chap11 solutions

chap11 solutions - CHAPTER ELEVEN SOLUTIONS If half of this...

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Unformatted text preview: CHAPTER ELEVEN SOLUTIONS If half of this goes into heat, Q=‘2-g--52—‘l 2 112.5 J, and ___Q__ , 112.5 1 _ , “"1116 “(5.00 x 10-3 kg)(128 J/kg°C) "176 C' Q 1674 I a _ _ a]: 74 = = = 11 6 Q—400c 16 J.soAT C (5.00 x 102 kgxzso J/kgoc) 146 C, and Tf=166“C. 1 1 .7 Consider a 1.00 kg mass of water: Q: APE: nigh = (1.00 kg)(9.80 m/sz)(50.0 m) = 490] Q 490 , A1 , Ar=—=———~————-l——-—-—=0.117 C. so Inc (1.00 kg)(4.186 x 103 J/kg°C) So, Tir=10.1C°. 11.3 (a) We use Q=chT=(0.85)-%mv2, or 0.35122 08513.0 m/sl2 _3, AT: = C. (b) The remaining energy is absorbed by the horizontal surface on which the block slides. 1 1.9 Let us find the heat extracted from the system in one minute. Q= mcupCcupA T4” mwateeraterA 7; 01‘ Q: {(0.20 kg)(900 J/kg”C) + (0.80 kg)(4186 J/kg°C)](1.5°C) : 5293 J. If this much heat is removed each minute, the rate of removal of heat is _Q‘_5293 _ _ P- t_ 60 S —88.2J/s-88W. 11.10 First, convert the change in temperature, AT, from Fahrenheit to Celsius: 5 5 Tci = §(TF1 - 32). and Tc2=§(TF2 - 32)- Thusv Ap-TCL'TCZ ' TC1=%ATF, and ifAT}: :2 1", thenATC=g°Q Now, use Q= 111621 T and the fact that 1 lbm = 0.4536 kg to evaluate the Btu: 1 Btu =(11bm)(4186J/kg °C)(1 ° F) = (0.4536 kg)(4186 J/kg "CH; “C) which yields 1 Btu = 1055 J. 1 1.1 1 The heat needed to raise the temperature of the water to 25“C is: Qweded = (0.500 kg)(4186 J/kg°C)(5.00°C) = 1.05 x 104 J. The heat received from each pellet is: qneuet=(10'3 kg)(128 J/kg°C)(175°C) = 22.4 J/pellet Thus, the number of pellets needed is: _ Queeded _ 1.05 x 104 _467 11 t n ‘ Tenet ‘22.4 J/pellet ’ Fe B 5' 1 1.12 Our heat loss = heat gain equation becomes: mgroncimnA Til-on: mwateerATw or, (0.40 kg)(448 J/kg°C)(SOO”C - 1) z (20 kg)(4186 J/kg°C)(T— 22°C). 154 I 11.13 11.14 11.15 11.16 11.17 11.18 11.19 CHAPTER ELEVEN SOLUTIONS From which, we find: T: 23°C. (heat gain = heat loss) becomes: mwatercwn Tw= 0.1301(1ch Tg. Thus, mwater(4186 J/kg°C)(25.0°C) = (3.00 kg)(129 J/kg°C)(50.0°C), or mwater = = g. (heat gain = heat loss) becomes: mcupCcA 7:: + 13waA Tw + mstirrerCsA 7s = mAgCAgA TAg J 01' mcup(900 J/kg°C)(5.0°C) + (0.225 kg)(4186 j/kg“C)(5.0“C) + (0.04 kg)(337 J/kg°C)(S°C) = (0.4 kg)(234 J/kg°C)(55°C). We find: mm = 30 x10' 3 kg = 30 g. Asttem = 0, since the system does not exchange heat with the environment. Asttem 2(0.200 kg)(4186 J/kg“C ) ( Tf- 10°C) + (0.300 kg)(900 J/kg°C)(Tf- 10°C) + (0.100 kg)(4186J/kg"C)(Tf—100°C): 0 Which gives Tf = 35°C. Asttem = 0, since the system does not exchange heat with the environment. Asttem =(0.1kg)(900]/kg”C)(20”C- 10°C) +(0.25 kg)(4186]/kg°C)(20“C~ 10°C) +(0.05 kg)(387 J/kg°C)(20°C- 80°C) + (0.07 kg)(cz)(20°C— 100°C). Which gives cz 2 1.8 x 103 J/kg“C = 0.44 cal/g°C. 500"l3 = 260°C, 100°F = 37.8°C, and 75°F = 233°C. (heat gain = heat loss) becomes: mironcironA Tiron = mwater‘C'W'aterA Twater mimn(448 j/kg°C)(260°C — 37.8”(3) = mwater(4186 j/kg°C)(37.8°C - 23.9“C) mwater miron cooled. = 1.7 Thus, 1.7 kg of water is required for every kg of iron First, we assume that both the water and aluminum will gain heat while the copper loses heat. mwaA Tw + malCalA Tal = mcuccuA Tcu 0.250(4186)(T- 20.0) + 0.400(900)(T— 26.0) = 0.100(387)(100- ’1) yielding T = 23.6“C, showing our assumption that the aluminum would gain heat to have been incorrect. Alternately, one could have assumed that the water would gain heat while the copper and aluminum would lose heat. Then the calculation would have been: mewA Tw 3 malcalA 1Val + mcuCcuA Tcu 0.250(4186)( T- 20.0) = 0.400(900)(26.0 — 1) + O.100(387)( 100 - 7) which also yields T: 216°C. heat loss = heat gain mu.ch Tcu =maica1A Ta1+ mewA Tw + msteeicsteeid Tsteel: but ATw= Anteel=0. (0.200 kg)(387 J/kg"C)(85“C — 25°C) 2 ma](900 J/kg°C)(25°C ~ 5°C) ma; = 0.26 kg = 260 g 155 CHAPTER ELEVEN SOLUTIONS (0.01 kg)(2.26x 106 J/kg) + (0.01 kg)(4186J/kg°C)(100°C — Tf) From which, Tf= 40°C. (b) Q = heat to melt all ice =1.67x104 J (see part (21)) Q =heat given up as steam condenses =(10‘3 kg)(2.26 x 106 ‘L kg? . 22.26 X J (3 = (heat given up as condensed steam cools to 0°C) =(10-3 kg)(4186 J/kg°C)(100°C) = 418 J Note that Q + Q <Q Therefore, the final temperature will be 0°C with some ice remaining. Let us find the mass of ice which must melt to condense the steam and cool the condensate to 0°C. mLf=Q2+Q312678J ‘ Thus, :11 = “fl;— ‘= 8.0 x 10'3 kg = 8.0g. 3.33 'x 105 J/kg AQ RAM!" 397 /s m“C 15 x 10-4 m2 30°C 11.30 (a)H=—A“{="—L--= I 0.080 m ‘ =220J/S (b) The steps are identical here as in (3) except k = 0.0234 J/s 111°C. ' The result is: H= 1.3 x10‘2 J/s. (c) k=0.10J/sm°C, and -H=s.6x10-2 J/s. MAT: (0.871%?)(046 m2)(1‘1_.1°C) L .- 3.0 x 10'3 m ‘ (0.8 —'L.-)(0.16 m2)(38.9"C) kAATfi—J—nl-Qmfinl 7x103 l (to L ‘ 3.0 x 10-3'm _‘ ' _s 7 1 1.3_1(a) H=i—%= =470£ (into house) _ A9. (b) H‘ At “ outdoors) 1 1.32 Rtota] =2Ri =Routside air film + Rshingles + Rsheating +_Rcellulose " + Rdry wall + Rinside air film- : (0.17 +0.87 + 1.32 + 3(3.7) + 0.45 + 0.17)ft2 F“ h/BTU = 14 82 13° h/BTU 11.33 For the Thermopane: ' . AQ‘ Apia H=—= At ( ki) _ - ' 1.0 m2 23.0°C - 0°C =~¥Mfig= /, 5.0 x 10-3 m 10-2 m. 5.0 x10'3 m 5“ S 0.8 j/srn°C + 0.0234 J/sm°C + 0.8 J/sm°C For the single pane: H__AQ_kAAT__ 0.8 /s 111°C 1 m2 23°C .At“ _ =1. 3 /. 1x10_2m 84x10 Js 1 1-34 A = Aend walls + Aends of-attic + Aside walls + Arno A = 2(8.00m x 5.00 m) + 2(2.00x%x (4.00 In) x (4.00 m)tan37°) 158 11.35 CHAPTER ELEVEN SOLUTIONS 4m ) c0537“ + 2(10.0m)x5.00m)+ 2(10mx A=.304m2. may: kAAT_(4.8 x 10-4 kW/ m°c2(304 m2)(25°C1 _ L - At_ 0.21 m = 17.4kW=4.15kcsal Thus, the heat lost per day = (4.15 kcal/s)(-86,400 s) = 3.59 x 105 kcal/day. 3.59 x 105 keel/dag: 3/ ' 9300 kcal/m3 39m day' The gas needed to replace this loss = {Let us call the copper rod object 1 and the aluminum rod object 2. At equilibrium, the flow rate through each must be the same. We have szzAT _ klAlAT “ L2 — L1 ' Since the cross-sectional areas of the rods are the same and the - temperature difference across the rods are also equal, we have 13 _ g; 233 {/3 m°C‘_ 397 [/s m°C . L2“L1’°" L2 ‘ 0.15m ' From which, we find L; = 9.0 x 10' 2 In: 9.0 cm. 1 1.36 The heat neededto melt 5.0 kg of ice = mLf= (5.0 kg)(3.33 x 105 J/kg) 11.37 11.38 11.39 11.40 = 1.67x106 J Thus, the required flow rate during the 8.0 h 6 ‘ £—-—“——*—ll'67 x 10 =57.8J/s. (st—2.83 x 104 s _ A_Q*kAAT _k(0.8 m2)('20°C) 5mm“ 2.0 x10‘2 m ’ , At- L .becomes: and we find k= 7.2 x 10-2 VS m°C. (2.33 x 104 3) period is: Then The area of the sphere is A: 4:51'2 =4n(0.060 In)2 =4.52 x 10‘2 m2. _ Pnet= er( Trjr - T04), and for a perfect radiator, e = 1. Thus, ' Pnetz = (5.67 x 10- 8 W/m2K4)(4.52 x 10-2 m2)((473 K)4 - (295 K)4)' = 110W. ‘ From Pnet= er( T4 - 111,4): For two identical objects, we have the ratio 0f the hotter to that from the colder is Q: (1200 K14 — (273 K14 = 2.07 x 1012 Fe (1100 104 — (273 104 1.46 x 1012 power emitted by the = 1.4 We use PM: aAe( T4 - T04): 25W: (5.67 x 10-8 W/m2K4)(2.5 x 10- 5 m2)(0.25)(74 - (295 104) From which, r: 2.9 x 103 K = 2.6 x103°C. The power radiated by an object is given by P: erT4 , where Tis the absolute temperature. The power output from Star X (at 6000 K) is therefore PX = an (1)(6000)4 , and the power output of Star Y (at 12000 K) is 159 ...
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chap11 solutions - CHAPTER ELEVEN SOLUTIONS If half of this...

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