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Unformatted text preview: 528 CHAPTER 19 TEMPERATURE, HEAT, AND... CHAPTER 19 Answer to Checkpoint Questions 1. 2. 3. 4. 5. 6. (a) all tie; (b) 50X, 50Y, 50W (a) 2 and 3 tie, then 1, then 4; (b) 3, 2, then 1 and 4 tie A c and e (a) zero; (b) positive b and d tie, then a, c Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 25 S , 25 O , 25 R (a) above; (b) yes (example: +5, +3, 4, 4, 6, +7) c, then the rest tie Celsius and Kelvin tie, then Fahrenheit B, then A and C tie (a) all tie; (b) 4, 3, 2, 1; (c) 4, 3, 2, 1 (a) both clockwise; (b) both clockwise (a) cycle 3; (b) cycle 2 c, a, b (a) all tie; (b) all tie upward (with liquid water on the exterior and at the bottom, T = 0 horizontally and downward) tie of A, B, and C, then D at the temperature of your ngers sphere, hemisphere, cube CHAPTER 19 TEMPERATURE, HEAT, AND... 529 15. 3, 2, 1 Solutions to Exercises & Problems The temperature to which max = 0:107 cm corresponds is T = 0:2898 cm K=0:107 cm = 2:71 K: The ratio is 373:15 K=273:16 K = 1:366.
3P 2E 1E Take p3 to be 80 kPa for both thermometers. According to Fig 19{6 the nitrogen thermometer gives 373:35 K for the boiling point of water. Use Eq. 195 to compute the pressure: 373 35 K p N = 273T K p3 = 273::16 K (80 kPa) = 109:343 kPa : :16 The hydrogen thermometer gives 373:16 K for the boiling point of water and 373 16 K p H = 273::16 K (80 kPa) = 109:287 kPa : The pressure in the nitrogen thermometer is higher than the pressure in the hydrogen thermometer by 0:056 kPa.
4P Let TL be the temperature and pL be the pressure in the lefthand thermometer. Let TR be the temperature and pR be the pressure in the righthand thermometer. According to the problem statement the pressure is the same in the two thermometers when they are both at the triple point of water. Take this pressure to be p3 . Write Eq. 19{5 for each thermometer: TL = (273:16 K)(pL =p3 ) and TR = (273:16 K)(pR =p3 ). Subtract the second equation from the rst to obtain TL TR = (273:16 K) pL p pR : 3 530 CHAPTER 19 TEMPERATURE, HEAT, AND... First take TL = 373:125 K (the boiling point of water) and TR = 273:16 K (the triple point of water). Then pL pR = 120 mm of mercury. Solve 120 mm of mercury 373:125 K 273:16 K = (273:16 K) for p3 . The result is p3 = 328 mm of mercury. Now let TL = 273:16 K (the triple point of water) and let TR be the unknown temperature. The pressure dierence is pL pR = 90:0 mm of mercury. Solve 90:0 mm of mercury 273:16 K TR = (273:16 K) 328 mm of mercury for TR . The result is TR = 348 K. Let the reading of the temperature be TF in Fahrenheit and TC in Celsius. Then 9 F (a) TF = 5 TC + 32 = 9 T2 + 32; or TF = 320 F: 5 9 (b) TF = 5 TC + 32 = 9 (2TF ) + 32; or TF = 12:3 F: 5 The doctor is probably using the Kelvin scale, in which case your temperature would be T = 310 K = (310 273) C = 37:0 C = 98:6 F. You should have nothing to worry about! (a) The temperature on Fahrenheit scale is 9 TF = 5 TC + 32 = 9 ( 71) + 32 = 96 F : 5 (b) The temperature on Celsius scale is 5 TC = 9 TF 17:8 = 5 (134) 17:8 = 56:6 C : 9 Use TC = TK 273 = (5=9)[TF 32]. You can easily make these conversions. The results are: (a) T = 10; 000 F; (b) T = 37:0 C; (c) T = 57 C; (d) T = 297 F; and (e) 27 C = 80 F (for example). (a) Fahrenheit and Celsius temperatures are related by TF = (9=5)TC + 32 . TF is numerically equal to TC if TF = (9=5)TF + 32 . The solution to this equation is TF = (5=4)(32 ) = 40 F.
9E 8E 7E 6E 5E p3 CHAPTER 19 TEMPERATURE, HEAT, AND... 531 (b) Fahrenheit and Kelvin temperatures are related by TF = (9=5)TC + 32 = (9=5)(T 273:15) + 32 . The Fahrenheit temperature TF is numerically equal to the Kelvin temperature T if TF = (9=5)(TF 273:15) + 32 . The solution to this equation is 5 9 TF = 4 5 (273:15) 32 = 575 F : (c) Since TC = T numerical value.
10P 273:15 the Kelvin and Celsius temperatures can never have the same Let the temperature be T X on the X scale. Then from the linearity of both scales we have T X ( 170 X) = 340 K 273 K : 53:5 X ( 170 X) 373 K 273 K
Solve for TX : TX = 91:9 X. (a) Changes in temperature take place by means of radiation, conduction, and convection. A can be reduced by placing the object in isolation, by surrounding it with a vacuum jacket, for example. This reduces conduction and convection. Absorption of radiation can be reduced by polishing the surface to a mirror nish. Clearly A depends on the condition of the surface and on the ability of the environment to conduct or convect energy to or from the object. A has the dimensions of reciprocal time. (b) Rearrange the equation to obtain 1 dT = A : T dt Now integrate with respect to time and recognize that
Z
11P 1 d T dt = Z 1 d(T ) : T dt T
Z Thus 1 d(T ) = Z t A dt : T0 T 0 The integral on the right side yields At and the integral on the left yields ln T jT0 = T ln(T ) ln(T0 ) = ln(T=T0 ), so T ln T = At : 0 T 532 CHAPTER 19 TEMPERATURE, HEAT, AND... Use each side as the exponent of e, the base of the natural logarithms, to obtain T = e At T0 or T = T0 e At : From T = T0 e At , we have T=T0 = e A1 t1 (before insulation) and T=T0 = e A2 t2 (after insulation). Thus the ratio is given by A2 =A1 = t1 =t2 = 1=2. The change in length of the rod is L = LT = (20 cm)(11 10 6 =C )(50 C 30 C) = 4:4 10 3 cm :
14E 13E 12P The change in length for the aluminum pole is ` = `0 Al T = (33 m)(23 10 6 =C )(15 C) = 0:011 m :
15E When the temperature changes from T to T + T the diameter of the mirror changes from D to D + D, where D = D T . Here is the coecient of linear expansion for Pyrex glass (3:2 10 6 =C , according to Table 192). The range of values for the diameters can be found by setting T equal to the temperature range. Thus D = (3:2 10 6 =C )(200 in.)(60 C) = 3:8 10 2 in.. The new diameter is
16E R = R0 (1 + Al T ) = (2:725 cm)[1 + (23 10 6 =C )(100:0 C 0:000 C)] = 2:731 cm :
17E (a) The coecient of linear expansion for the alloy is = L=LT = (10:015 cm 10:000 cm)=[(10:01 cm)(100 C 20:000 C)] = 1:8810 5 =C . Thus from T = 100 C to T = CHAPTER 19 TEMPERATURE, HEAT, AND... 533 0 C we have L = LT = (10:015 cm)(1:88 10 5 =C )( 100 C) = 1:88 10 2 cm: The length at T = 0 C is therefore L0 = L + L = 10:015 cm 0:0188 cm = 9:996 cm: (b) Let the temperature be Tx . Then from 20 C to Tx we have L = 10:009 cm 10:000 cm = LT = (1:88 10 5 =C )(10:000 cm)T; giving T = 48 C. Thus Tx = 20 C + T = 20 C + 48 C = 68 C: (a) Since TC = (5=9)TF ,
18E Al = 23 10 6 =C = 23 10 6 =[(9=5) F] = 13 10 6 =F :
(b) The change in length of the rod is L = LT = (20 ft)(12 in./ft)(13 10 6 =F )(95 F 40 F) = 0:17 in. : The radius of the Earth has increased by approximately Re = T = (6:4 103 km)(3:0 10 5 = K)(3000 K 300 K) = 1:7 103 km : 19E The diameter in the dry ice is 20E D0 = D(1 + T ) = (80:00 mm)[1 + (19 10 6 =C )( 57:00 C 43:00 C)] = 79:85 mm :
21E If the window is L1 high and L2 wide at the lower temperature and L1 + L1 high and L2 + L2 wide at the higher temperature then its area changes from A1 = L1 L2 to A2 = (L1 + L1 )(L2 + L2 ) L1 L2 + L1 L2 + L2 L1 , where the term L1 L2 has been omitted because it is much smaller than the other terms if the changes in the lengths are small. Thus the change is area is A = A2 A1 = L1 L2 + L2 L1 . If T is the change in temperature then L1 = L1 T and L2 = L2 T , where is the coecient of linear expansion. Thus A = (L1 L2 + L1 L2 ) T = 2L1 L2 T = 2(9 10 6 =C )(30 cm)(20 cm)(30 C) = 0:32 cm2 : 534 CHAPTER 19 TEMPERATURE, HEAT, AND... The increase in the surface area of the brass cube whose side length is L before heating is A = 6(L + L)2 6L2 12LL = 12b L2 T = 12(19 10 6 =C )(30 cm)2 (75 C 20 C) = 11 cm2 :
23E 22E Since a volume is the product of three lengths, the change in volume due to a temperature change T is given by V = 3V T , where V is the original volume and is the coecient of linear expansion. The proof is similar to the proof given in the solution to 21E. Since V = (4=3)R3 , where R is the original radius of the sphere, 4 R3 T = (23 10 6 =C )(4)(10 cm)3 (100 C) = 29 cm3 : V = 3 3
24E The volume at 30 C is given by V 0 = V (1 + T ) = V (1 + 3T ) = (50 cm3 )[1 + 3(29 10 6 =C )(30 C 60 C)] = 49:87 cm3 ; where we have used = 3. The increase in volume is V = s V T = 3s V T = 3(23 10 6 =C )(5:00 cm)3 (60:0 C 10:0 C) = 0:43 cm2 : If Vc is the original volume of the cup, a is the coecient of linear expansion of aluminum, and T is the temperature increase, then the change in the volume of the cup is Vc = 3a Vc T . If is the coecient of volume expansion for glycerin then the change in the volume of glycerin is Vg = Vc T . Note that the original volume of glycerin is the same as the original volume of the cup. The volume of glycerin that spills is Vg Vc = ( 3a )Vc T = (5:1 10 4 =C ) 3(23 10 6 =C ) (100 cm3 )(6 C) = 0:26 cm3 :
26E 25E CHAPTER 19 TEMPERATURE, HEAT, AND... 535 Use L = LT and jF=Aj = E L=L. Here the minus sign indicates that L < 0. The ultimate strength for steel is (F=A)rupture = 400 106 N/m2 . Thus the rod will rupture if the temperature change exceeds T = E F A 1 27E 400 106 N/m2 = (200 109 N/m2 )(11 10 6 =C ) = 182 C : rupture The temperature at which the rod will rupture is therefore T = 25:0 C + ( 182 C) = 157 C. The change in length for the section of the steel ruler between its 20:05 cm mark and 20:11 cm mark is Ls = Ls s T = (20:11 cm)(11 10 6 =C )(270 C 20 C) = 0:055 cm : Thus the actual change in length for the rod is L = (20:11 cm 20:05 cm) + 0:055 cm = 0:115 cm. The coecient of thermal expansion for the material of which the rod is made of is then L 0:115 = T = 270 C cm C = 23 10 6 =C : 20
29P 28E After the change in temperature the diameter of the steel rod is Ds = Ds0 + s Ds0 T and the diameter of the brass ring is Db = Db0 + b Db0 T , where Ds0 and Db0 are the original diameters, s and b are the coecients of linear expansion, and T is the change in temperature. The rod just ts through the ring if Ds = Db . This means Ds0 + s Ds0 T = Db0 + b Db0 T . Solve for T : D D0 T = Ds0 bD b b0 s s0 :000 cm 2:992 = (19 10 6 =C )(23992 cm) (11 cm 6 =C )(3:000 cm) = 335 C : : 10
The temperature is T = 25 C + 335 C = 360 C. The change in area for the plate is A = (a + a)(b + b) ab ab + ba = 2abT = 2AT :
30P 536 CHAPTER 19 TEMPERATURE, HEAT, AND... For small we have 31P d d dT T = dT m T = V m V2 dV T = T ; dT where in the last step we used dV=dT = V and m=V = . (a) Use = m=V and = (m=V ) = m(1=V ) ' mV=V 2 = (V=V ) = 3(L=L): The percent change in density is = 3 L = 3(0:23%) = 0:69%: L (b) Since = L=(LT ) = 0:23 10 2 =(100 C 0:0 C) = 23 10 6 =C ; the metal is aluminum (see Table 192). The change in volume of the liquid is given by V = V T . If A is the crosssectional area of the tube and h is the height of the liquid, then V = Ah is the original volume and V = A h is the change in volume. Since the tube does not change the crosssectional area of the liquid remains the same. Therefore A h = Ah T or h = h T . (a) The area increases by 2(0:18%) = 0:36%. (b) The thickness increases by 0:18%: (c) The volume increases by 3(0:18%) = 0:54%: (d) The mass is unchanged (i.e., it increases by 0:00%): (e) The coecient of linear expansion is L = LT = 0:18 10 100 C
35P 34P 33P 32P 2 = 18 10 6 =C : Use = 2 L=g for the period of the pendulum. The change in due to temperature change T is then 2 = 2 L ' p ddLL g g s p p dL T = L T = T : dT g 2 s CHAPTER 19 TEMPERATURE, HEAT, AND... 537 The percent change in period is then = = T=2. Thus at 0:0 C, the clock will be faster than normal by (3600 s)(19 10 6 =C )(20 C)=2 = 0:68 s each hour. Consider half the bar. Its original length is `0 = L0 =2 and its length after the temperature increase is ` = `0 + `0 T . The old position of the halfbar, its new position, and the distance x that one end is displaced form a right triangle, with a hypotenuse of length `, one side of length `0 , and the other side of length x. The Pythagorean theorem yields x2 = `2 `2 = `2 (1 + T )2 `2 . Since the change in length is small we may approximate 0 0 0 (1 + T )2 by 1 + 2 T , where the small term ( T )2 was neglected. Then
36P x2 = `2 + 2`2 T `2 = 2`2 T 0 0 0 0
and x = `0 2 T = 3:77 m 2(25 10 6 =C )(32 C) = 7:5 10 2 m : 2
p p (a) The total change in length for the composite bar is L = L1 + L2 = 1 L1 T + 2 L2 T = (1 L1 + 2 L2 )T: Comparing this with L = LT , we get = (1 L1 + 2 L2 )=L: (b) Now
8 < 37P = 13:0 10 : Ls + Lb = 52:4 cm : 6 =C = (11 10 6 =C )Ls + (19 10 6 =C )Lb ; Ls + Lb Solve for Ls and Lb : Ls = 39:3 m, Lb = 13:1 cm. The work needed is W = mLF = (1:00 g)(333 J/g) = 333 J:
39E 38E (a) The specic heat is given by c = Q=m(Tf Ti ), where Q is the heat added, m is the mass of the sample, Ti is the initial temperature, and Tf is the nal temperature. Thus 314 J c = (30:0 10 3 kg)(45:0 C 25:0 C) = 523 J/kg K : 538 CHAPTER 19 TEMPERATURE, HEAT, AND... (c) The molar specic heat is given by (b) If N is the number of moles of the substance and M is the mass per mole, then m = NM , so m 30: 3 kg N = M = 50 010 10kg/mol = 0:600 mol : 3 314 cm = N (T Q T ) = (0:600 mol)(45:0JC 25:0 C) = 26:2 J/mol K : f i Use Q = cmT: The amount of water needed is (1:00 6 V = m = CQ T = (1:00 103 kg/m3 )(110 kcal/da)(5 da)C 22:0 C) = 35:7 m3 : :00 kcal/kg)(50:0 40E Use Q = cmT . The mass m of the water that must be consumed is 41E Q 103 m = cT = (1 g/cal 3500 :0 Ccal0:00 C) = 94:6 104 g ; C)(37
which is equivalent to 9:46 104 g=(103 g/liter) = 94:6 liters of water. This is way too much to drink in a single day! The heat needed is
42E Q = V LF = mLF = (10%)(1:00 103 kg/m3 )(200; 000 m3 )(333 kJ/kg) = 6:7 1012 J : The amount of water m which is frozen is Q 50 2 kJ m = L = 333 :kJ/kg = 0:151 kg = 151 g : F Therefore the amount of water which remains unfrozen is 260 g 151 g = 109 g. 43E CHAPTER 19 TEMPERATURE, HEAT, AND... 539 44E The melting point of silver is 1235 K, so the temperature of the silver must rst be raised from 15:0 + 273 = 288 K to 1235 K. This requires heat Q = cm(Tf Ti ) = (236 J/kg K)(0:130 kg)(1235 C 288 C) = 2:91 104 J :
Now the silver at its melting point must be melted. If LF is the heat of fusion for silver this requires Q = mLF = (0:130 kg)(105 103 J/kg) = 1:36 104 J : The total heat required is 2:91 104 J + 1:36 104 J = 4:27 104 J. The heat Q which is added to the room in 1 h is
45E Q = 4(100 W)(90%)(1:00 h)(3600 s=1:00 h) = 1:30 106 J : The work the man has to do to climb to the top of Mt. Everest is given by W = mgh = (160 lb)(29; 000 ft) = (160 lb)(4:45 N/lb)(29; 000 ft)(1:00 m=3:28 ft) = 6:29 106 J: Thus the amount of butter needed is
6 J)(1:00 cal m = (6:29 106000 cal/g =4:18 J) = 250 g : 46E The rate of energy dissipation for the athlete is
3 :18 J/cal) = 194 W ; P = 4000 Cal/da = (4000 10 cal)(4s/da) (1 da)(86400 47E which is 1.94 times as much as the power of a 100W light bulb. Use Q = cmT = K = 1 mv2 . The speed of the water is then 2
48E v = 2cT = 2(4180 J/kg C )(78 68)(5=9) C = 2:2 102 m/s : p p 540 CHAPTER 19 TEMPERATURE, HEAT, AND... 49E (a) The heat generated is the power output of the drill multiplied by the time: Q = Pt. Use 1 hp = 2545 Btu/h to convert the given value of the power to Btu and 1 min = (1=60) h to convert the given value of the time to hours. Then Q = (0:400 hp)(2545 Btu/h)(2:00 min) = 33:9 Btu : 60 min/h
(b) Use 0:750Q = cm T to compute the rise in temperature. Here c is the specic heat of copper and m is the mass of the copper block. Table 1 gives c = 386 J/kg K. Use 1 J = 9:481 10 4 Btu and 1 kg = 6:852 10 2 slug to show that :481 c = (386 J/kg K)(910 2 10 Btu/J) = 5:341 Btu/slug K : 6:852 slug/kg
4 The mass of the block is its weight W divided by g: m = W=g = (1:60 lb)=(32 ft/s2 ) = 0:0500 slug. Thus (0:750)(33:9 Btu) T = 0:750Q = (5:341 Btu/slug K)(0:0500 slug) = 95:3 K : cm This is equivalent to (9=5)(95:3 K) = 172 F . Consider the object of mass m1 falling through a distance h. The loss of its mechanical energy is E = m1 gh: This amount of energy is then used to heat up the temperature of water of mass m2 : E = m1 gh = Q = m2 cT . Thus the maximum possible water temperature is
50E Tf = Ti + T = Ti + m1 gh m2 c (6:00 kg)(9 80 2 = 15:0 C + (0:600 kg)(4::18 m/s 3)(50:0m) ) = 1:17 C : 10 J/kg C
51E (a) The water (of mass m) releases energy in two steps, rst by lowering its temperature from 20 C to 0 C, and then by freezing into ice. Thus the total energy transferred from the water to the surroundings is Q = cw mT + LF m = (4190 J/kg K)(125 kg)(20 C) + (333 kJ/kg)(125 kg) = 5:2 107 J :
(b) Before all the water freezes, the lowest temperature possible is 0 C, below which the water must have already turned into ice. CHAPTER 19 TEMPERATURE, HEAT, AND... 541 Mass m (= 0:100 kg) of water, with specic heat c (= 4190 J/kgK), is raised from an initial temperature Ti (= 23 C) to its boiling point Tf (= 100 C). The heat input is given by Q = cm(Tf Ti ). This must be the power output of the heater P multiplied by the time t; Q = Pt. Thus 100 kg)(100 C 23 C) = 160 s : f t = Q = cm(TP Ti ) = (4190 J/kg K)(0:200 W P (a) The amount of water which could be vaporized is 1 mv2 2 mw = 2 L = (2200 kg)[(65:0 mi/h)(1609 m/mi)(1 h=3600 s)] = 0:412 kg = 412 g : 2(539 103 cal/kg)(4:18 J/cal)
53E 52E (b) The amount of energy in KW h is 1 E = 1 mv2 = 2 (2200 kg)[(165:0 mi/h)(1609 m/mi)(1 h=3600 s)]2 (1 KW h=3600 103 J) 2 = 0:26 KW h : Thus it would cost you (0:26 KW h)(12 cents= KW h) = 3:1 cents. (a) The heat transferred to the water of mass m1 is Qw = cw mw T + LV ms = (1 cal/g C )(220 g)(100 C 20:0 C) + (539 cal/g)(5:00 g) = 20:3 kcal : (b) The heat transferred to the bowl is Qb = cb mb T = (0:0923 cal/g C )(150 g)(100 C 20:0 C) = 1:11 kcal : (c) Let the orginal temperature of the cylinder be Ti , then Qw + Qb = cc mc (Ti Tf ), which gives Ti = Qw + Qb + Tf = (0:20:3 kcal +1:11 kcal g) + 100 C = 873 C : cm 0923 cal/g C )(300
c c
54E Let mw = 14 kg; mc = 3:6 kg, mm = 1:8 kg, Ti1 = 180 C, Ti2 = 16:0 C, and Tf = 18:0 C. The specic heat cm of the water then satises (mw cw + mc cm )(Tf Ti2 ) + mm cm (Tf Ti1 ) = 0 ; 55P 542 CHAPTER 19 TEMPERATURE, HEAT, AND... which we solve for cm : T f cm = m (T mw cw)(+i2 TT ) T ) Ti2 mm ( f i1 c f (14 kg)(4:18 kJ/kg K)(16:0 C 18:0 C) = (3:6 kg)(18:0 C 16:0 C) + (1:8 kg)(18:0 C 180 C) = 0:41 kJ/kg C : Let the initial water temperature be Twi and the initial thermometer temperature be Tti . Then ct mt (Tf Tti ) = cw mw (Twi Tf ). Solve for Twi : T = ct mt (Tf Tti ) + T
wi 56P = 45:5 C :
57P K)(44 4 = (0:0550 kg)(0:837 kJ/kg)(0:300 :kg) 15:0) K + 44:4 C (4:18 kJ/kg C cw mw f Use Q = Pt = cmT . The time t required is cmT = (103 cal/kg C )(40 gal)(103 kg=264 gal)(100 F 70 F)(5 C =9 F ) t= P (2:0 105 Btu/h)(252:0 cal/Btu)(1 h=60 min) = 3:0 min : (a) Let the number of weight liftings needed be n, then nmgh = Q, or Q n = mgh = (80(3500 Cal)(41802J/Cal)m) = 18; 700 : :0 kg)(9:80 m/s )(1:00 (b) The time it takes is t = (18; 700)(2:00 s)(1:00 h=3600 s) = 10:4 h :
2 The deceleration a of the car is given by vf vi2 = vi2 = 2ad, or 3 m/km)(1 2 a = [(90 km/h)(10 2(80 m) h=3600 s)] = 3:9 m/s2 :
59P 58P CHAPTER 19 TEMPERATURE, HEAT, AND... 543 The time t it takes for the car to stop is then 3 t = vf a vi = (90 km/h)(10 3:m/km)(1 h=3600 s) = 6:4 s: 9 m/s2 The average rate at which thermal energy is produced is then 1 mv2 2 i = (1500 kg)[(90 km/h)(103 m/km)(1 h=3600 s)]2 = 7:3 104 W: = t 2(6:4 s) Mass m of water must be raised from an initial temperature Ti (= 59 F = 15 C) to a nal temperature Tf (= 100 C). If c is the specic heat of water then the energy required is Q = cm(Tf Ti ). Each shake supplies energy mgh, where h is the distance moved during the downward stroke of the shake. If N is the total number of shakes then Nmgh = Q. If t is the time taken to raise the water to its boiling point then (N=t)mgh = Q=t. Notice that N=t is the rate R of shaking (30 shakes/min). Thus Rmgh = Q=t. The distance h is 1:0 ft = 0:3048 m. Hence t = Q = cm(Tf Ti ) = c(Tf Ti )
60P Rmgh = This is 2:8 da.
61P (30 shakes/min)(9:8 m/s2 )(0:3048 m) (4190 J/kg K)(100 C 15 C) Rmgh Rgh = 3:97 103 min : The mechanical energy which is converted to heat is E = 1 mvi2 = 1 (50:0 kg)(5:38 m/s)2 = 2 2 7:24 102 J: The amount of ice melted is then 102 E m = L = 7:24 J/g J = 2:17 g : 333 F The heat need is
62P Q= Z Tf Ti cm dT = m
Z Z Tf = (2:09) 15:0 C 0 C Ti c dT 5: (0:20 + 0:14T + 0:023T 2 ) dT = (2:0)(0:20T = 82 cal : 15:0 2 + 0:00767T 3 ) (cal) + 0:070T 5:0 544 CHAPTER 19 TEMPERATURE, HEAT, AND... The power consumed by the system is 1 cmT P = 20% t 3 cm3 )(1 g/cm3 1 = 20% (4:18 J/g C)(200 :10h)(3600 s/h) )(40 C 20 C) (1 0 = 2:3 104 W : The area needed is then :3 104 A = 2700 W/mW = 33 m2 : 2 Denote thhe ice with subscript I and the coee with c, respectively. Let the nal temperature be Tf . The heat absorbed by the ice is QI = F mI + mI cw (Tf 0 C); and the heat given away by the coee is Qc = mw cw (TI Tf ). Let QI = Qc and solve for Tf :
wc Tf = m(mw TI m F mI I + c )cw C )(80:0 (79 = (130 g)(1:00 cal/g + 130 g)(1C) cal/g:7 cal/g)(12:0 g) (12:0 g :00 C ) = 66:5 C : Thus the temperature of the coee will cool by jT j = 80:0 C 66:5 C = 13:5 C :
65P 64P 63P Let the mass of the steam be ms and that of the ice be mi . Then LF mc +cw mc (Tf 0:0 C) = Ls ms + cw ms (100 C Tf ), where Tf = 50 C is the nal temperature. Solve for ms : m + mc ( f ms = LF L c+ ccw(100TC 0:0) C) Tf s w (79:7 cal/g)(150 g) + (1 cal/g C)(150 g)(50 C 0:0 C) = 539 cal/g + (1 cal/g C)(100 C 50 C) = 33 g : Let the nal temperature be Tf . The heat absorbed by the ice is QI = F mI + mI cw (Tf 0 C); while the heat given away by the water is Qw = cw mw (Ti Tf ): Let Qw = QI and solve for Tf : wc Tf = m(mw Ti m )F mI : + c
I c w
66P CHAPTER 19 TEMPERATURE, HEAT, AND... 545 (a) Now Ti = 90 C so (79 cal/g)(500 g) = 5:3 C : Tf = (500 g)(1:00 cal/g C )(90 C) cal/g:7C ) (500 g + 500 g)(1:00 (b) If you were to use the formula above for Tf in this case, you would get Tf < 0, which is of course impossible. In fact, not all the ice has melted in this case. The amount of ice melted is given by )(500 T m0I = cw mw (i 0 C) = (1:00 cal/g C7 cal/g g)(70 C) = 440 g: 79: F Therefore there are mI = mI m0I = 500 g 440 g = 60 g of ice left, and the nal temperature is Tf = 0 C (because the system is an icewater mixture in thermal equilibrium). (a) There are three possibilities: 1. None of the ice melts and the waterice system reaches thermal equilibrium at a temperature that is at or below the melting point of ice. 2. The system reaches thermal equilibrium at the melting point of ice, with some of the ice melted. 3. All of the ice melts and the system reaches thermal equilibrium at a temperature at or above the melting point of ice. First suppose that no ice melts. The temperature of the water decreases from Twi (= 25 C) to some nal temperature Tf and the temperature of the ice increases from TIi (= 15 C) to Tf . If mw is the mass of the water and cw is its specic heat then the water rejects heat Q = cw mw (Twi Tf ) : If mI is the mass of the ice and cI is its specic heat then the ice absorbs heat Q = cI mI (Tf TIi ) : Since no energy is lost these two heats must be the same and cw mw (Twi Tf ) = cI mI (Tf TIi ) : The solution for the nal temperature is T +c m Tf = cw mwmwi + c Im I TIi cw w I I = (4190 J/kg K)(0:200 kg)(25 C) + (2220 J/kg K)(0:100 kg)( 15 C) (4190 J/kg K)(0:200 kg) + (2220 J/kg K)(0:100 kg)
67P = 16:6 C : This is above the melting point of ice, so at least some of the ice must have melted. The calculation just completed does not take into account the melting of the ice and is in error. 546 CHAPTER 19 TEMPERATURE, HEAT, AND... Now assume the water and ice reach thermal equilibrium at Tf = 0 C, with mass m (< mI ) of the ice melted. The heat rejected by the water is Q = cw mw Twi
and the heat absorbed by the ice is cI mI TIi + mLF ;
where LF is the heat of fusion for water. The rst term is the energy required to warm all the ice from its initial temperature to 0 C and the second term is the energy required to melt mass m of the ice. The two heats are equal, so cw mw Twi = cI mI TIi + mLF :
This equation can be solved for the mass m of ice melted: m = cw mw Twi + cI mI TIi = (4190 J/kg K)(0:200 kg)(25 C) + (2220 J/kg K)(0:100 kg)( 15 C) 333 103 J/kg = 5:3 10 2 kg = 53 g : LF Since the total mass of ice present initially was 100 g, there is enough ice to bring the water temperature down to 0 C. This is the solution: the ice and water reach thermal equilibrium at a temperature of 0 C with 53 g of ice melted. (b) Now there is less than 53 g of ice present initially. All the ice melts and the nal temperature is above the melting point of ice. The heat rejected by the water is Q = cw mw (Twi Tf )
and the heat absorbed by the ice and the water it becomes when it melts is Q = cI mI ( TIi ) + cw mI Tf + mI LF :
The rst term is the energy required to raise the temperature of the ice to 0 C, the second term is the energy required to raise the temperature of the melted ice from 0 C to Tf , and the third term is the energy required to melt all the ice. Since the two heats are equal cw mw (Twi Tf ) = cI mI ( TIi ) + cw mI Tf + mI LF :
The solution for Tf is Substitute given values to obtain Tf = 2:5 C. c Tf = cw mw Twi +mI mI TIi ) mI LF : c ( +m
w w I CHAPTER 19 TEMPERATURE, HEAT, AND... 547 If the ring diameter at 0:000 C is Dr0 then its diameter when the ring and sphere are in thermal equilibrium is Dr = Dr0 (1 + c Tf ) ; where Tf is the nal temperature and c is the coecient of linear expansion for copper. Similarly, if the sphere diameter at Ti (= 100:0 C) is Ds0 then its diameter at the nal temperature is Ds = Ds0 [1 + a (Tf Ti )] ; where a is the coecient of linear expansion for aluminum. At equilibrium the two diameters are equal, so 68P Dr0 (1 + c Tf ) = Ds0 [1 + a (Tf Ti )] :
The solution for the nal temperature is T = Dr0 Ds0 + Ds0 a Ti
f Ds0 a Dr0 c 00000 in. 1:00200 in. in.)(23 10 6 = = 1:(1:00200 in.)(23 10 6+(1:00200:00000 in.)(17 C )(100C) / C) (1 10 6 = C ) = 50:38 C : Q = cc mr Tf ; where cc is the specic heat of copper and mr is the mass of the ring. The heat rejected
two heats are equal, and The heat absorbed by the ring is up by the sphere is Q = ca ms (Ti Tf ) ; where ca is the specic heat of aluminum and ms is the mass of the sphere. Since these cc mr Tf = ca ms (Ti Tf ) :38 C) f ms = c cc mr TT ) = (386 J/kg K)(0:0200 kg)(5038 C) = 8:71 10 3 kg : C 50: (900 J/kg K)(100 a (Ti f Denote the density of the liquid as , the rate of liquid owing in the calorimeter as , the specic heat of the liquid as c, the rate of heat ow as P , and the temperature change as T . Consider a time duration dt, during this time interval, the amount of liquid being heated is dm = dt. The energy required for the heating is dQ = Pdt = c(dm)T = cTdt. Thus 250 c = P T = (8:0 10 6 m3 =s)(0:85W 103 kg/m3 )(15 C) = 2:5 103 J/kg C : 69P 548 CHAPTER 19 TEMPERATURE, HEAT, AND... (a) Let the rate of ow of energy into the system be dQ=dt. Then 70P dQ = cm dT ; dt dt
which gives c as a function of T : ( c(T ) = mdQ=dt)) : (dT=dt (b) In this case t / T 1=3 , and c(T ) / dT dt
71E 1 3 / dt dt 1 /t 2/T 2=3 : One part of path A represents a constant pressure process. The volume changes from 1:0 m3 to 4:0 m3 while the pressure remains at 40 Pa. The work done is W = p V = (40 Pa)(4:0 m3 1:0 m3 ) = 120 J :
The other part of the path represents a constant volume process. No work is done during this process. The total work done over the entire path is 120 J. To nd the work done over path B we need to know the pressure as a function of volume. R Then we can evaluate the integral W = p dV . According to the graph the pressure is a linear function of the volume, so we may write p = a + bV , where a and b are constants. In order for the pressure to be 40 Pa when the volume is 1:0 m3 and 10 Pa when the volume is 4:00 m3 the values of the constants must be a = 50 Pa and b = 10 Pa=m3 . Thus p = 50 Pa (10 Pa=m3 )V and W= Z 4 = 50V 4 1 1 p dV = Z 4 1 2 4 = 200 J 5V 1 (50 10V ) dV 50 J 80 J + 5 J = 75 J : One part of path C represents a constant pressure process in which the volume change from 1:0 m3 to 4:0 m3 while p remains at 10 Pa. The work done is p V = (10 Pa)(4:0 m3 1:0 m3 ) = 30 J :
The other part of the process is at constant volume and no work is done. The total work is 30 J. CHAPTER 19 TEMPERATURE, HEAT, AND... 549 Notice that the work is dierent for dierent paths. The net work done in process BA is WBA = 1 (40 Pa 10 Pa)(4:0 m3 1:0 m3 ) = 45 J : 2 The net work done in process BC is 1 WBC = 2 (40 Pa 10 Pa)(4:0 m3 1:0 m3 ) = 45 J : (a) W = 200 J: (b) Q = 70:0 cal = 293 J: (c) Eint = Q W = 293 J ( 200 J) = 93 J: (a) A ! B : Q(+); W (+); B ! C : W = 0; Eint (+); C ! A: Q( ); W ( ); Eint ( ): (b) WABCA = 1 (40 Pa 20 Pa)(3:0 m3 1:0 m3 ) = 20 J: 2
75E 74E 73E 72E Since the process is a complete cycle (beginning and ending in the same thermodynamic state) the change in the internal energy is zero and the heat absorbed by the gas is equal to the work done by the gas: Q = W . In terms of the contributions of the individual parts of the cycle QAB + QBC + QCA = W and QCA = W QAB QBC = +15:0 J 20:0 J 0 = 5:0 J. This means 5:0 J of energy leaves the gas in the form of heat. where the coecients were chosen so that p = 10 Pa when V = 1:0 m3 and p = 30 Pa when V = 4:0 m3 . The work done by the gas during this portion of the cycle is Z 4 Z 4 10 + 20 V dV WAB = p dV = 3 3 1 1 4 4 = 10 V 1 + 10 V 2 1 = 40 + 160 10 10 = 60 J : 3 3 3 3 3 3 The work W done over the cycle can easily be calculated. Since the internal energy is the same at the beginning and end the heat absorbed Q equals the work done: Q = W . Over the portion of the cycle from A to B the pressure p is a linear function of the volume V and we may write 10 Pa + 20 Pa/m3 V ; p= 3 3 76E 550 CHAPTER 19 TEMPERATURE, HEAT, AND... The BC portion of the cycle is at constant pressure and the work done by the gas is WBC = p V = (30 Pa)(1:0 m3 4:0 m3 ) = 90 J. The CA portion of the cycle is at constant volume, so no work is done. The total work done by the gas is W = WAB + WBC + WCA = 60 J 90 J + 0 = 30 J and the total heat absorbed is Q = W = 30 J. This means the gas loses 30 J of energy in the form of heat. Use Q = F mice = W + Eint : In this case Eint = 0. Since T = 0 for the idea gas, thus the work done on the gas is
77P W 0 = W = F mi = (333 J/g)(100 g) = 33:3 kJ :
78P (a) The change in internal energy E int is the same for path iaf and path ibf . According to the rst law of thermodynamics E int = Q W , where Q is the heat absorbed and W is the work done by the system. Along iaf E int = Q W = 50 cal 20 cal = 30 cal. Along ibf W = Q E int = 36 cal 30 cal = 6 cal. (b) Since the curved path is traversed from f to i the change in internal energy is 30 cal and Q = E int + W = 30 cal 13 cal = 43 cal. (c) Let E int = E int,f E int,i . Then E int,f = E int + E int,i = 30 cal + 10 cal = 40 cal. (d) The work Wbf for the path bf is zero, so Qbf = E int,f E int,b = 40 cal 22 cal = 18 cal. For the path ibf Q = 36 cal so Qib = Q Qbf = 36 cal 18 cal = 18 cal. Use H = kA(TH TC )=L. The temperature TH at a depth of 35:0 km is
3 W/m2 )(35 3 TH = HL + TC = (54:0 10 2:50 W/m K:0 10 m) + 10:0 C = 766 C : kA
79E Denote the polyurethane foam with subscript p and silver with subscript s, respectively. Use Eq 1931 to nd L = kR. (a) From Table 196 we nd kp = 0:024 W/m K so 80E Lp = kp Rp = (0:024 W/m K)(30 ft2 F h/Btu)(1 m=3:281 ft)2 (5 C =9 F )(3600 s=h)(1 Btu=1055 J) = 0:13 m ;
where the conversion factors are found in Appendix D. CHAPTER 19 TEMPERATURE, HEAT, AND... 551 (b) For silver ks = 428 W/m K, so k Ls = ks Rs = ks Rs Lp = 0428(30) (0:13 m) = 2:3 103 : :024(30) p Rp
(a)
81E k = 2:9 10 3 cal/cm C s 10 3 cal)(4: 3 = (1:0 cm)(3:3 (2:9 ft/cm)(1:0 C0 )(910 =5Btu/cal) s)(1 h=3600 s) F C )(1:0 10 2 = 0:70 Btu/ft F h ; k = 2:9 10 3 cal/cm C s (2:9 10 3 cal)(4:18 = (1:0 cm)(0:010 m/cm)(1:0 s)(1:J/cal) K=C ) 0 C )(1 = 1:2 W/m K :
(b) The Rvalue is in.)(1:0 ft R = L = (0:2570 Btu/ft =12hin.) = 0:030 ft2 F h/Btu : k 0: F
82E (a) The rate of heat ow is m2 )(33 H = kA(TH TC ) = (0:040 W/m K)(1:810 2 m C 1:0 C) = 2:3 102 J/s : L 1:0 (b) The new rate of heat ow is K)(230 H 0 = k kH = (0:600W/mW/m K J/s) = 3:5 103 J/s ; :040
which is about 15 times as fast as the original heat ow.
83E 0 The rate of heat ow is given by H = kA TH L TC ; 552 CHAPTER 19 TEMPERATURE, HEAT, AND... where k is the thermal conductivity of copper (401 W/m K), A is the crosssectional area (in a plane perpendicular to the ow), L is the distance along the direction of ow between the points where the temperature is TH and TC . Thus
4 m2 H = (401 W/m K)(90:0 :10 m )(125 C 10:0 C) = 1:66 103 J/s : 0 250 (a) The rate of heat ow is 84E kA(TH TC ) = (401 W/m K)(4:8 10 4 m2 )(100 C ) = 16 J/s : H= L 1:2 m
(b) The rate at which ice melts is dm = H = 16 J/s = 0:048 J/g : dt LF 333 J/g
85E The rate of heat ow through the warmer slab is H2 = AK2 (TH TX ) = A(THR TX ) ; L
2 2 while that through the cooler one is H1 = AK1 (TX TC ) = A(TX TC ) : L R
1 1 Let H1 = H2 to obtain (TH TX )=R2 = (TX TC )=R1 ; which we solve for TX : R +T TX = TH R1 + RC R2 :
1 2
86E (a) We estimate the surface area of the average human body to be about 2 m2 and the skin temerature to be about 300 K (somewhat less than the internal temperature of 310 K). Then from Eq. 1937 Pr = "AT 4 (5:6703 10 8 W/m2 K4 )(0:9)(2:0 m2 )(300 K)4 = 8 102 W : CHAPTER 19 TEMPERATURE, HEAT, AND... 553 (b) The energy lost is given by E = Pr t = (8 102 W)(30 s) = 2 104 J : With arrangement (a), the rate of the heat ow is 87E With arrangement (b), we use Eq. 1935 to nd the rate of heat ow: Ha = H1 + H2 = Ak1 (TH TC ) + Ak2 (TH TC ) 2L 2L A k (T T ) ; = 2L a H C where ka = 4K1 + k2 . TC 2A Hb = (L=k(TH (L=k) ) = 2A kb (TH TC ) ; )+ L
1 2 where kb = fk1 k2 =(k1 + k2 ). Since k1 6= k2 , we see that (k1 k2 )2 = (k1 + k2 )2 4k1 k2 > 0, or kb = 4k1 + k2 < 0 : k (k + k )2 Thus Hb < Ha ; i.e., arrangement (b) would give the lower heat ow.
88P a 1 2 Use H = kAT=L / A=L: Comparing cases (a) and (b) in Fig. 1942, we have Hb = (Ab La )=(Aa Lb )Ha = 4Ha : Thus it would take 2:0 min=4 = 0:5 min for the same amount of heat to be conducted through the rods welded as shown in Fig. 1942(b). (a) The rst door of area A can be thought of as a parallel construction of two parts: part 1, of a single aluminum panel of area A1 = 25%A; and part 2, of a composite structure with aluminum and glass, of area A2 = 75%A. Thus Ha =H1 + H2 = ka A1 T + (L =kA)2+ TL =k ) La a a ( b b (0:25)(235) + 0:75 = 1:5 10 3 m 1:5 10 3 m=235 + 3:0 10 3 m=1:0 (W/m K) (2:0 m)(0:75 m)(33 C ) =2:0 106 W :
89P 554 CHAPTER 19 TEMPERATURE, HEAT, AND... (b) For the second door made of pine, m)(0:75 p Hb = kLA T = (0:11 W/m K)(2:010 2 m m)(33 C ) = 220 W : 2:5 p The heat conducted is
90P Q = Ht = kAtT L : m)2 (5 = (67 W/m K)(=4)(1572 10:03min)(60 s/min)(2:3 C ) : m 7 J: = 2:0 10 H = kA TH L TC ; where H is the heat ow through an area A, between points a distance L apart with a temperature dierence of TH TC . Since the conductivity of glass is given in Table 4 as 1:0 W/mK we use the Celsius scale to give the values of the temperature. A temperature of 20 F is equivalent to 28:9 C and a temperature of 72 F is equivalent to 22:2 C. Thus H = k TH TC = (1:0 W/m K) 22:2 C + 28:9 C = 1:7 104 W/m2 : A L 3:0 10 3 m (a) Use 91P (b) The energy now passes in succession through 3 layers, one of air and two of glass. The heat ow H is the same in each layer and is given by P H = A(TH TC ) ; where the sum in the denominator is over the layers. If Lg is the thickness of a glass layer, La is the thickness of the air layer, kg is the thermal conductivity of glass, and ka is the thermal conductivity of air, then the denominator is XL = 2Lg + La = 2Lg ka + La kg and L=k k kg ka ka kg H = (TH TC )ka kg A 2Lg ka + La kg :2 + 28 9 C)(0:026 W/m :0 W/m K) = 2(3:0(2210 C m)(0::026 W/m K) + (0K)(1m)(1:0 W/m K) 3 :075 2: = 18 W/m CHAPTER 19 TEMPERATURE, HEAT, AND... 555 Use Eqs. 1937 and 1938. Note that the surface area of the sphere is given by A = 4r2 , where r = 0:500 m is the radius. (a) The temperature of the spere is T = 273 + 27:0 = 300 K. Thus Pr = "AT 4 = (5:6703 10 8 W/m2 K4 )(0:850)(4)(0:500 m2 )(300 K)4 = 1:23 103 W : (b) Now Tenv = 273 + 77:0 = 350 K so 4 Pa = "ATenv = (5:6703 10 8 W/m2 K4 )(0:850)(4)(0:500 m2 )(350 K)4 = 2:27 103 W : (c) From Eq. 1939 Pn = Pa Pr = 2:27 103 W 1:23 103 W = 1:04 103 W : Use Eq. 1939. Here T = 273 + ( 100) = 173 K, Tenv = 273 + ( 150) = 123 K, and A = 6(6:0 10 6 m)2 = 2:16 10 10 m2 . Thus 4 Pn = "A(Tenv T 4 ) = (5:6703 10 8 W/m2 K4 )(0:75)(2:16 10 10 m2 )[(123 K)4 (173 K)4 ] = 6:1 10 9 W :
2 (a) The surface are of the cylinder is given by A1 = 2r1 + 2r1 h1 = 2(2:5 10 2 m)2 + 2(2:5 10 2 m)(5:0 10 2 m) = 1:18 10 2 m2 , its temperature is T1 = 273+30 = 303 K, and the temerature of the environment is Tenv = 273 + 50 = 323 K. Thus from Eq. 1939 4 P1 = "A1 (Tenv T 4 ) = (5:6703 10 8 W/m2 K4 )(0:85)(1:18 10 2 m2 )[(323 K)4 (303 K)4 ] = 1:39 W : (b) Let the new height of the cylinder be h2 . Since the volume V of the cylinder is xed, we 2 2 must have V = r1 h1 = r2 h2 . Solve for h2 : h2 = (r1 =r2 )2 h1 = (2:5 cm=0:50 cm)2 (5:0 cm) = 125 cm = 1:25 m. The corresponding new surface area A2 of the cylinder is A2 = 2 2r2 + 2r2 h2 = 2(0:50 10 2 m)2 + 2(0:50 10 2 m)(1:25 m) = 3:94 10 2 m2 . Thus P2 = A2 = 3:94 10 2 m2 = 3:3 : P1 A1 1:18 10 2 m2
94P 93P 92P 556 CHAPTER 19 TEMPERATURE, HEAT, AND... Let h be the thickness of the slab and A be it area. Then the rate of heat ow through the slab is H = kA(THh TC ) ; where k is the thermal conductivity of ice, TH is the temperature of the water (0 C), and TC is the temperature of the air above the ice ( 10 C). The heat leaving the water freezes it, the heat required to freeze mass m of water being Q = LF m, where LF is the heat of fusion for water. Dierentiate with respect to time and recognize that dQ= dt = H to obtain H = LF dm : dt Now the mass of the ice is given by m = Ah, where is the density of ice, so dm=dt = A(dh=dt) and H = LF A dh : dt Equate the two expressions for H and solve for dh=dt: dh = k(TH TC ) : Since 1 cal = 4:186 J and 1 cm = 1 10 2 m, the thermal conductivity of ice has the SI value k = (0:0040 cal/s cmK)(4:186 J/cal)=(1 10 2 m/cm) = 1:674 W/mK. The SI value for the density of ice is = 0:92 g/cm3 = 0:92 103 kg/m3 . Thus dh = (1:674 W/m K)(0 C + 10 C) = 1:1 10 6 m/s = 0:40 cm/h : 3 J/kg)(0:92 103 kg/m3 )(0:050 m) dt (333 10 Refer to the gure to the right. Let the thickness of the ice be x. In a steady state, the rate of heat ow in ice is equal to that in water: H = ki A(T2 T3 ) = kw A(T1 T2 ) : Note that T2 = 0 C. Substitute the numerical values to the equation for H above ans solve for x to obtain x = 1:1 m.
97P 96P 95P dt LF h T 3=5.0 oC x H ice T2 Lx H water T 1=4.0 oC x L x Denote TH = 100 C; TC = 0 C, the temperature of the copperaluminum junction by T1 and that of the aluminumbrass junction by T2 . Then
c b H = kLA (TH T1 ) = ka A (T1 T2 ) = kLA (T2 Tc ) : L CHAPTER 19 TEMPERATURE, HEAT, AND... 557 Solve for T1 and T2 to obtain
C T1 = TH + 1 + k Tk +TH)=k k ( a kb a b c 0:00 C 100 C = 100 C + 1 + 401(235 + 109)=[(235)(109)] = 84:3 C and
H T2 = Tc + 1 + k Tk + TC)=k k ( k b c a = 0:00 C + 1 + 109(235 + 401)=[(235)(401)] = 57:6 C : 100 C 0:00 C c a ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.
 Fall '07
 GROUPTEST
 Heat

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