so The number of atoms in a pure substance can be found by dividing the mass of the substance by the mass of a single atom. Take the atomic masses of gold and silver from the periodic table. 2.75x10'2 kg NAu _ (196.96655 u/an.mi)(1.66> N = 0,543N NAg 2.75x10'2 kg 196.96655 A\" Ag (107.8682 u/atcmi)(1.66>[T(F)-32] = -[136-32] = 57.8.C Low: T(C)= [T(OF)-32]= $[-129-32]= -89.4.C 5. I(OF) = $I(C)+32=-(38.9.C)+32 = 102.0.F 6. (a) I(OF) = =I(C)+32 = =(-18)+32 =-0.4OF = 0OF (b) I(C)= [I(OF)-32]= [-18-32]=-27.78.C =-28.C 7. The two temperatures, as given by the conversions, are to be the same. T(F) ='T(C)+32 -> x=2x+32 - -32=4x -> x= (-32)2 =-40 Thus, 40 C = -40OF9. Take the 300-111 height to be the height in January. Then the increase in the height of the tower as the temperature rises is given by Eq. 131a. At = wear = (12 > AT=~ > T=Q+=25C+L=8145C=1 810C ate ate 19x10'6/c0 21. Use the relationship that T(K) = [T( F) -32]+273.15. T(K) = [I(OF)-32]+273.15 I(F) = ;[T(K)-273.15]+32 =[0 K-273.15]+32= -459.67 F 22. Use the relationship that T(K) = T(C) + 273.15. (a) T(K) = T(C)+273.15= 4270 K = 4300 K ; T(K) = T(C) + 273.15 = 15x100 K AT 273.15 (b) % error = X100 = x100 T(K) T(K) 4000.C: 273.15 15x106 .C: 273.15 X100 = 6.4% X100 = 1.8x10-3% 4270 15x106 23. Assume the gas is ideal. Since the amount of gas is constant, the value of PV/T is constant. PV _ P2 2 - V2 = 1- P1 12 = (3.50 m3) 1.00 atm (273 + 38.0) K 1.25 m I1 I2 P2 I1 3.20 atm 273 K 24. Assume the air is an ideal gas. Since the amount of air is constant, the value of PV/T is constant. RV1 _ P2/2 - 12 = 1 P2 V2 = (293 K) 40 atm = 1302 K = 1029.C = 1000.C I2 P V1 1 atm25. (a) Assume that the helium is an ideal gas and then use the ideal gas law to calculate the volume. Absolute pressure must be used, even though gauge pressure is given. nRT (16.00 mol)(8.314J/mol . K)(283.15 K) PV = nRT - V = = 0.2754 m P (1.350 atm)(1.013x10' Pa/atm) (b) Since the amount of gas is not changed, the value of PV/T is constant. PV1_ P2 /2 - 12 = 1 1 P2 2 = (283.15 K) 2.00 atm = 210 K = -63.C I2 P V1 1.350 atmAssume that the nitrogen and carbon dioxide are ideal gases and that the volume and tempera hue are . P RT . constant for the two gases. From the ideal gas law. the value of = 7 1s constant. Also note that n concerning the ideal gas law. the identity of the gas is unimportant. as long as the number of moles is considered. 21.6 kg ('02 3 Pl P; n, 44.01x10 kg C 03/1110] =\" > P =P\"= 3.45 atm + = 2.20at111 1:1 \"2 2 11:1 ( ) 21.6 kg N:2 - 28.01 x10'3 kg N2f11101 ((1) Assume the nitrogen is an ideal gas. The number of moles of nitrogen is found from the atomic (5) weight. and then the ideal gas law is used to calculate the volume of the gas. 1 mole N2 _3=101Tmol 28.01x10 kg n = (28.5 kg) P 1.013x105 Pa Hold the volume and temperature constant and again use the ideal gas law. ) 1 mole N2 3=2167 11101 28.01x10' kg )1 = (38.5 kg+32.2 k rIQ PV = \"RT > 7 . P: %= W: 216x105 pa = 213 am 22.79 m3 38. At STP, 1 mole of ideal gas occupies 22.4 L. 1 mole 6.02x1025 molecules IL = 2.69x1025 molecules/m 22.4 L mole 103 m3 39. We assume that the water is at 4 C, so its density is 1000 kg/m'. 1.000 L 10 m3 1000 kg 1 mol 55.51 mol IL (15.9994 + 2x1.00794)x10- kg 55.51 mol 6.022x102 molecules = 3.343x10- molecules 1 mol 40. (a) Since the average depth of the oceans is very small compared to the radius of the Earth, the ocean's volume can be calculated as that of a spherical shell with surface area AT REarth and a thickness Ay. Then use the density of sea water to find the mass and the molecular weight of water to find the number of moles. Volume = 0.75(4/t REarth )Ay = 0.75(47)(6.38x10 m) (3x103 m) = 1.15x1018 m3 1.15x1018 m3 1025 kg 1 mol = 6.55x10-2 moles = 7x10- moles m 18x10 kg (b) 6.55x10-2 moles(6.02x102 molecules/1 mol) = 4x10* molecules41. Assume the gas is ideal at those low pressures and use the ideal gas law. N P 1x10-12 N/m2 PV = NKT 3x108 molecules 10-6 m3 V KT (1.38X10-25 J/K)(273 K) m 1 cm' 300 molecules/cm 42. We assume an ideal gas at STP. Example 13-9 shows that the molar volume of this gas is 22.4 L. We calculate the actual volume of one mole of gas particles, assuming a volume of Co, and then find the ratio of the actual volume of the particles to the volume of the gas. molecules _ (6.02x1025 molecules)((3.0x10-10 m) /molecule) 7.3x10-4 gas (22.4 L)(1x10-3 m3 /1L) The molecules take up less than 0.1% of the volume of the gas. 43. (a) The average translational kinetic energy of a gas molecule is 3 KT. KEavg =KI = (1.38X10-25 J/K)(273 K) = 5.65x10-21 J (b) The total translational kinetic energy is the average kinetic energy per molecule times the number of molecules. 6.02x10- molecules KE total = N(KEavg) = (1.0 mol) 1 mol (1.38x10-23 J/K) (298 K) = 3700 J1. The kilocalorie is the heat needed to raise 1 kg of water by 1 Co. Use this relation to find the change in the temperature. 1 kcal (8200 J) (1 kg)(109) 1 = 0.653 Co 4186 J 1 kcal 3.0 kg Thus, the final temperature is 10.0 C + 0.653 C = 10.7 C 2. The kilocalorie is the heat needed to raise 1 kg of water by 1 Co. Use the definition to find the heat needed. 1 kcal 4186 J (34.0 kg)(95.C-15 C) =1.139x10' J = 1.1x107 J (1 kg)(1 09) 1 kcal 3. Find the mass of warmed water from the volume of water and its density of 1025 kg/m'. Then use the fact that 1 kcal of energy raises 1 kg of water by 1 Co and that the water warms by 25 Co. V = At = m - m= pAt = (1025 kg/m3)(1.0 m2)(0.5x10 3 m) = 0.5125 kg (0.5125 kg)(25 Co) (1 kcal) 1 bar = 12.8 kcal; 12.8 kcal = 0.043 bars = 0.04 bars (1 kg)(1 09) 300 kcalE (a) 2500Ca1 3 = -1.0> a[860m] (c) At 10 cents per day. the food energy costs $0.29 per day . It would be impossible to feed yourself in the United States on this amount of money. 6. The energy generated by using the brakes must equal the car's initial kinetic energy, since its final kinetic energy is 0. 9=2mug =(1300 kg) (95 km/h) 1 m/s = 4.526x10# J = 4.5x104 J 3.6 km/h 4.526x104 J 1 kcal = 108.1 kcal = 110 kcal 4186 J 7. The energy input is causing a certain rise in temperature, which can be expressed as a number of joules per hour per Co. Convert that to mass using the definition of kcal, relates mass to heat energy. 3.2x10' J/ 1 kcal (1 kg)(1 Co) = 254.8 kg/h = 250 kg/h 30 Co 4186 J 1 kcal 8. The wattage rating is 375 joules per second. Note that 1 L of water has a mass of 1 kg. (2.5x10-] L) ( 60 (9 ) 1 kcal 4186 J 1 s = 167 s = 170 s = 2.8 min (1kg)(1 Co) (kcal 375 J 9. The heat absorbed can be calculated from Eq. 14-2. Note that 1 L of water has a mass of 1 kg. Q = mcAT = (18L) 1x10 3 m' 1.0x10' kg (4186 J/kg . Co)(95.C-15.C) = 6.0x106 J IL 1 m'10. The specific heat can be calculated from Eq. 14-2. O 1.35x10' J Q = mcAT - C= = 1715 J/kg . Co = 1700 J/kg . Co MAT (4.1 kg)(37.2.C-18.0.C)11. (a) The heat absorbed can be calculated from Eq. 14-2. Note that 1 L of water has a mass of 1 kg. 1x10 m3 Q = mcAT = (1.0L) 1.0x10' kg (4186 J/kg . Co)(100 C -20 C) IL = 3.349x10' J = 3.3x105 J (b) Power is the rate of energy usage. AE 3.349x10' J P = - At= = = 5582 s = 5600 s = 93 min At At P 60 W14. The heat lost by the copper must be equal to the heat gained by the aluminum and the water. The aluminum and water have the same temperature change. mcuccu (Ticu - Teq) = MAICAl(Teq - TiAl) + MH,OCH,O(Teq - IjH,0) [(0.145 kg)(900 J/kg . Co) (0.265 kg)(390 J/kg . Co)(245.C-Teq) = +(0.825 kg)(4186 J/kg . Co) (Teq - 12.0.C) Teq = 18.532.C = 18.5.C 15. The heat gained by the glass thermometer must be equal to the heat lost by the water. m glass glass (Teq - Tglass) = MH,OCH20 (THO - Teq) (31.5 g)(0.20 cal/g . Co)(41.8.C-23.6.C) = (135 g)(1.00 cal/g . C)(TH,O -41.8.C) THO = 42.6.C17. The heat lost by the iron must be the heat gained by the aluminum and the glycerin. mFeCFe (TiFe - Teq) = MAICAl(Teq - TiAl) + melyCgly (Teq - Tigly) (0.290 kg)(450 J/kg . Co)(142 Co) = (0.095 kg)(900 J/kg . Co)(28 Co) + (0.250 kg)cely (28 Co) Cgly = 2305 J/kg . Co = 2300 J/kg . Co 18. (a) Since O= meAT and O = CAT, equate these two expressions for O and solve for C . O = mcAT = CAT - C = mc (b) For 1.0 kg of water: C=mc = (1.0 kg)(4186 J/kg . Co) = 4.2x103 J/Co (c) For 45 kg of water: C = mc = (45 kg)(4186 J/kg . Co) = 1.9x10' J/Co 19. We assume that all of the kinetic energy of the hammer goes into heating the nail. KE = Q - 8(mhammer hammer ) = mailFeAT -> 8 5m hammer hammer 4(1.20 kg)(7.5m/s)2 AT = = 42.86 Co = 43 CO mnail Fe (0.014 kg)(450 J/kg . Co)20. The heat lost by the substance must be equal to the heat gained by the aluminum, water, and glass. mxx (Tix - Teq) = MAICAl (Teq - TiAl) + MH,OCH,O (Teq - TH,O ) + mglass Cglass (Teq - Tiglass) mAICAl(Teq -TiAl) + MH,OCH,O (Teq - TiH,O) + mglass glass (Teq - Tiglass) mx (Tix - Teq) (0.105 kg)(900 J/kg . Co) + (0.185 kg)(4186 J/kg . Co) (35.0.C-10.5.C) +(0.017 kg)(840 J/kg . Co) = (0.215 kg)(330 C-35.0.C) = 341.16 J/kg . Co = 341 J/kg . Co 21. 65% of the original potential energy of the aluminum goes to heating the aluminum. 0.65 PE = 0 - 0.65mAigh = mAICAlAT - AT 0.65 gh 0.65(9.80 m/s2) (55 m) 0.39 Co CAI (900 J/kg . Co)24. The oxygen is all at the boiling point, so any heat added will cause oxygen to evaporate (as opposed to raising its temperature). We assume that all the heat goes to the oxygen and none to the flask. Q = mLy -> m= 2_ 3.40x10' J 1.6 kg Ly 2.1x10' J/kg 25. The silver must be heated to the melting temperature and then melted. Q = Cheat + melt = mcAT + mLF = (23.50 kg)(230 J/kg . C)(961.C- 25.C) + (23.50 kg)(0.88x10' J/kg) = 7.1x10 J 26. Assume that the heat from the person is only used to evaporate the water. Also, we use the heat of vaporization at room temperature (585 kcal/kg), since the person's temperature is closer to room temperature than 100 C. 185 kcal Q = mLy = m= 0.316 kg = 316 mL Lv 585 kcal/kg 27. The heat lost by the steam condensing and then cooling to 30 C must be equal to the heat gained by the ice melting and then warming to 30 C. I steam [ Ly + CH,O (Tisteam - Teq )] = mice [LF + CH,O (Teq - Tice)] [LF + CH,O (Teq - Tice)] [3.3310' J/kg + (4186 J/kg . Co)(30.C)] m steam = mice (1.00 kg) [Ly + CH,O (Tisteam - Teq )] [22.6X10' J/kg + (4186 J/kg . Co)(30.C)] 0.18 kg