17.1: From Eq.         F.1.134327.5659b) F.0.81328.6259a) ,1.17     F.0.88321.3159c)  17.2: From Eq.         C.7.413210795b) C.0.5320.4195a) ,2.17     C.8.2732189/5c)  17.3:  F0.720.40so ,FC1 5 9 F140.2F0.70 12  TT 17.4: a) C.6.55)440.56(9)5(b) C.2.27))0.4((45.0)95(  17.5:    F,4.104322.4059(17.1),Eq.Froma)  which is cause for worry.    F54or F,6.53321259b)  to two figures. 17.6:     F2.218.1159 17.7:  FC1K1 5 9 , so a temperature increase of 10 K corresponds to an increase of 18 F . Beaker B has the higher temperature. 17.8: For    C0.10(a),for Then .C0.10),b( 5 9 C 5 9 FKC TTTT .F0.18  17.9: Combining Eq. (17.2) and Eq. (17.3),   ,15.27332 9 5  FK TT and substitution of the given Fahrenheit temperatures gives a) 216.5 K, b) 325.9 K, c) 205.4 K. 17.10: (In these calculations, extra figures were kept in the intermediate calculations to arrive at the numerical results.)  )85.126)(5/9(C,12715.273400a) FC TT F.1079.232)1055.1)(5/9(C,1055.115.2731055.1c) F.28932)15.178)(5/9(C,17815.27395b)F.26032 77 F 77 C FC   TT TT 17.11: From Eq. K.23.2715.273)C92.245(),3.17( K T 17.12: From Eq. C.1769273.15K2042.14K)3.16(7.476)(27),4.17(  17.13: From Eq.   mm.444)mm0.325(),4.17( K273.16 K15.373  17.14: On the Kelvin scale, the triple point is 273.16 K, so R.491.69K5(9/5)273.1R  One could also look at Figure 17.7 and note that the Fahrenheit scale extends from F32 toF460  and conclude that the triple point is about 492 R. 17.15: From the point-slope formula for a straight line (or linear regression, which, while perhaps not appropriate, may be convenient for some calculators), C,33.282 Pa1080.4Pa106.50 Pa1080.4 )C0.100()C01.0( 44 4     which is C282 to three figures. b) Equation (17.4) was not obeyed precisely. If it were, the pressure at the triple point would be   Pa.1076.4)16.273( 4 15.373 Pa1050.6 4   P 17.16:             C,168m1.62C104.2m1025 1 52 0    αLLT so the temperature is C183 . 17.17: .TL m39.0C)5.0)(Cm)(18.01410)()C(102.1( 15 0   17.18: )1( Tαddd  C)))78.0(C0.23)()C(10(2.4cm)(14500.0( 15   mm.4.511cm4511.0  17.19: a) cm,101.4C)(28.0cm)90.1)()C(106.2( 315 0  TαD so the diameter is 1.9014 cm. b) cm,106.3 3 0  TαD so the diameter is 1.8964 cm. 17.20: .102.9C)19.5C)(5.00)C(100.2( 415  Tα 17.21:           C0.25m10125.40m103.2)()( 24 0 TLLα   .C103.2 1 5    17.22: From Eq. (17.8), C.4.49soC,4.29 15 3 0 K101.5 1050.1       TT β VV 17.23        L,11C0.9L1700C1075 1 5 0    TVβ so there is 11 L of air. 17.24: The temperature change is .C0.14C0.32C0.18 T The volume of ethanol contracts more than the volume of the steel tank does, so the additional amount of ethanol that can be put into the tank is   TVββVV  0ethanolsteelethanolsteel     15 1 5 )C(1075C106.3         33 m0280.0C0.14m80.2  17.25: The amount of mercury that overflows is the difference between the volume change of the mercury and that of the glass;         .C107.1 0.55cm1000 cm95.8 K100.18 1 5 3 3 15 glass      C β 17.26: a) .222, 0 22 ALLLALA L L L L   But ,Tα L L   and so   .22 00 TAαTAαA  b)         .m104.1C5.12)m275.())C(104.2(22 24 2 15   πTAαA  17.27: a)   .cm431.1cm350.1 44 2 2 2 0  ππD A b)             .cm437.1C150C1020.121cm431.121 252 0   TαAA 17.28: (a ) No, the brass expands more than the steel. (b) call  D the inside diameter of the steel cylinder at BRST :C150At C20 DD    cm026.25 )C130)()C(102.1(1 )C130)()C(10(2.01cm)25( 1 )cm(125 )cm25(cm25T Dcm000.25 15 15 ST BR BRST BRST             Tα Tα D T αDαD DD    17.29: The aluminum ruler expands to a new length of cm048.20)]C100)()C(10(2.4cm)[10.20()1( 15 0   TαLL The brass ruler expands to a new length of cm040.20)]C100)()C(10(2.0cm)[10.20()1( 15 0   TαLL The section of the aluminum ruler will be longer by 0.008 cm 17.30: From Eq. (17.12), N.100.4 )m10C)(2.01110)()C(10Pa)(2.0109.0( 4 241511     TAYαF 17.31: a)   .)C(102.3)Cm)(400(1.50m)109.1()( 152 0   TLLα b) Pa.102.5m)(1.50m)10Pa)(1.9100.2( 9211 0   LLYTYα 17.32: a) m.105.0m)K)(12.00.35)(K102.1( 315   TLαL b) Using absolute values in Eq. (17.12),  A F Pa.108.4K)0.35)(K10Pa)(1.2100.2( 71511   TYα 17.33: a) J38)KkgJ1020()Lkg103.1)(L50.0))(C20(C37( 3   b) There will be 1200 breaths per hour, so the heat lost is J.104.6J)38)(1200( 4  17.34: s,104.1 ) W1200( )C7)(KkgJ3480)(kg70( 3      P Tmc P Q t about 24 min. 17.35: Using Q=mgh in Eq. (17.13) and solving for Τ gives .C53.0 )kg.KJ4190( )m225)(sm80.9( 2  c gh T 17.36: a) The work done by friction is the loss of mechanical energy, J.1054.1 )sm50.2( 2 1 9.36sin)m00.8)(sm80.9()kg0.35()( 2 1 3 222 2 2 1         vvmmgh b) Using the result of part (a) for Q in Eq. (17.13) gives       .C1021.1KkgJ3650kg0.35J1054.1 23   T 17.37:          J.1003.3KkgJ470kg30.0KkgJ910kg60.1C20C210 5  17.38: Assuming   ,1060.0 KQ             .C1.45 KkgJ910kg1000.8 sm80.7kg80.16 61060.0 3 2 2 1 2 2 1     mc MV mc K T 17.39:          KkgJ4190kg80.1KkgJ910kg50.1C20.0C0.85  J.1079.5 5  17.40: a)     J.1005.8K0.60KkgJ4190kg320.0 4  TmcQ b) s.402 W200 J1005.8 4   P Q t 17.41: a)       K.kgJ1051.2 C55.18C54.22kg780.0 W0.65s120 3      Tm Q c b) An overstimate; the heat Q is in reality less than the power times the time interval. 17.42: The temperature change is soK,0.18T       K.kgJ240 K0.18N4.28 J1025.1sm80.9 42        Tw gQ Tm Q c 17.43: a) KkgJ470 ,  cTmcQ We need to find the mass of 3.00 mol:     kg1675.0molkg10845.55mol00.3 3   nMm         C114K114KkgJ470kg1675.0J8950mcQT b) For  C35.6 kg,00.3 mcQTm c) The result of part (a) is much larger; 3.00 kg is more material than 3.00 mol. 17.44: (a)    kgJ000,30 kg50.0 min5.1minJ000,10 melt  m Q L F (b) Tm Q cTmcQLiquid   :          CkgJ000,1 C30kg50.0 min5.1minJ000,10 c            CkgJ1300 C15kg50.0 min0.1minJ000,10 : Tm Q cSolid 17.45: a) 0 metalwater  QQ 0 metalmetalmetalwaterwaterwater  TcmTcm         0C0.78kg500.0C0.2KkgJ4190kg00.1 metal  c KJ/kg215  metal c b) Water has a larger specific heat capacity so stores more heat per degree of temperature change. c) If some heat went into the styrofoam then metal Q should actually be larger than in part (a), so the true metal c is larger than we calculated; the value we calculated would be smaller than the true value. 17.46: a) Let the man be designated by the subscript m and the “‘water” by w, and T is the final equilibrium temperature. wwwmmm TCmTCm      wwwmmm TTCmTTCm      wwwmmm TTCmTTCm  Or solving for T, . wwmm wwwmmm CmCm TCmTCm T    Inserting numbers, and realizing we can change K to C , and the mass of water is .355 kg, we get )kgJ4190(kg)355.0()kg.J3480(kg)0.70( )0.12()kgJ4190(kg)355.0()0.37()KkgJ3480(kg)0.70( CC CCC T    Thus, .85.36 CT  b) It is possible a sensitive digital thermometer could measure this change since they can read to .C1.  It is best to refrain from drinking cold fluids prior to orally measuring a body temperature due to cooling of the mouth. 17.47: The rate of heat loss is   numbers,gInterestin.or ,. )( t Q tmC t TmC t Q ttQ         6.7or d,005.0 dayJ107 C)C)(0.15kg.Jkg)(3480355.70( 6    tt minutes. This may acount for mothers taking the temperature of a sick child several minutes after the child has something to drink. 17.48: )( f LTcmQ    Btu.136kcal34.2J1043.1 kgJ10334K)0.18)(KkgJ(4190kg)350.0( 5 3   17.49: )( Vwaterwaterficeice LTcLTcmQ  Btu.34.5kcal8.69J1064.3 kgJ102256K)kgJ)(4190C100( kgJ10334)CK)(10.0kgJ(2100 kg)100.12( 4 3 3 3              17.50: a) .min7.21 )minJ (800 K) K)(15.0kgJkg)(2100550.0(      P Tmc P Q t b) min,230 )minJ(800 )kgJ10kg)(334550.0( 3 f   P mL so the time until the ice has melted is min.252min230min 7.21  17.51:   hr.Btu102.40kW01.7 s)(86,400 )kgJ10334()kglb2.205lb)4000( 4 3   17.52: a)     kgJ102256K)K)(66.0kgJ(4190kg)100.25()( 33 v LTcm J.106.91K)K)(66.0kgJkg)(4190100.25(b) J.1033.6 334   Tmc c) Steam burns are far more severe than hot-water burns. 17.53: With KQmvKLTcmQ  setting,)2/1(and)( 2 f and solving for v gives   .sm357)kgJ105.24CK)(302.3KgJ130(2 3 v 17.54: a) g.101 )kgJ10(2.42 K)K)(1.00kgJkg)(34800.70( 6 v sweat       L TMc m b) This much water has a volume of 101 ,cm 3 about a third of a can of soda. 17.55: The mass of water that the camel saves is kg,45.3 )kgJ10(2.42 K)K)(6.0kgJkg)(3480400( 6 v      L TMc which is a volume of 3.45 L. 17.56: For this case, the algebra reduces to C.1.35 ).KkgJ4190)(kg240.0( )KkgJ390)(kg1000.3)(200(( )C0.20)(kgJ4190)(kg240.0( )C0.100)(KkgJ390))(kg1000.3)(200(( 3 3                         T 17.57: The algebra reduces to C5.27 )KkgJkg)(470(0.250 ))KkgJkg)(4190(0.170)KkgJkg)(390500.0(( C))(85.0KkgJkg)(470(0.250 C))(20.0KkgJkg)(4190(0.170)KkgJkg)(390500.0((                       T 17.58: The heat lost by the sample is the heat gained by the calorimeter and water, and the heat capacity of the sample is )Ckg)(73.9(0.0850 )C))(7.1KkgJkg)(390(0.150)KkgJkg)(4190200.0((      Tm Q c = K,kgJ1010  or KkgJ1000  to the two figures to which the temperature change is known. 17.59: The heat lost by the original water is J,10714.4)C0.45)(KkgJ4190)(kg250.0( 4  Q and the mass of the ice needed is waterwaterfice ice TcLTc Q m ice    )C0.30)(KkgJ4190()kgJ10334()C0.20(K)kgJ(2100 J)10714.4( 3 4    g.94.0kg1040.9 2   17.60: The heat lost by the sample (and vial) melts a mass m, where g.08.3 )kgJ10(334 K)))(19.5KkgJg)(2800(6.0)KkgJg)(22500.16(( 3 f     L Q m Since this is less than the mass of ice, not all of the ice melts, and the sample is indeed cooled to C.0 Note that conversion from grams to kilograms was not necessary. 17.61: kg.10.2 kg)J10334( )CK)(750kgJkg)(23400.4( 3    17.62: Equating the heat lost by the lead to the heat gained by the calorimeter (including the water-ice mixtue), .)()200( ficecucuicewPb b P LmTcmTcmmTCcm w  Solving for the final temperature T and using numerical values, C.4.21 K)kgJkg)(390100.0( K)kgJkg)(4190178.0( kgK)Jkg)(130750.0( kg)J10kg)(334018.0( )CK)(255kgJkg)(130750.0( 3                          T (The fact that a positive Celsius temperature was obtained indicates that all of the ice does indeed melt.) [...]...  10.0C 3.0  10 2 m  11 W m 2 17. 68: a) From Eq 17. 21,  H  0.040 W m  K  1040 m 2   140 K    196 W, 4.0  10 m 2 or 200 W to two figures b) The result of part (a) is the needed power input 17. 69: From Eq 17. 23, the energy that flows in time t is Ht    AT 125 ft 2 34F t  5.0 h   708 Btu  7.5  105 J R 30 ft 2  F  h Btu   17. 70: a) The heat current will be the... Solving for T gives T  90.2C 17. 74: From Eq (17. 25), with e  1, a) (5.67  108 W m 2  K 4 )(273 K) 4  315 W m 2 b) A factor of ten increase in temperature results in a factor of 10 4 increase in the output; 3.15  10 6 W m 2 17. 75: Repeating the calculation with Ts  273 K  5.0 C  278 K gives H  167 W 17. 76: The power input will be equal to H net as given in Eq (17. 26); P  Aeσ (T 4  Ts4 )... are [ W][m] W  2 [m ][K] m  K 17. 65: 100 a) 0.450Km  222 K m b)(385 W m  K)(1.25  10-4 m 2 )(400 K m)  10.7 W c)100.0C  (222 K m)(12.00  102 m)  73.3C 17. 66: Using the chain rule, H  dQ  Lf dm and solving Eq (17. 21) for k, dt dt dm L k  Lf dt AT (8.50  10 3 kg) (60.0  10 2 s)  (334  103 J kg ) (600 s) (1.250  10 4 m 2 )(100 K)  227 W m  K 17. 67: (Although it may be easier... 105 (C) 1  1.2  105 (C) 1 )(120C)  1.92  108 Pa 17. 90: In deriving Eq (17. 12), it was assumed that L  0; if this is not the case when there are both thermal and tensile stresses, Eq (17. 12) becomes F   L  L0  αT   AY   For the situation in this problem, there are two length changes which must sum to zero, and so Eq (17. 12) may be extended to two materials a and b in the form... cold spell 17. 114: Equation (17. 21) becomes H  kA T x a) H  (380 J kg  K)(2.50  104 m 2 )(140 C m)  13.3 W b) Denoting the points as 1 and 2, H 2  H1  dQ dt  mc T Solving for t T x at 2, T T mc T   x 2 x 1 kA t The mass m is ρAx, so the factor multiplying T in the above expression is t cρ k x  137 s m Then, T x  140 C m  (137 s m)(0.250 C s)  174 C m 2 17. 115: The... s) (50.2 W m  K )(0.150 m 2 ) and the temperature of the bottom of the pot is 100 C  6 C  106 C 17. 71: 17. 72: Q T  kA t L W   300 K   150 J s   50.2  A  m K   0.500 m   A  4.98  10 3 m 2 D A  πR  π   2 2 2 D  4A π  4(4.98  10 3 m 2 ) π  8.0  10 2 m  8.0 cm 17. 73: H a  H b (a  aluminum, b  brass) A(150.0C  T ) A(T  0 C) H a  ka , H b  kb La Lb (It has... C)(37 C) (70 kg)(9.8 m s 2 )   1.08  105 m  108 km 17. 99: a) (90)(100 W)(3000 s)  2.7  10 7 J Q 2.7  107 J Q    6.89 C, cm cρV (1020 J kg  K)(1.20 kg m 3 )(3200 m 3 ) or 6.9 C to the more apropriate two figures c) The answers to both parts (a) and (b) are multiplied by 2.8, and the temperature rises by 19.3 C b) ΔT  17. 100: See Problem 17. 97 Denoting C by C  a  bT , a and b independent... 0.408 kg of water 17. 108: Solving Eq (17. 21) for k, Τ (3.9  102 m)  (180 W) kH  5.0  10 2 W m  K 2 (2.18 m )(65.0 K) ΑΤ   Τ 28.0 C  (0.120 J mol.K) (2.00  0.95 m 2 )  5.0  10 2 m  1.8  10 2 m   L    93.9 W b) The flow through the wood part of the door is reduced by a factor of ( 0.50 ) 2 1  ( 2.000.95)  0.868 to 81.5 W The heat flow through the glass is 17. 109: a) H  kΑ... 45.0 W,    81.5  45.0 and so the ratio is 93.9  1.35 17. 110: R1  L k1 , R2  L k2 , H1  H 2 , and so T1  H A R1 , Τ 2  H A R2 The temperature difference across the combination is Τ  Τ1  T2  H H ( R1  R2 )  R, A A so, R  R1  R2 17. 111: The ratio will be the inverse of the ratio of the total thermal resistance, as given by Eq (17. 24) With two panes of glass with the air trapped in... 1.00 m   5.39 W b) The length of the steel rod may be found by using the above value of H in Eq 17. 21 and solving for L2 , or, since H and A are the same for the rods, L2  L 50.2 W m  K 65.0 K   0.242 m k2 T2  1.00 m  385.0 W m  K 35.0 K  k T Using H  Lv dm (see Problem 17. 66) in Eq (17. 21), dt dm L T  Lv dt kA (0.390 kg) (0.85  102 m) 3  (2256  10 J kg)  5.5 C, (180 s) (50.2 . F.28932)15 .178 )(5/9(C ,178 15.27395b)F.26032 77 F 77 C FC   TT TT 17. 11: From Eq. K.23.2715.273)C92.245(),3 .17( K T 17. 12: From. T 17. 12: From Eq. C .176 9273.15K2042.14K)3.16(7.476)(27),4 .17(  17. 13: From Eq.   mm.444)mm0.325(),4 .17( K273.16 K15.373  17. 14: On the Kelvin scale,