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### ch1309

Course: CHEM 101, Spring 2009
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Principles Giancoli Physics: with Applications, 6th Edition (c) For concrete, repeat the calculation with the expansion coefficient and elastic modulus for concrete. Stress FA ET 12 10 6 Co 20 109 N m2 60 C o 1.4 107 N m2 6 2 The ultimate tensile strength of concrete is 2 10 N m , and so the concrete will fracture . 25. (a) Calculate the change in temperature needed to increase the diameter of the iron...

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Principles Giancoli Physics: with Applications, 6th Edition (c) For concrete, repeat the calculation with the expansion coefficient and elastic modulus for concrete. Stress FA ET 12 10 6 Co 20 109 N m2 60 C o 1.4 107 N m2 6 2 The ultimate tensile strength of concrete is 2 10 N m , and so the concrete will fracture . 25. (a) Calculate the change in temperature needed to increase the diameter of the iron band so that it fits over the barrel. Assume that the barrel does not change in dimensions. L L0 T L L0 L0 T T0 T T0 L L0 L0 20o C 134.122 cm 134.110 cm 12 10 6 C o 134.110 cm 27.457 o C 27 o C (b) Since the band cannot shrink while cooling, the thermal stress must compensate in order to keep the length at a constant 132.122 cm. E is Youngs modulus for the material. L Stress F A ET F AE AE T L0 7.4 10 2 m 6.5 10 3 m 100 109 N m 2 12 10 6 Co 32 7.457Co 273.15 . 4.3 103 N 26. Use the relationships T K ) (a T K (b ) T K (c) T K (d ) T K (e) T K T o C 273.15 and T K 359 K 273.15 5 9 T o F T 5 9 o C o 273.15 86 273.15 F 32 273.15 5 9 T o 78 32 299 K T T 5 9 C C o 273.15 100 273.15 173 K 5773 K 273.15 0.37 K o 273.15 5500 273.15 F 32 273.15 5 9 5 9 T 459 32 32 27. Use the relationship that T K T o F 273.15 . TK T o 5 9 9 5 T o F 32 273.15 32 9 5 F TK 273.15 0 273.15 o 32 459.67o F 28. Use the relationship that T K (a ) T K (b) % error T C 273.15 . 4300 K ; T K T o T o C T 273.15 100 100 4270 K 273 100 C 273.15 15 106 K TK 273 4000 TK 7% 4000o C: 15 106 o C: 273 15 10 6 100 2 10 3% 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 323
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Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory29. Assume the gas is ideal. Since the amount of gas is constant, the value ofPV Tis constant.PV1 1 T1P2V2 T2V2V1P T2 1 P2 T13.00m31.00 atm 3.20 atm273 38 K 273 K1.07 m330. Assume the air is an ideal g
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition34. (a) Assume that the helium is an ideal gas, and then use the ideal gas law to calculate the volume. Absolute pressure must be used, even though gauge pressure is given. 18.75 mol 8.315 J mol
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic TheoryOne factor limiting the maximum altitude would be that as the balloon rises, the density of the air decreases, and thus the temperature required gets higher. Eventually the air would be too hot and the balloon fab
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(b) 6.55 1022 moles6.02 1023 molecules 1 mol4 1046 molecules44. The net force on each side of the box will be the pressure difference between the inside and outside of the box, times the area
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory48. The rms speed is given by Equation 13-9, vrms3kT m .vrms vrms2 13kT2 m 3kT1 mT2 T1373 K 273 K1.1749. The rms speed is given by Equation 13-9, vrms 3kT m . Since the rms speed is proportional to the squ
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition55. The temperature of the nitrogen gas is found from the ideal gas law, and then the rms speed is found from the temperature. 2.1 atm 1.013 105 Pa atm 8.5 m3 PV PV nRT T 167.3 K nR 1300 mol 8.31
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory59. The pressure can be stated in terms of the ideal gas law, PNkT V . Substitute for the temperaturefrom the expression for the rms speed, vrms 3kT m T mvr2ms 3k . The mass of the gas is the mass of a molecule
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition66. The relative humidity of 40% with a partial pressure of 530 Pa of water gives a saturated vapor pressure of 530 Pa 0.40 Psaturated 530 Pa Psaturated 1325 Pa 0.40 From Table 13-3, the temperat
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory71. From Example 13-19, we have an expression for the time to diffuse a given distance. Divide the distance by the time to get the average speed.t C C x Dx2 1 21.00 0.40 mol m3 1.00 0.40 mol m315 10 6 m 95 10
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionThus 6 290.20721% of the original gas remains in the cylinder.76. Assume the air is an ideal gas, and that the pressure is 1.0 atm. PV NkTN PV kT 1.013 105 Pa 6.5 3.1 2.5 m3 1.38 1023J K 27
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory80. The temperature can be found from the rms speed by Equation 13-9, vrmsvrms 3kT m23kT m .T2 mvrms28 1.66 10 kg274 10 km h2341m s 3.6 km h3k3 1.38 10JK1.4 105 K81. From the ideal gas law, if the
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionM N2 4 REarth P46.38 106 m21.01 105 Pag 5.27 1018 kg 1 mole9.80 m s 2 29 10 3 kg 1 mole5.27 1018 kg 1.1 1044 molecules6.02 1023 molecules84. The temperature of the nitrogen gas is found
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory87. (a) Assume that a mass M of gasoline with volume V0 at 0oC is under consideration, and so its density is volume V0M V0 . At a temperature of 32oC, the same mass has aT.M V0 1 T 10V0 1M V0.68 103 kg m3
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionbelow the surface of the water is discussed in chapter 10, and is given by PP0 is atmospheric pressure andP0gh , whereis the density of the sea water.P2V2 T2 V2PV1 1 T1 11.3 LV2V1P T2 1
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theorydsurfaced depthPdepth Psurface1/ 33.00 cm1.01 105 Pa1.0 103 kg m3 1.01 105 Pa9.8 m s 2 14.0 m1/ 33.99 cm95. (a) At a temperature of 30oC, the saturated vapor pressure, from Table 13-6, is 4.24 103 Pa . I
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionButVV0 T , and so V293 K , 1 T0 1 293 KV0 T0TV0 T1 T0For T03.4 10 3 K , which agrees well with Table 13-1.(b) Assume the temperature and amount of gas are held constant, and so P0V0 nRT
Arcadia University - CHEM - 101
CHAPTER 14: Heat Answers to Questions1. The work goes primarily into increasing the temperature of the orange juice, by increasing the average kinetic energy of the molecules comprising the orange juice. When a hot object warms a cooler object, energy is
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition9.The potatoes will not cook faster if the water is boiling faster. The boiling water is the same temperature whether it is boiling fast or slow.10. An ordinary fan does not cool the air direct
Arcadia University - CHEM - 101
Chapter 14Heat17. When the garment is fluffed up, it will have the most air trapped in its structure. The air has a low thermal conductivity, and the more the garment can be fluffed, the more air it will trap, making it a better insulator. The loft valu
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition24. Shiny surfaces absorb very little of the radiation that is incident on them they reflect it back towards the source. Thus the liner is silvered to reduce radiation energy transfer (both into
Arcadia University - CHEM - 101
Chapter 14Heat2.The kcal is the heat needed to raise 1 kg of water by 1 Co. Use that definition to find the temperature change. Then the final temperature can be found. o 7700 J 1 kcal 1kg 1C 0.61Co Final Temperature 10.6o C 3.0 kg 4186 J 1 kcal3.(a)
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition10. The heat absorbed by all three substances is given by Eq. 14-2, Q mc T . Thus the amount of Q mass can be found as m . The heat and temperature change are the same for all three cT substances
Arcadia University - CHEM - 101
Chapter 14Heat15. The heat must warm both the water and the pot to 100oC. The heat is also the power times the time. Q Pt mAl cAl mH O cH O TH O2 2 2tmAl cAl 425 smH O cH O2 2TH O20.36 kg 900 J kg Co0.75 kg 4186 J kg Co 750 W92CoP 7 min16. T
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition22. Assume that the heat from the person is only used to evaporate the water. Also, we use the heat of vaporization at room temperature (585 kcal/kg), since the persons temperature is closer to r
Arcadia University - CHEM - 101
Chapter 14Heat28. The heat lost by the steam condensing and then cooling to 20oC must be equal to the heat gained by the ice melting and then warming to 20oC. msteam Lvap cH O Ti steam Teq mice Lfus cH O Teq Ti ice2 2msteammiceLfus LvapcH O Teq Ti
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition34. The heat conduction rate is given by Eq. 14-4.Q t kA T1 T2 l 0.84 J s m Co3.0 m215.0o C35o C3.2 10 m1.6 104 W35. (a) The power radiated is given by Eq. 14-5. The temperature of the
Arcadia University - CHEM - 101
Chapter 14Heat40. (a) The cross-sectional area of the Earth, perpendicular to the Sun, is a circle of radius REarth , and2 so has an area of REarth . Multiply this area times the solar constant to get the rate at which the Earth is receiving solar ener
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionQ l1 t k1 A Q t AQ l2 t k2 AQ l3 t k3 A T2 T1Tx T1 Ty Tx T2 TyQ l1 t k1l2 k2l31k3 AT2 T1l1 k1 l2 k2 l3 k3(b) For n materials placed next to one another, the expression would beQ tAT
Arcadia University - CHEM - 101
Chapter 14Heat49. (a) Use Eq. 14-5 for total power radiated. Q 2 e AT 4 e 4 RSunT 4 1.0 5.67 10 8 W m2 K 4 4 t7.0 108 m25500 K43.195 1026 W 3.2 1026 W (b) Assume that the energy from the Sun is distributed symmetrically over a spherical surface wit
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition56. (a) To calculate heat transfer by conduction, use Eq. 14-4 for all three areas walls, roof, and windows. Each area has the same temperature difference. Qconduction kA kA kA T1 T2 t l walls l
Arcadia University - CHEM - 101
Chapter 14HeatQmc TQ t21000 W m2 eA cos 1000 W m 2 0.85 40 cm2 1 m21 1 104 cm2 2.3 Co s 4186 J t mc 4.5 10 4 kg 0.80 kcal kg Co 1 kcal (b) We assume that the rate of heat loss by radiation must equal the rate of heat absorption of solar energy. Not
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition61. (a) We consider just the 30 m of crust immediately below the surface of the Earth, assuming that all the heat from the interior gets transferred to the surface, and so it all passes through t
Arcadia University - CHEM - 101
Chapter 14Heat64. The bodys metabolism (blood circulation in particular) provides cooling by convection. If the metabolism has stopped, then heat loss will be by conduction and radiation, at a rate of 200 W, as given. The change in temperature is relate
Arcadia University - CHEM - 101
CHAPTER 15: The Laws of Thermodynamics Answers to Questions1. If water vapor condenses on the outside of a cold glass of water, the internal energy of the water vapor has decreased, by an amount equal to the heat of vaporization of the water vapor. Heat
Arcadia University - CHEM - 101
Chapter 15The Laws of Thermodynamics8.This definition of efficiency is not useful, because with this definition, if the exhaust heat QL is less than the work done W (which is possible), the efficiency would exceed unity. Efficiency should be comparing
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition15. 1 kg of liquid iron has more entropy, because the atoms in liquid iron are less ordered than those in solid iron. Also, heat had to be added to solid iron in order to melt it, and S Q T . 16.
Arcadia University - CHEM - 101
Chapter 15The Laws of Thermodynamics21. In an action movie, seeing a building or car go from an exploded state to an un-exploded state. In a movie with vehicle crashes, seeing two collided vehicles separate from each other, becoming unwrecked as they se
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition5.Segment A is the isothermal expansion. Since the temperature and the amount of gas are constant, the quantity PV nRT is constant. Since the pressure is reduced by a factor of 4.5, the volume w
Arcadia University - CHEM - 101
Chapter 15The Laws of Thermodynamics11. (a,c) See diagram. (b) The work done is found from Eq. 15-3. W PV 455 N m2 8.00 m3 2.00 m3500 400 300 200 100P N m21 2A2.73 103 J The change in internal energy depends on the temperature change, which can be
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(e) Since U a U b10 JUbU a 10 J , we have the following.U bcUc UbUcU a 10 JU ac 10 J25 J 10 J15 J .Use the first law of thermodynamics to find Qbc .U bcQbc WbcQbcU bc Wbc15 J 015
Arcadia University - CHEM - 101
Chapter 15The Laws of ThermodynamicsTotal Power Exhaust PowerActual Power max. eff. operating eff.1.3 GW 0.306 0.755.664 GWTotal Power Actual Power 5.664 GW 1.3 GW 4.364 GW 4.364 109 J s 3600 s h 1.6 1013 J h22. Calculate the Carnot efficiency for
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editione2 1TL2 THTL2TH 1 e21021 K 1 0.49521 K248o C250o C28. For each engine, the efficiency is given by e 0.60eCarnot . Thuse1 e20.6eC 0.6eC10.60 1TL1 TH1 TL20.60 1440 273 K 670 273 K 29
Arcadia University - CHEM - 101
Chapter 15The Laws of Thermodynamics(a ) W (b ) WQH 1 QH 1TL TH TL TH2800 J 1 2800 J 10 273 22 273 15 273 22 273210 J 350 J33. The heat to be removed, QL , is the latent heat of fusion for the ice, so QLmLfusionVH2OQL WH2OLfusion .The work d
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editioninto that great deal of ice. Since it is a large quantity of ice, we assume that its temperature does not change during the processes. 1.00 103 kg 3.33 105 J kg mLfusion Q1 S1 1.2198 106 J K T1 T
Arcadia University - CHEM - 101
Chapter 15The Laws of Thermodynamics1.0 kg 4186 J kg Co 15Co1 37.5 273 K1 52.5 273 K9.3 J K43. The equilibrium temperature is found using calorimetry, from chapter 14. The heat lost by the aluminum is equal to the heat gained by the water.mAl cAl T
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editiondecrease speed by depressing it (friction will slow the car). Finally, shutting the car off can be used to decrease its speed. Any change in speed or direction means that an object is acceleratin
Arcadia University - CHEM - 101
Chapter 5Circular Motion; Gravitation. Thus the mass of the Earth determines the T 42 ratio R 3 T 2 . If the mass of the Earth were doubled, then the ratio R 3 T 2 would double, andGM Earth M Moon R242M Moon R T 2R32GM Earthso R 3 T2R 3 T 2
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionFon planetGM sun M planet r2 Sun to planetaplanetFon planet M planetGrsun 2 Sun to planetM18. In order to orbit, a satellite must reach an orbital speed relative to the center of the Ea
Arcadia University - CHEM - 101
Chapter 5Circular Motion; Gravitationtimes. So when the Earth is close to the Sun, it must move faster to sweep out a given area than when the Earth is far from the Sun. Thus the Earth is closer to the Sun in January. 24. Let the mass of Pluto be M, the
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition5.The orbit radius will be the sum of the Earth s radius plus the 400 km orbit height. The orbital period is about 90 minutes. Find the centripetal acceleration from these data. 60 sec r 6380 km
Arcadia University - CHEM - 101
Chapter 5Circular Motion; GravitationNotice that the result is independent of the car s mass . 10. In the free-body diagram, the car is coming out of the paper at the reader, and the center of the circular path is to the right of the car, in the plane o
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition14. (a) A free-body diagram of the car at the instant it is on the top of the hill is shown. Since the car is moving in a circular path, there must be a net centripetal force downward. Write Newt
Arcadia University - CHEM - 101
Chapter 5Circular Motion; GravitationIf the tension is to be zero, then3 . 28 m s m The bucket must move faster than 3.28 m/s in order for the rope not to go slack.17. The centripetal acceleration of a rotating object is given by aRv aR r 1.15 105 g
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionFyFT sinmg0FTmg sinThe force moving the ball in a circle is the horizontal portion of the tension. Write Newton s 2nd law for that radial motion.FR FT cos maR mv r2Substitute the express
Arcadia University - CHEM - 101
Chapter 5Circular Motion; Gravitationv2 rscosg si n v2v2 r gg tan v226.39 m s 88 m 9.8 m s229.8 m s 26.39 m s 88 m2tan 26.7 o 0.22 tan 26.7o2g cosrsi nrtan22. The car moves in a horizontal circle, and so there must be a net horizont
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition23. If the masses are in line and both have the same frequency, then they will always stay in line. Consider a free-body FN1 FN2 diagram for both masses, from a side view, at the instant that FT2
Arcadia University - CHEM - 101
Chapter 5Circular Motion; GravitationThe total acceleration is given by the Pythagorean combination of the tangential and centripetal accelerations. atotal2 aRat2an . If static friction is to provide the total acceleration, thenFfrm a t o t al2 m a
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editiong MoonGM Moon r2 Moon6.67 1011N m 2 kg 27.35 1022 kg 1.74 10 m6 21.62 m s 231. The acceleration due to gravity at any location on or above the surface of a planet is given by g planet G
Arcadia University - CHEM - 101
Chapter 5Circular Motion; Gravitation36. The acceleration due to gravity at any location at or above the surface of a star is given by g star G M star r 2 , where r is the distance from the center of the star to the location in question.g starGM star
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionJupiter and Saturn will exert a rightward force, while Venus will exert a leftward force. Take the right direction as positive. M Earth M Jupiter MM MM FEarth- G G Earth2 Saturn G Earth2 Venus 2
Arcadia University - CHEM - 101
Chapter 5Circular Motion; Gravitation44. The shuttle must be moving at orbit speed in order for the satellite to remain in the orbit when released. The speed of a satellite in circular orbit around the Earth is given byvGM Earth rGM Earth REarth 65