K -1 )(30.0 K) (50.0  10 6 m 2 )  4.68  10 2 m.
  19. 17.87: To save some intermediate calculation, let the third rod be made of fractions f 1 and f 2 of the original rods; then f1  f 2  1 and f1 (0.0650)  f 2 (0.0350)  0.0580. These two equations in f 1 and f 2 are solved for 0.0580  0.0350 f1  , f 2  1  f1 , 0.0650  0.0350 and the lengths are f1 (30.0 cm)  23.0 cm and f 2 (30.0 cm)  7.00 cm 17.88: a) The lost volume, 2.6 L, is the difference between the expanded volume of the fuel and the tanks, and the maximum temperature difference is V T  ( βfuel  βA1 )V0 (2.6  103 m3 )  (9.5  10 4 (C) 1  7.2  10 5 (C) 1 )(106.0  103 m3 )  2.78 C, or 28C to two figures; the maximum temperature was 32C. b) No fuel can spill if the tanks are filled just before takeoff. 17.89: a) The change in length is due to the tension and heating L F F ΔL  L0  AY  αT . Solving for F A, A  Y L0  αΔT . b) The brass bar is given as “heavy” and the wires are given as “fine,” so it may be assumed that the stress in the bar due to the fine wires does not affect the amount by which the bar expands due to the temperature increase. This means that in the equation preceding Eq. (17.12), L is not zero, but is the amount αbrass L0 T that the brass expands, and so F  Ysteel (αbrass  αsteel )T A  20  1010 Pa)(2.0  105 (C) 1  1.2  105 (C) 1 )(120C)  1.92  108 Pa.
  20. 17.90: In deriving Eq. (17.12), it was assumed that L  0; if this is not the case when there are both thermal and tensile stresses, Eq. (17.12) becomes  F  L  L0  αT  .  AY  For the situation in this problem, there are two length changes which must sum to zero, and so Eq. (17.12) may be extended to two materials a and b in the form  F   F  L0a  αa T     L0b  αb T      0.  AYa   AYb   Note that in the above, T , F and A are the same for the two rods. Solving for the stress F A, F αa L0a  αb L0b  T A (( Loa Ya )  ( L0b Yb )) (1.2  10-5 (C) 1 )(0.350 m)  (2.4  105 (C) 1 )(0.250 m)  (60.0 C) ((0.350 m) 20  1010 Pa)  (0.250 m 7  1010 Pa))  1.2  108 Pa to two figures. ( 0.0020 in.) 17.91: a) T  R αR0  (1.210  5 (C  ) 1 ( 2.5000 in.)  67 C to two figures, so the ring should be warmed to 87C. b) the difference in the radii was initially 0.0020 in., and this must be the difference between the amounts the radii have shrunk. Taking R0 to be the same for both rings, the temperature must be lowered by an amount R T  αbrass  αsteel R0  0.0020 in.  100 C  2.0  10 5 C  1.2  10 5 C 2.50 in. 1 1  to two figures, so the final temperature would be  80C. 17.92: a) The change in volume due to the temperature increase is βVT , and the change in volume due to the pressure increase is  V p Eq. 11.13. Setting the net B p change equal to zero, βVT  V B , or p  BβV . b) From the above,   p  1.6  10 Pa 3.0  10 11 5 K 1 15.0 K   8.64  10 7 Pa.
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