physics - ch 12 solutions
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physics - ch 12 solutions

Course Number: PHY 2054, Spring 2008

College/University: University of Florida

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Chapter 12 The Laws of Thermodynamics Quick Quizzes 1. (b). The work done on a gas during a thermodynamic process is the negative of the area under the curve on a PV diagram. Processes in which the volume decreases do positive work on the gas, while processes in which the volume increases do negative work on the gas. The work done on the gas in each of the four processes shown is: Wa = - 4.00 10 5 J , Wb = +...

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12 The Chapter Laws of Thermodynamics Quick Quizzes 1. (b). The work done on a gas during a thermodynamic process is the negative of the area under the curve on a PV diagram. Processes in which the volume decreases do positive work on the gas, while processes in which the volume increases do negative work on the gas. The work done on the gas in each of the four processes shown is: Wa = - 4.00 10 5 J , Wb = + 3.00 10 5 J , Wc = - 3.00 10 5 J , and Wd = + 4.00 10 5 J Thus, the correct ranking (from most negative to most positive) is a,c,b,d. 2. 3. A is isovolumetric, B is adiabatic, C is isothermal, D is isobaric. (c). The highest theoretical efficiency of an engine is the Carnot efficiency given by ec = 1 - Tc Th . Hence, the theoretically possible efficiencies of the given engines are: eA = 1 - 500 K 300 K 700 K = 0.300 , eB = 1 - = 0.375 , and eC = 1 - = 0.500 800 K 600 K 1 000 K and the correct ranking (from highest to lowest) is C, B, A. 4. Qr = 0 and Q = 0 in an adiabatic process. If the process was reversible, but not T adiabatic, the entropy of the system could undergo a non-zero change. However, in that case, the entropy of the system's surroundings would undergo a change of equal magnitude but opposite sign, and the total change of entropy in the universe would be zero. If the process was irreversible, the total entropy of the universe would increase. (b). S = The number 7 is the most probable outcome because there are six ways this could occur: 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1. The numbers 2 and 12 are the least probable because they could only occur one way each: either 1-1, or 6-6. Thus, you are six times more likely to throw a 7 than a 2 or 12. 5. 431 432 CHAPTER 12 Answers to Even Numbered Conceptual Questions 2. Either statement can be considered an instructive analogy. We choose to take the first view. All processes require energy, either as energy content or as energy input. The kinetic energy which it possessed at its formation continues to make Earth go around. Energy released by nuclear reactions in the core of the Sun drives weather on Earth and essentially all processes in the biosphere. The energy intensity of sunlight controls how lush a forest or jungle can be and how warm a planet is. Continuous energy input is not required for the motion of the planet. Continuous energy input is required for life because energy tends to be continuously degraded, as energy is transferred by heat into lowertemperature sinks. The continuously increasing entropy of the Universe is the index to energy-transfers completed. Shaking opens up spaces between the jelly beans. The smaller ones have a chance of falling down into spaces below them. The accumulation of larger ones on top and smaller ones on the bottom implies an increase in order and a decrease in one contribution to the total entropy. However, the second law is not violated and the total entropy of the system increases. The increase in the internal energy of the system comes from the work required to shake the jar of beans (that is, work your muscles must do, with an increase in entropy accompanying the biological process) and also from the small loss of gravitational potential energy as the beans settle together more compactly. Temperature = A measure of molecular motion. Heat = the process through which energy is transferred between objects by means of random collisions of molecules. Internal energy = energy associated with random molecular motions plus chemical energy, strain potential energy, and an object's other energy not associated with center of mass motion or location. A higher steam temperature means that more energy can be extracted from the steam. For a constant temperature heat sink at Tc and steam at Th , the maximum efficiency of the T - Tc T = 1 - c and is maximized for high Th . power plant goes as h Th Th 4. 6. 8. 10. 12. emax = T 80 K = 22% (Assumes atmospheric temperature of 20C.) Th 373 K An analogy due to Carnot is instructive: A waterfall continuously converts mechanical energy into internal energy. It continuously creates entropy as the organized motion of the falling water turns into disorganized molecular motion. We humans put turbines into the waterfall, diverting some of the energy stream to our use. Water flows spontaneously from high to low elevation and energy is transferred spontaneously from high to low temperature by heat. Into the great flow of solar radiation from Sun to Earth, living things put themselves. They live on energy flow. A basking snake diverts high-temperature energy through itself temporarily, before it is inevitably lost as low-temperature energy radiated into outer space. A tree builds organized cellulose molecules and we build libraries and babies who look like their grandmothers, all out of a thin diverted stream in the universal flow of energy crashing down to disorder. We do not violate the second law, for we build local reductions in the entropy of one thing within the inexorable increase in The Laws of Thermodynamics 433 the total entropy of the Universe. Your roommate's exercise increases random molecular motions within the room. 14. Even at essentially constant temperature, energy must be transferred by heat out of the solidifying sugar into the surroundings. This action will increase the entropy of the environment. The water molecules become less ordered as they leave the liquid in the container to mix with the entire atmosphere. A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the floor and shatters. Any process is irreversible if it looks funny or frightening when shown in a videotape running backward. At fairly low speeds, air resistance is small and the flight of a projectile is nearly reversible. 16. 434 CHAPTER 12 Answers to Even Numbered Problems 2. (a) 610 J (d) 0 (a) (c) (b) 0 (e) 200 J (b) (c) 410 J 4. 6. 8. 10. 12. 14. 16. 18. 31 m s 0.17 More work is done in (a) because of higher pressure during the expansion. -465 J (a) 12.0 MJ (b) 12.0 MJ 6P0V0 U = 0, Q<0, W >0 (a) 12 kJ (b) 12 kJ QAB >0, WAB <0, U AB > 0 QBC <0, WBC =0, U BC < 0 QCA <0, WCA >0, UCA < 0 (a) (a) (b) 20. 22. 8.24 J (b) 12.4 J (c) 20.6 J WIAF = -76.0 J, WIBF = -101 J, WIF = -88.7 J QIAF = 165 J, QIBF = 190 J, QIF = 178 J 24. 26. 28. 30. 19.7% (a) 10.7 kJ (b) 0.533 s 13.7C (a) 0.0512 (or 5.12%) (b) 5.27 1012 J (c) Such engines would be one way to harness solar energy as conventional energy sources become more expensive. However, this engine could produce unacceptable thermal pollution in the deep ocean waters. (a) 32. 34. 36. 38. 9.10 kW (b) 11.9 kJ 453 K 6.06 kJ K 2.7 kJ K The Laws of Thermodynamics 435 40. (a) End Result All R 1G, 2R 2G, 1R All G Possible Draws RRR RRG, RGR, GRR GGR, GRG, RGG GGG Total Number of Same Result 1 3 3 1 (b) End Result All R 1G, 4R 2G, 3R 3G, 2R 4G, 1R All G Possible Draws RRRRR RRRRG, RRRGR, RRGRR, RGRRR, GRRRR RRRGG, RRGRG, RGRRG, GRRRG, RRGGR, RGRGR, GRRGR, RGGRR, GRGRR, GGRRR GGGRR, GGRGR, GRGGR, RGGGR, GGRRG, GRGRG, RGGRG, GRRGG, RGRGG, RRGGG GGGGR, GGGRG, GGRGG, GRGGG, RGGGG GGGGG Total Number of Same Result 1 5 10 10 5 1 42. 44. 46. 48. 50. (a) 1 52 (b) 1 13 (c) 14 3.0 Sh = -16.0 J K , Sc = 26.7 J K , SUniverse = 10.7 J K 0.55 kg (a) 251 J (d) 104 J on the gas (b) 314 J (e) zero in both cases (c) 104 J by the gas 52. 54. 56. 58. 5.97 10 4 kg s (a) (a) (a) (c) 4 P0V0 -4.9 10 -2 J 21 nRT0 2 0.190 (or 19.0%) + 335 J 2.4 10 6 J (b) (b) (b) 4 P0V0 16 kJ (c) (c) 9.07 kJ 16 kJ 17 nRT0 2 (d) 0.833 (or 83.3%) (b) (b) 60. 62. (a) (a) 2.09 10 3 J 1.6 106 J (c) 2.8 10 2 J 436 CHAPTER 12 Problem Solutions 12.1 From kinetic theory, the average kinetic energy per molecule is KEmolecule = 3 3 R kBT = T 2 2 NA For a monatomic ideal gas containing N molecules, the total energy associated with random molecular motions is U = N KEmolecule = 3 N 3 N RT = 2 nRT 2 A Since PV = nRT for an ideal gas, the internal energy of a monatomic ideal gas is found 3 to be given by U = PV . 2 (a) Wab = Pa (Vb - Va ) P (atm) -3 3 12.2 10 m = 3 ( 1.013 10 5 Pa ) ( 3.0 L - 1.0 L ) = 610 J 1L (b) Wbc = P (Vc - Vb ) = 0 (c) Wcd = Pc (Vd - Vc ) 10 -3 m 3 = 2 ( 1.013 10 5 Pa ) ( 1.0 L - 3.0 L ) = - 410 J 1L (d) Wda = P (Va - Vd ) = 0 (e) Wnet = Wab + Wbc + Wcd + Wda = +610 J + 0 - 410 J + 0 = + 200 J 3 2 1 0 a d b c V (L) 0 1 2 3 The Laws of Thermodynamics 437 12.3 The number of molecules in the gas is N = nN A and the total internal energy is 3 3 U = N KE = nN A k B T = nRT 2 2 = 3 J 4 ( 3.0 mol ) 8.31 ( 303 K ) = 1.1 10 J 2 mol K ( ) Alternatively, use the result of Problem 12.1, U= 3 3 PV = nRT , just as found above. 2 2 12.4 (a) The work done by the gas on the projectile is given by the area under the curve in the PV diagram. This is Wby gas = ( triangular area ) + ( rectangular area ) = 1 P0 - Pf 2 ( )(V f - V0 + Pf Vf - V0 = ) ( ) 1 P0 + Pf 2 ( )(V f - V0 ) = 1 m3 1 ( 11 + 1.0 ) 10 5 Pa ( 40.0 - 8.0 ) cm 3 6 10 cm 3 2 = 19 J 1 mv 2 - 0 where W is the work done 2 on the projectile by the gas. Thus, the speed of the emerging projectile is From the work-energy theorem, W = KE = v= 2W = m 2 (19 J ) = 31 m s 40.0 10 -3 kg (b) The air in front of the projectile would exert a retarding force of Fr = Pair A = 1.0 10 5 Pa 1.0 cm 2 1 m 2 10 4 cm 2 = 10 N on the projectile as it moves down the launch tube. The energy spent overcoming this retarding force would be Wspent = Fr s = (10 N )( 0.32 m ) = 3.2 J and the needed fraction is Wspent W = 3.2 J = 0.17 19 J ( )( )( ) 438 12.5 CHAPTER 12 In each case, the work done on the gas is given by the negative of the area under the path on the PV diagram. Along those parts of the path where volume is constant, no work is done. Note that 1 atm = 1.013 10 5 Pa and 1 Liter = 10 -3 m 3 . (a) WIAF = WIA + WAF = - PI (VA - VI ) + 0 = - 4.00 1.013 10 5 Pa ( 4.00 - 2.00 ) 10 -3 m 3 = - 810 J (b) WIF = - ( triangular area ) - ( rectangular area ) =- 1 1 ( PI - PB )(VF - VB ) - PB (VF - VB ) = - 2 ( PI + PB )(VF - VB ) 2 ( ) 1 = - ( 4.00 + 1.00 ) 1.013 10 5 Pa ( 4.00 - 2.00 ) 10 -3 m 3 2 = - 507 J (c) WIBF = WIB + WBF = 0 - PB (VF - VI ) = - 1.00 1.013 10 5 Pa ( 4.00 - 2.00 ) 10 -3 m 3 = - 203 J 12.6 ( ) ( ) The sketches for (a) and (b) are shown below: P1 P2 (a) V1 V2 P P1 P2 P V (b) V1 V2 V (c) As seen from the areas under the paths in the PV diagrams above, the higher pressure during the expansion phase of the process results in more work done by the gas in (a) than in (b). The constant pressure is P = (1.5 atm ) 1.013 10 5 Pa atm = 1.52 10 5 Pa and the work done on the gas is W = - P ( V ) . (a) V = + 4.0 m 3 and W = - P ( V ) = - 1.52 10 5 Pa + 4.0 m 3 = - 6.1 10 5 J 12.7 ( ) ( )( ) The Laws of Thermodynamics 439 (b) V = - 3.0 m 3 , so W = - P ( V ) = - 1.52 10 5 Pa - 3.0 m 3 = + 4.6 10 5 J ( )( ) 12.8 As the temperature increases, while pressure is held constant, the volume increases by V = Vf - Vi = nRTf P - nRTi nR ( T ) = P P where the change in absolute temperature is T = TC = 280 K . The work done on the gas is W = - P ( V ) = - nR ( T ) = - ( 0.200 mol ) ( 8.31 J mol K ) ( 280 K ) = - 465 J 12.9 (a) From the ideal gas law, nR = PVf Tf = PVi Ti . With pressure constant this gives Vf Tf = Ti = ( 273 K )( 4 ) = 1.09 10 3 K Vi (b) The work done on the gas is W = - P ( V ) = - PVf - PVi = - nR Tf - Ti ( ) ( ) = - ( 1.00 mol ) ( 8.31 J mol K )( 1 092 J - 273 K ) = -6.81 10 3 J = - 6.81 kJ 12.10 (a) The work done on the fluid is the negative of the area under the curve on the PV diagram. Thus, Wif = - 6.00 10 6 Pa ( 2.00 - 1.00 ) m 3 {( ) + 1 ( 6.00 - 2.00 ) 10 6 Pa ( 2.00 - 1.00 ) m 3 2 + 2.00 10 6 Pa ( 4.00 - 2.00 ) m 3 ( ) } Wif = -1.20 107 J = - 12.0 MJ 440 CHAPTER 12 (b) When the system follows the process curve in the reverse direction, the work done on the fluid is the negative of that computed in (a), or W fi = -Wif = + 12.0 MJ 12.11 (a) Because the volume is held constant, W = 0 . Energy is transferred by heat from the burning mixture, so Q < 0 . The first law then gives U = Q + W = Q , so U < 0 . (b) Again, since volume is constant, W = 0 . Energy is transferred by heat from the burning mixture to the water, so Q > 0 . Then, U = Q + W = Q gives U > 0 . 12.12 The work done on the gas is the negative of the area under the curve on the PV diagram, or 1 3 W = - P0 ( 2V0 - V0 ) + ( 2P0 - P0 )( 2V0 - V0 ) = - P0V0 2 2 From the result of Problem 1, U = 3 3 3 3 9 Pf Vf - PVi = ( 2P0 )( 2V0 ) - P0V0 = P0V0 i 2 2 2 2 2 9 3 P0V0 - - P0V0 = 6 P0V0 2 2 Thus, from the first law, Q = U - W = 12.13 1.013 10 5 Pa 10 -3 m 3 (a) W = - P ( V ) = - ( 0.800 atm )( -7.00 L ) 1 L = 567 J 1 atm (b) U = Q + W = -400 J+567 J = 167 J The Laws of Thermodynamics 441 12.14 The work done on the gas is the negative of the area under the curve on the PV diagram, so 1 3 W = - P0 (V0 - 2V0 ) + ( 2P0 - P0 )(V0 - 2V0 ) = + P0V0 , or W > 0 2 2 From the result of Problem 1, U = 3 3 3 3 Pf Vf - PVi = ( 2P0 )(V0 ) - ( P0 )( 2V0 ) = 0 i 2 2 2 2 3 3 P0V0 = - P0V0 , or Q < 0 2 2 Then, from the first law, Q = U - W = 0 - 12.15 (a) Along the direct path IF, the work done on the gas is W = - ( area under curve ) 1 = - (1.00 atm ) ( 4.00 L - 2.00 L ) + ( 4.00 atm - 1.00 atm ) ( 4.00 L - 2.00 L ) 2 1.013 10 5 Pa 10 -3 m 3 W = - ( 5.00 atm L ) 1 L = -506.5 J 1 atm Thus, U = Q + W = 418 J - 506.5 J= - 88.5 J (b) Along path IAF, the work done on the gas is 1.013 10 5 Pa 10 -3 m 3 W = - ( 4.00 atm ) ( 4.00 L - 2.00 L ) 1 L = -810 J 1 atm From the first law, Q = U - W = - 88.5 J - ( - 810 J ) = 722 J 442 12.16 CHAPTER 12 (a) In a cyclic process, U = 0 and the first law gives U = Q + W = 0 , or Q = -W The total work done on the gas is WABC = WAB + WBC + WCA , and on each step the work is the negative of the area under the curve on the PV diagram, or WAB = - ( 2.0 kPa ) ( 10 m 3 - 6.0 m 3 ) 1 + ( 8.00 kPa - 2.0 kPa ) ( 10 m 3 - 6.0 m 3 ) = -20 kJ 2 WBC = 0 , and WcA - ( 2.0 kPa ) 6.0 m 3 - 10 m 3 = + 8.0 kJ ( ) Thus, WABC = -20 kJ + 0 + 8.0 kJ = -12 kJ , and Q = -W = 12 kJ (b) If the cycle is reversed, WCBA = -WABC = - ( -12 kJ ) = 12 kJ and Q = -W = - 12 kJ 12.17 (a) The change in the volume occupied by the gas is V = Vf - Vi = A L f - Li = ( 0.150 m 2 ) ( -0.200 m ) = -3.00 10 -2 m 3 ( ) and the work done by the gas is Wby gas = + P ( V ) = ( 6 000 Pa ) -3.00 10 -2 m 3 = - 180 J ( ) (b) The first law of thermodynamics is U = Qinput + Won gas = -Qoutput - Wby gas . Thus, if U = -8.00 J , the energy transferred out of the system by heat is Qoutput = - U - Wby gas = - ( -8.00 J ) - ( -180 J ) = + 188 J 12.18 Volume is constant in process BC, so WBC = 0 . Given that QBC < 0 , the first law shows that U BC = QBC + WBC = QBC + 0 . Thus, U BC < 0 . For process CA, VCA = VA - VC < 0 , so W = - P ( V ) shows that WCA > 0 . Then, given that UCA < 0 , the first law gives QCA = UCA - WCA and QCA < 0 . The Laws of Thermodynamics 443 In process AB, the work done on the system is W = - ( area under curve AB) where ( area under curve AB) = PA (VB - VA ) + ( PB - PA )(VB - VA ) > 0 Hence, WAB < 0 . For the cyclic process, U = U AB + U BC + UCA = 0 , so, U AB = - ( U BC + UCA ) . This gives U AB > 0 , since both U BC and UCA are negative. 1 2 Finally, from the first law, Q = U - W shows that QAB > 0 since both U AB and - WAB are positive. 12.19 (a) W = - P ( V ) 1 m3 = - 1.013 10 5 Pa 1.09 cm 3 - 1.00 cm 3 6 = - 9.12 10 -3 J 3 10 cm ( )( ) (b) To freeze the water, the required energy transfer by heat is Q = - mL f = - 1.00 10 -3 kg 3.33 10 5 J kg = -333 J The first law then gives U = Q + W = -333 J - 9.12 10 -3 J = - 333 J ( )( ) 12.20 Treating the air as an ideal gas at constant pressure, the final volume is Vf = Vi Tf Ti , or the change in volume is Tf - Ti V = Vf - Vi = Vi Ti 10 -3 m 3 310 K - 273 K 3 -5 = ( 0.600 L ) = 8.13 10 m 273 K 1 L (a) The work done on the lungs by the air is Wby gas = + P ( V ) = 1.013 10 5 Pa 8.13 10 -5 m 3 = 8.24 J (b) Using the result of Problem 1, the change in the internal energy of the air is U = 3 3 P ( V ) = 1.013 10 5 Pa 8.13 10 -5 m 3 = 12.4 J 2 2 ( ) ( )( ) ( )( ) 444 CHAPTER 12 (c) The energy added to the air by heat is Q = U - W = 3 5 P ( V ) - - P ( V ) = P ( V ) 2 2 or, Q = 5 1.013 10 5 Pa 8.13 10 -5 m 3 = 20.6 J 2 ( )( ) 12.21 (a) The original volume of the aluminum is V0 = m = 5.0 kg = 1.85 10 -3 m 3 2.70 10 3 kg m 3 and the change in volume is V = V0 ( T ) = ( 3 ) V0 ( T ) , or -1 V = 3 24 10 -6 ( C) 1.85 10 -3 ( 70C ) = 9.3 10 -6 m 3 ( ) The work done by the aluminum is then Wby system = + P ( V ) = 1.013 10 5 Pa 9.3 10 -6 m 3 = 0.95 J (b) The energy transferred by heat to the aluminum is Q = mcAl ( T ) = ( 5.0 kg )( 900 J kg C ) ( 70C ) = 3.2 10 5 J (c) The work done on the aluminum is W = -Wby system = -0.95 J , so the first law gives U = Q + W = 3.2 10 5 J - 0.95 J= 3.2 10 5 J ( )( ) 12.22 (a) The work done on the gas in each process is the negative of the area under the process curve on the PV diagram. For path IAF, WIAF = WIA + WAF = 0 + WAF , or 10 -3 m 3 Pa WIAF = - (1.50 atm ) 1.013 10 5 ( 0.500 L ) atm 1 L = - 76.0 J The Laws of Thermodynamics 445 For path IBF, WIBF = WIB + WBF = WIB + 0 , or 10 -3 m 3 Pa WIBF = - ( 2.00 atm ) 1.013 10 5 ( 0.500 L ) atm 1 L = - 101 J For path IF, WIF = WAF - ( triangular area) , or WIF = -76.0 J - = - 88.7 J 10 -3 m 3 1 Pa 0.500 atm ) 1.013 10 5 0.500 L ) ( ( 2 atm 1 L (b) Using the first law, with U = U F - U A = (180 - 91.0 ) J = 89.0 J , for each process gives QIAF = U - WIAF = 89.0 J - ( -76.0 J ) = 165 J QIBF = U - WIBF = 89.0 J - ( -101 J ) = 190 J QIF = U - WIF = 89.0 J - ( -88.7 J ) = 178 J 12.23 The maximum efficiency possible is that of a Carnot engine operating between the given reservoirs. ec = Th - Tc T 293 K = 1- c = 1- = 0.489 ( or 48.9% ) Th Th 573 K 12.24 The maximum possible efficiency of an engine is the Carnot efficiency, T - Tc T ec = h = 1 - c , where Th and Tc are the absolute temperatures of the reservoirs the Th Th engine operates between. For the given engine, the temperatures of the reservoirs are 300F = 149C = 422 K and 150F = 65.6C = 339 K so the maximum efficiency is ec == 1 - 339 K = 0.197 or 19.7% 422 K 446 CHAPTER 12 12.25 (a) e Weng Qh Weng Qh = Weng 3Weng Qc Qh = 1 = 0.333 or 33.3% 3 Qc Qh = 1- e = 1- 1 2 = 3 3 (b) e = 1- , so 12.26 (a) e From Weng Qh Qc = 1- Qc Qh , the energy intake each cycle is Qh = 1- e = 8 000 J = 10 667 J = 10.7 kJ 1 - 0.250 = e Qc t , the time for one cycle is (b) From = Weng t = t= e Qc ( 0.250) (10 667 J ) 5.00 10 3 W = 0.533 s 12.27 (a) The maximum efficiency possible is that of a Carnot engine operating between the specified reservoirs. ec = Th - Tc T 703 K = 1- c = 1- = 0.672 2 143 K Th Th Weng Qh = ( or 67.2% ) (b) From e = , we find Weng = e Qh = 0.420 1.40 10 5 J = 5.88 10 4 J 5.88 10 4 J = 5.88 10 4 W = 58.8 kW 1.00 s ( ) so = Weng t The Laws of Thermodynamics 447 12.28 The work done by the engine equals the change in the kinetic energy of the bullet, or Weng = 1 1 2 mb v 2 - 0 = 2.40 10 -3 kg ( 320 m s ) = 123 J f 2 2 Weng Qh = Weng Weng + Qc where Qc is ( ) Since the efficiency of an engine may be written as e = the exhaust energy from the engine, we find that 1 1 - 1 = 1.10 10 4 J Qc = Weng - 1 = (123 J ) e 0.0110 This exhaust energy is absorbed by the 1.80-kg iron body of the gun, so the rise in temperature is T = Weng Qh Qc mgun ciron = 1- Qc Qh = 1.10 10 4 J = 13.7 C (1.80 kg )( 448 J kg C) 1 200 J = 0.294 1 700 J 12.29 (a) e = 1- ( or 29.4% ) (b) Weng = Qh - Qc = 1 700 J - 1 200 J = 500 J (c) = Weng t = 500 J = 1.67 10 3 W = 1.67 kW 0.300 s 12.30 (a) The Carnot efficiency represents the maximum possible efficiency. With Th = 20.0C = 293 K and Tc = 5.00C = 278 K , this efficiency is given by ec = 1 - Tc 278 K = 1- = 0.0512 ( or 5.12% ) Th 293 K (b) The efficiency of an engine is e = Weng Qh , so the minimum energy input by heat each hour is Qh = Weng emax 6 Pt ( 75.0 10 J s ) ( 3 600 s ) = = = 5.27 1012 J emax 0.0512 min (c) As fossil-fuel prices rise, this could be an attractive way to use solar energy. However, the potential environmental impact of such an engine would require serious study. The energy output, Qc = Qh - Weng = Qh (1 - e ) , to the low temperature reservoir (cool water deep in the ocean) could raise the temperature of over a million cubic meters of water by 1 C every hour. 448 CHAPTER 12 12.31 The actual efficiency of the engine is e = 1 - Qc Qh = 1- 300 J = 0.400 500 J If this is 60.0% of the Carnot efficiency, then ec = Tc , we find Th e 0.400 2 = = 0.600 0.600 3 Thus, from ec = 1 - Tc 2 1 = 1 - ec = 1 - = Th 3 3 12.32 (a) The Carnot efficiency is ec = 1 - output is max = Tc 353 K = 1- = 0.433 , so the maximum power Th 623 K (W ) eng max t Qc Qh = e c Qh t = 0.433 ( 21.0 kJ ) = 9.10 kW 1.00 s (b) From e = 1 - , the energy expelled by heat each cycle is Qc = Qh (1 - e ) = ( 21.0 kJ )(1 - 0.433) = 11.9 kJ 12.33 From = Weng t = e Qh t , the energy input by heat in time t is Qh = t e Thus, from e = Qh - Qc Qh , the energy expelled in time t is t 1 Qc = Qh (1 - e ) = (1 - e ) = t - 1 e e The Laws of Thermodynamics 449 In time t, the mass of cooling water used is m = 1.0 106 kg s t , and its rise in temperature is ( ) T = Qc mc = 1 -1 (1.0 10 kg s ) t c e 6 6 t (1 000 10 J s ) 1 = - 1 (1.0 10 kg s ) ( 4 186 J kg C ) 0.33 6 or T = 0.49C 12.34 The actual efficiency of the engine is e actual = (W ) eng actual Qh = 1 2 2 mtrain vactual Qh while the theoretically possible efficiency (the Carnot efficiency) is ec = (W ) eng theoretical Qh = 1 2 2 mtrain vtheoretical Qh The energy input from the high temperature reservoir is the same in the two cases since it is specified that the same amount of fuel is consumed in both cases. Thus, we find ec e actual (W ) = (W ) eng eng theoretical actual 6.50 m s v v = theoretical or ec = e actual theoretical = ( 0.200 ) = 0.338 vactual vactual 5.00 m s Tc T 300 K , giving Th = c = = 453 K 1 - ec 1 - 0.338 Th 2 2 2 But, we also know that ec = 1 - 12.35 The energy transferred from the water by heat, and absorbed by the freezer, is J 5 Q = mL f = ( V ) L f = 10 3 kg m 3 1.0 10 -3 m 3 3.33 10 5 = 3.3 10 J kg ( )( ) Thus, the change in entropy of the water is 450 CHAPTER 12 (a) Swater = ( Qr )water T = - 3.3 10 5 J J = - 1.2 10 3 = - 1.2 kJ K 273 K K and that of the freezer is (b) S freezer = ( Qr ) freezer T = + 3.3 10 5 J = + 1.2 kJ K 273 K 12.36 The energy added to the water by heat is Qr = mLv = (1.00 kg ) 2.26 10 6 J kg = 2.26 106 J so the change in entropy is S = Qr 2.26 10 6 J J = = 6.06 10 3 = 6.06 kJ K 373 K K T ( ) 12.37 The potential energy lost by the log is transferred away by heat, so Q = mgh = ( 70 kg ) 9.80 m s 2 ( 25 m ) = 1.7 10 4 J and the change in entropy is S = Qr 1.7 10 4 J = = 57 J K 300 K T ( ) 12.38 The total momentum before collision is zero, so the combined mass must be at rest after the collision. The energy dissipated by heat equals the total initial kinetic energy, of 2 1 Q = 2 mv 2 = ( 2 000 kg )( 20 m s ) = 8.0 10 5 J = 800 kJ 2 The change in entropy is then S = Qr 800 kJ = = 2.7 kJ K T 296 K The Laws of Thermodynamics 451 12.39 A quantity of energy, of magnitude Q, is transferred from the Sun and added to Earth. -Q +Q Thus, SSun = , so the total change in entropy is and SEarth = TSun TEarth Stotal = SEarth + SSun = Q Q - TEarth TSun 1 1 = ( 1 000 J ) - = 3.27 J K 290 K 5 700 K 12.40 (a) End Result All R 1G, 2R 2G, 1R All G Possible Draws RRR RRG, RGR, GRR GGR, GRG, RGG GGG Total Number of Same Result 1 3 3 1 (b) End Result All R 1G, 4R 2G, 3R 3G, 2R 4G, 1R All G Possible Draws RRRRR RRRRG, RRRGR, RRGRR, RGRRR, GRRRR RRRGG, RRGRG, RGRRG, GRRRG, RRGGR, RGRGR, GRRGR, RGGRR, GRGRR, GGRRR GGGRR, GGRGR, GRGGR, RGGGR, GGRRG, GRGRG, RGGRG, GRRGG, RGRGG, RRGGG GGGGR, GGGRG, GGRGG, GRGGG, RGGGG GGGGG Total Number of Same Result 1 5 10 10 5 1 452 12.41 CHAPTER 12 (a) The table is shown below. On the basis of the table, the most probable result of a toss is 2 H and 2 T . Total Number of Same Result 1 4 6 4 1 End Result All H 1T, 3H 2T, 2H 3T, 1H All T Possible Tosses HHHH HHHT, HHTH, HTHH, THHH HHTT, HTHT, THHT, HTTH, THTH, TTHH TTTH, TTHT, THTT, HTTT TTTT (b) The most ordered state is the least likely. This is seen to be all H or all T . (c) The least ordered state is the most likely. This is seen to be 2H and 2T . 12.42 (a) There is only one ace of spades out of 52 cards, so the probability is 4 1 = 52 13 13 1 = 52 4 1 52 (b) There are four aces out of 52 cards, so the probability is (c) There are 13 spades out of 52 cards, so the probability is 12.43 The maximum efficiency is that of a Carnot engine and is given by ec = 1 - Tc 100 K = 1- = 0.50 , or emax = 50% . The claim is invalid 200 K Th The Laws of Thermodynamics 453 12.44 Operating between reservoirs having temperatures of Th = 100C = 373 K and Tc = 20C = 293 K , the theoretical efficiency of a Carnot engine is ec = 1 - Tc 293 K = 1- = 0.21 373 K Th If the temperature of the hotter reservoir is changed to Th = 550C = 823 K , the theoretical efficiency of the Carnot engine increases to ec = 1 - Tc 293 K = 1- = 0.64 823 K Th The factor by which the efficiency has increased is ec 0.64 = = 3.0 ec 0.21 12.45 (a) The entropy change of the hot reservoir, with an energy output of magnitude Qh , is Sh = - Qh Qr = Th Th (b) For the cold reservoir, with an energy input of magnitude Qc , the change in entropy is Sc = + Qc Qr = Tc Tc (c) The engine has an energy input of magnitude Qh from a reservoir at temperature Th and an energy output of magnitude Qc to a reservoir at temperature Tc . The net change in entropy for the engine is Qh Q Qr Seng = - c = T Th Tc (d) For the isolated system consisting of the engine and the two reservoirs, the change in entropy is Sisolated = Sh + Seng + Sc = - system Qh Q Q Q + h - c + c = 0 Th Th Tc Tc 454 12.46 CHAPTER 12 In this case, Qh = Qc = 8 000 J . The change in entropy of the hot reservoir is Sh = - Qh Th = - 8 000 J = - 16.0 J K 500 K + Qc Tc = 8 000 J = 26.7 J K 300 K For the cold reservoir, Sc = The net entropy change for this irreversible process is Suniverse = Sh + Sc = ( -16.0+26.7 ) J K = 10.7 J K > 0 12.47 The energy output to the river each minute has magnitude Q J Qc = (1 - e ) Qh = (1 - e ) h t = (1 - 0.30 ) 25 10 8 ( 60 s ) = 1.05 1011 J s t so the rise in temperature of the 9.0 10 6 kg of cooling water used in one minute is T = Qc mc = 1.05 1011 J = 2.8 C ( 9.0 106 kg ) ( 4 186 J kg C ) 12.48 The energy exhausted from a heat engine is Qc = Qh - Weng = Weng e 1 - Weng = Weng - 1 e where Qh is the energy input from the high temperature reservoir, Weng is the useful work done, and e = Weng Qh is the efficiency of the engine. For a Carnot engine, the efficiency is ec = 1 - Tc Th = ( Th - Tc ) Th Th Tc Qc = Weng - 1 = Weng Th - Tc Th - Tc so we now have The Laws of Thermodynamics 455 Thus, if Th = 100 C = 373 K and Tc = 20 C=293 K , the energy exhausted when the engine has done 5.0 10 4 J of work is 293 K Qc = 5.0 10 4 J = 1.83 10 5 J 373 K - 293 K ( ) The mass of ice (at 0C) this exhaust energy could melt is m= Qc L f , water = 1.83 10 5 J = 0.55 kg 3.33 10 5 J kg 12.49 The work output from the engine in an interval of one second is Weng = 1 500 kJ . Since the efficiency of an engine may be expressed as e= Weng Qh = Weng Weng + Qc 1 1 the exhaust energy each second is Qc = Weng - 1 = (1 500 kJ ) - 1 = 4.5 10 3 kJ e 0.25 The mass of water flowing through the cooling coils each second is m = V = 10 3 kg m 3 ( 60 L ) 10 -3 m 3 1 L = 60 kg so the rise in the temperature of the water is T = Qc mcwater = 4.5 10 6 J = 18C ( 60 kg )( 4 186 J kg C) ( ) ( ) 12.50 (a) From the first law, U1 3 = Q123 + W123 = + 418 J + ( -167 J ) = 251 J (b) The difference in internal energy between states 1 and 3 is independent of the path used to get from state 1 to state 3. Thus, and U1 3 = Q143 + W143 = 251 J , Q143 = 251 J - W143 = 251 J - ( - 63.0 J ) = 314 J 456 CHAPTER 12 (c) W12341 = W123 + W341 = W123 + ( -W143 ) = -167 J - ( -63.0 J ) = -104 J or 104 J of work is done by the gas in the cyclic process 12341. (d) W14321 = W143 + W321 = W143 + ( -W123 ) = -63.0 J - ( -167 J ) = +104 J or 104 J of work is done on the gas in the cyclic process 14321. (e) The change in internal energy is zero for both parts (c) and (d) since both are cyclic processes. 12.51 (a) The work done by the system in process AB equals the area under this curve on the PV diagram. Thus, Wby system = ( triangular area ) + ( rectangular area ) , or Wby 1 = ( 4.00 atm )( 40.0 L ) 2 Pa 10 -3 m 3 + (1.00 atm )( 40.0 L ) 1.013 10 5 atm L = 1.22 10 4 J = 12.2 kJ Note that the work done on the system is WAB = -Wby = -12.2 kJ for this process. system system (b) The work done on the system (that is, the work input) for process BC is the negative of the area under the curve on the PV diagram, or Pa 10 -3 m 3 WBC = - (1.00 atm )(10.0 L - 50.0 L ) 1.013 10 5 atm 1 L = 4.05 kJ (c) The change in internal energy is zero for any full cycle, so the first law gives Qcycle = U cycle - Wcycle = 0 - ( WAB + WBC + WCA ) = 0 - ( -12.2 kJ + 4.05 kJ + 0 ) = 8.15 kJ The Laws of Thermodynamics 457 12.52 The efficiency of the plant is e = ec = 1 - Qc Qh = Qc 1- e Qc 0.600 Tc 300 K = 1- = 0.400 Th 500 K Also, e = 1 - , so Qh = = From = Weng t e Qh t 0.400 Qc = , 0.600 t the rate of energy transfer to the river by heat is Qc t = 1.50 = 1.50 ( 1 000 MW ) = 1.50 109 J s The flow rate in the river is then ( Qc t ) = 1.50 109 J s m = = 5.97 10 4 kg s t cwater ( T )river ( 4 186 J kg C )( 6.00 C ) 12.53 (a) The change in length, due to linear expansion, of the rod is -1 L = L 0 ( T ) = 11 10 -6 ( C ) ( 2.0 m ) ( 40C - 20C ) = 4.4 10 -4 m The load exerts a force F = mg = ( 6 000 kg ) ( 9.80 m s 2 ) = 5.88 10 4 N on the end of the rod in the direction of movement of that end. Thus, the work done on the rod is W = F L = 5.88 10 4 N 4.4 10 -4 m = 26 J (b) The energy added by heat is J 5 Q = mc ( T ) = (100 kg ) 448 ( 20 C) = 9.0 10 J kg C (c) From the first law, U = Q + W = 9.0 10 5 J+26 J = 9.0 10 5 J ( )( ) 12.54 (a) The work done by the gas during each full cycle equals the area enclosed by the cycle on the PV diagram. Thus Wby gas = ( 3P0 - P0 )( 3V0 - V0 ) = 4 P0V0 458 CHAPTER 12 (b) Since the work done on the gas is W = -Wby gas = -4P0V0 and U = 0 for any cyclic process, the first law gives Q = U - W = 0 - ( - 4P0V0 ) = 4P0V0 (c) From the ideal gas law, P0V0 = nRT0 , so the work done by the gas each cycle is J Wby gas = 4 nRT0 = 4 ( 1.00 mol ) 8.31 ( 273 K ) mol K = 9.07 10 3 J = 9.07 kJ 12.55 (a) The energy transferred to the gas by heat is J Q = mc ( T ) = (1.00 mol ) 20.79 (120 K ) mol K = 2.49 10 3 J = 2.49 kJ (b) Treating the neon as an ideal gas, the result of Problem 1 gives the change in internal energy as U = 3 3 3 Pf Vf - PVi = nRTf - nRTi = nR ( T ) , or i 2 2 2 3 J 3 (1.00 mol ) 8.31 (120 K ) = 1.50 10 J = 1.50 kJ 2 mol K ( ) ( ) U = (c) From the first law, the work done on the gas is W = U - Q = 1.50 10 3 J - 2.49 10 3 J = - 990 J The Laws of Thermodynamics 459 12.56 (a) The change in volume of the aluminum is V = V0 ( T ) = ( 3 ) ( m ) ( T ) , or 1.0 kg -1 -7 3 V = 3 24 10 -6 ( C) 3 3( 2.70 10 kg m 18 C ) = 4.8 10 m so the work done on the aluminum is W = - P ( V ) = - 1.013 10 5 Pa 4.8 10 -7 m 3 = - 4.9 10 -2 J (b) The energy added by heat is Q = mc ( T ) = (1.0 kg )( 900 J kg C)(18 C ) = 1.6 10 4 J = 16 kJ (c) The first law gives the change in internal energy as U = Q + W = 1.6 10 4 J - 4.9 10 -2 J=1.6 10 4 J = 16 kJ ( )( ) 12.57 (a) The energy input by heat from the molten aluminum is Qh = mAl L f = 1.00 10 -3 kg 3.97 10 5 J kg = 397 J and the energy output to the frozen mercury is Qc = mHg L f = 15.0 10 -3 kg 1.18 10 4 J kg = 177 J The efficiency of the heat engine is given by e= Weng Qh = Qh - Qc Qh = 1- Qc Qh = 1- 177 J = 0.554 or 55.4% 397 J ( )( ) ( )( ) (b) Th = 660 C = 933 K and Tc = -38.9 C = 234 K . The Carnot efficiency for a heat engine operating between these two reservoirs is ec = 1 - Tc 234 K = 1- = 0.749 or 74.9% Th 933 K 460 12.58 CHAPTER 12 (a) From the result of Problem 1, U AC = 3 3 15 15 ( PCVC - PAVA ) = 2 ( 3P0 )( 2V0 ) - P0V0 = 2 P0V0 = 2 nRT0 2 The work done on the gas in process ABC equals the negative of the area under the process curve on the PV diagram, or WABC = - ( 3P0 )( 2V0 - V0 ) = - 3P0V0 = - 3nRT0 The total energy input by heat, QABC = Q1 + Q2 , is found from the first law as QABC = U AC - WABC = 15 21 nRT0 - ( -3nRT0 ) = nRT0 2 2 (b) For process CDA, the work done on the gas is the negative of the area under curve CDA, or WCDA = - ( P0 )(V0 - 2V0 ) = + P0V0 = + nRT0 . The change in internal energy is UC A = -U AC = - 15 nRT0 . Thus, the energy input by heat for this process is 2 15 17 nRT0 - nRT0 = - nRT0 2 2 QCDA = UC A - WCDA = - The total energy output by heat for the cycle is 17 17 Q3 + Q4 = - QCDA = - - nRT0 = nRT0 2 2 (c) The efficiency of a heat engine using this cycle is e = 1- Qoutput Qinput = 1- (17 nRT 2) = 1 - 17 = 21 ( 21nRT 2) 0 0 0.190 or 19.0% (d) ec = 1 - ( PAVA nR) = 1 - P0V0 = 5 = 0.833 or 83.3% TA = 1- TC ( 3P0 )( 2V0 ) 6 ( PCVC nR) 12.59 The mass of coal consumed in time t is given by M = Qh Qcoal where Qh is the required energy input and Qcoal is the heat of combustion of coal. Thus, if is the power output and e is the efficiency of the plant, M = Qh Qcoal = (W eng e Qcoal )= t e Qcoal The Laws of Thermodynamics 461 (a) The coal used each day is t M = = e Qcoal (150 10 6 J s ) ( 86 400 s d ) cal 10 3 g 4.186 J g 1 kg 1 cal ( 0.15 ) 7.8 10 3 = 2.6 10 6 kg d = 2.6 10 3 metric ton d (b) The annual fuel cost is: cost = ( coal used yearly ) ( rate ) , or t cost = ( $8.0 ton ) e Qcoal = 1 ton $8.0 6 = $7.7 10 yr 3 10 g 4.186 J 10 kg ton cal ( 0.15) 7.8 10 3 g 1 kg 1 cal 6 3 (150 10 J s 3.156 107 s yr )( ) (c) The rate of energy transfer to the river by heat is Qc t = Qh - Weng t = (W eng e - Weng t ) 1 = - 1 e Thus, the flow required is Qc t m 1 = = - 1 t cwater ( T ) cwater ( T ) e = 150 10 6 J s 1 - 1 = 4.1 10 4 kg s ( 4186 J kg C ) ( 5.0C ) 0.15 12.60 (a) The energy transfer by heat required to raise the temperature of the water to the boiling point is Q = mc ( T ) = 1.00 10 -3 kg ( 4 186 J kg C) (100C - 20.0C) = 335 J We neglect the very small volume expansion (and associated work done) by the water while in the liquid state. The first law of thermodynamics then gives the change in internal energy as U = Q + W = 335 J + 0 = 335 J ( ) 462 CHAPTER 12 (b) To completely evaporate the water, the required energy input by heat is Q = mLv = 1.00 10 -3 kg 2.26 106 J kg = 2.26 10 3 J The work done on the water in this process is 1 m3 W = - P Vf - Vi = - 1.013 10 5 Pa 1 671 cm 3 - 1.00 cm 3 6 = -169 J 10 cm 3 ( )( ) ( ) ( )( ) so the change in internal energy is U = Q + W = 2.26 10 3 J - 169 J = 2.09 10 3 J 12.61. (a) The work done on the gas during this isobaric expansion process is W = - P ( V ) = - P A ( x ) = - ( 8 000 Pa ) 0.10 m 2 + 4.0 10 -2 m = -32 J The first law of thermodynamics then gives the change in the internal energy of the system as U = Q + W = 42 J - 32 J = 10 J (b) If the piston is clamped in a fixed position, then V = 0 and the work done on the gas is zero In this case, the first law gives U = Q + W = 42 J + 0 = 42 J ( )( ) 12.62. (a) The energy transferred from the water by heat as it cools is Qh = mc T = ( V ) c T g 10 3 cm 3 cal 4.186 J = 1.0 (1.0 L ) 3 1.0 g C 1 cal ( 570C - 4.0C) cm 1 L or Qh = 2.4 106 J The Laws of Thermodynamics 463 (b) The maximum efficiency of a heat engine is the Carnot efficiency. Thus, ec = 1 - Tc ( 4.0 + 273) K = 1 - 277 K = 0.67 = 1- 843 K Th ( 570+273) K The maximum useful work output is then (W ) eng max = ec Qh = ( 0.67 ) 2.4 106 J = 1.6 10 6 J ( ) (c) The energy available from oxidation of the hydrogen sulfide in 1.0 L of this water is mol J 3 2 U = n ( 310 kJ mol ) = 0.90 10 -3 (1.0 L ) 310 10 = 2.8 10 J L mol 12.63. The work that you have done is step in 2.54 10 -2 m 4.448N Weng = mg ( h ) = (150 lb) 90.0 ( 30.0 min ) 8.00 1 lb min step 1 in or Weng = 3.66 10 5 J 4 186 J If the energy input by heat was Qh = ( 600 kcal ) = 2.51 106 J , your efficiency 1 kcal has been e= Weng Qh = 3.66 10 5 J = 0.146 or 14.6% 2.51 106 J If the actual efficiency was e = 0.180 or 18.0% , the actual energy input was Qh = Weng e actual = 1 kcal 3.66 10 5 J = 2.03 106 J = 486 kcal 0.180 4 186 J actual ( ) 464 CHAPTER 12

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CS1371 - Computing for Engineers Test 1 Version CSeptember 19th, 2007Georgia Institute of Technology College of Computing CS 1371 Computing for Engineers Test 1 Version C - Fall Semester 20070 1 2 3 4 5 6 7 8 9 A B C D E F G H I J K L M N O P Q
Georgia Tech - PSYC - 1101
Learning: Classical ConditioningSleep Circadian Rhythms Neural Basis of Sleep Stages of Sleep Theories of Sleep Sleep Disorders DreamingDreams: TheoriesWish Fulfillment Sigmund Freud proposed that people fulfill ungratified needs from wak
Georgia Tech - CHEM - 2312
CHEM 2312 Fall 2007 Exam 1 version 2Name:_ Roll Number _1. (32 points, 4 points each) Circle the correct answer. There is only one correct answer. a. Which of the following compounds would be the most stable?IIIIIIIVA) B) C) D) E)I II
Georgia Tech - CHEM - 2312
CHEM 2312 Fall 2007 Exam 3 version 2Name:_ Roll Number _1. (50 points, 5 points each) Circle the correct answer. There is only one correct answer. a. What is the major organic product obtained from the following reaction?O O 1. NaBH4 H 2. H3O+ O
Georgia Tech - PSYC - 1101
Language&quot;Language shapes the way we think and determines what we can think about.&quot; Benjamin Lee WhorfMemory III Forgetting Memory Distortions Biological Basis of MemoryWhere are Memories Stored?1Where are Memories Stored? No where in p
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CHEM 2312 Fall 2007 Exam 2 version 2 TanName:_ Roll Number _1. (32 points, 4 points each) Circle the correct answer. There is only one correct answer.a.Which of the following is the reactive intermediate formed in the electrophilic nitration
Georgia Tech - CS - 1371
CS1371Matlab API reference:Previous Exam QuestionsDecember, 2003Constants: pi, eps, inf, NaN, realmax, realmin Math: cos(), sin(), tan(), log(), log10(), log2(), exp(), sqrt(), abs(), angle(), conj(), imag(), real(), complex(re,im) Round &amp; rem
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Memory I&quot;Everything in life is memory, save for the thin edge of the present &quot;- Gazzaniga, 2000MemoryMemory is the process by which we observe, store, and recall informationMemoryThree key processes involved in memory:1. Encoding Forming
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Motivation: SleepCharacteristics of SleepBehavioral Characteristics Minimal movement Stereotyped prone posture Require a high degree of stimulation to arouse organismPhysiological Characteristics Brain wave activity Paralysis of muscles Ca
Georgia Tech - PSYC - 1101
Learning IIIGeneralization &amp; Discrimination In operant conditioning, behavior is controlled by its consequences. BUT! Stimuli preceding a response also influence the behavior CONTEXT! In operant conditioning, discrimination and generalization ar
Georgia Tech - CHEM - 2312
CHEM 2312 Fall 2007 Exam 4 version 2Name:_ Roll Number _1. (40 points, 5 points each) Circle the correct answer. There is only one correct answer. a. What is the major organic product obtained from the following sequence of reactions?Ph PhPh
Georgia Tech - CHEM - 2312
CHEM 2312 Fall 2007 Exam 4 version 1Name:_ Roll Number _1. (40 points, 5 points each) Circle the correct answer. There is only one correct answer. a. In which of the following sequences are the compounds listed in order of decreasing acidity? (mo
Georgia Tech - CHEM - 2312
CHEM 2312 practice final Version iiThe following standardized final examination covers the entire introductory year of organic chemistry (CHEM 2311 and 2312 at Georgia Tech) . The exam consists of 70 multiple choice questions, each with four choice
Georgia Tech - CHEM - 4511
Georgia Tech - CHEM - 2312
CHEM 2312 Fall 2007 Exam 1 version 1Name:_ Roll Number _1. (32 points, 4 points each) Circle the correct answer. There is only one correct answer. a.does not undergo the Diels-Alder reaction because: A) B) C) D) E) ring systems cannot function
Georgia Tech - CHEM - 4511
Georgia Tech - CHEM - 2312
CHEM 2312 Fall 2007 Exam 2 version 1 BlueName:_ Roll Number _1. (32 points, 4 points each) Circle the correct answer. There is only one correct answer.a.What is the major organic product obtained from the following reaction?1 CH3 O C Cl H3C
Georgia Tech - PSYC - 1101
Thought &amp; Knowledge IIMental RotationNormal Mirror ReversedIdentify whether the `R' is normal or mirror reversed.Mental RotationThe greater the rotation of the stimulus from 0 the longer it takes for the subject to react.1Analogical Repr
Georgia Tech - CS - 1371
CS1371Answers to Previous QuestionsDecember, 20031. Matlab Concepts and Syntax(A) For each of the following questions, write a single Matlab statement to accomplish what is requested. (i) Create a row vector containing the sequence of numbers:
Georgia Tech - PSYC - 1101
Memory II&quot;Everything in life is memory, save for the thin edge of the present &quot;- Gazzaniga, 2000The Stage Theory of MemorySensory Memory Working MemoryUnder the proper conditions, information from Working Memory can be moved into Long-Term Me
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Selected major legislation of the New Deal First New Deal Emergency Banking Act (1933)-Stabilized the nation's private banking system; provided loans to private banks. Civilian Conservation Corps Act (1933)-Provided unemployment relief through conser