chap11 solutions

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

• Homework Help
• PresidentHackerCaribou10582
• 4

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

• Fall '01
• Shapiro

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern