Physics Solution Manual for 1100 and 2101

From equation 126 this implies that the partial

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Unformatted text preview: ed by the unknown material is equal to the heat lost by the glycerin and the calorimeter. The heat gained by the unknown material is used to melt the material and then raise its temperature from the initial value of –25.0 °C to the final equilibrium temperature of Teq = 20.0 °C . SOLUTION Qgained by = Qlost by unknown m u Lf + c u mu ∆Tu + Qlost by glycerine = calorimeter c gl m gl ∆ Tgl + c al m al ∆Tal Taking values for the specific heat capacities of glycerin and aluminum from Table 12.2, we have (0.10 kg) L f + [160 J/(kg ⋅ C ° )](0.10 kg)(45.0 C ° ) = [2410 J/(kg ⋅ C °)](0.100 kg)(7.0 C °) + [9.0 × 10 2 J/(kg ⋅ C° )](0.150 kg)(7.0 C °) Solving for Lf yields, 4 Lf = 1.9 × 10 J/kg ______________________________________________________________________________ Chapter 12 Problems 669 70. REASONING The amount Q of heat required to melt an iceberg at 0 °C is equal to mLf, where m is its mass and Lf is the latent heat of fusion for water (see Table 12.3). The mass is related to the density ρ and the volume V of the ice by Equation 11.1, m = ρ V. SOLUTION a. The amount of heat required to melt the iceberg is Q = mLf = ρ V Lf ( = 917 kg/m 3 (12.5) ) 144444424444443 (3.35 × 10 (120 × 10 m )(35 × 10 m ) ( 230 m ) 3 3 5 J/kg ) Volume = 3.0 × 1020 J b. The number of years it would take to melt the iceberg is equal to the energy required to melt it divided by the energy consumed per year by the U.S. 3.0 × 10 J = 2.7 years 20 1.1 × 10 J/y ______________________________________________________________________________ 20 Number of years = 71. SSM REASONING In order to melt, the bullet must first heat up to 327.3 °C (its melting point) and then undergo a phase change. According to Equation 12.4, the amount of heat necessary to raise the temperature of the bullet to 327.3 °C is Q = cm(327.3 °C − 30.0 °C) , where m is the mass of the bullet. The amount of heat required to melt the bullet is given by Qmelt = mLf , where Lf is the latent heat of fusion of lead. The lead bullet melts completely when it comes to a sudden halt; all of the kinetic energy of the bullet is converted into heat; therefore, KE = Q + Qmelt 1 mv 2 2 = cm (327.3 ° C − 30.0 °C) + mL f The value for the specific heat c of lead is given in Table 12.2, and the value for the latent heat of fusion Lf of lead is given in Table 12.3. This expression can be solved for v, the minimum speed of the bullet for such an event to occur. SOLUTION Solving for v, we find that the minimum speed of the lead bullet is v = 2 L f + 2 c (327.3 °C − 30.0 °C) 670 TEMPERATURE AND HEAT v = 2(2.32 × 10 J/kg) + 2[128 J/(kg ⋅ C°)] (327.3 °C − 30.0 °C) = 3.50 × 10 m/s 4 2 ______________________________________________________________________________ 72. REASONING The mass mremaining of the liquid water that remains at 100 °C is equal to the original mass m minus the mass mvaporized of the liquid water that has been vaporized. The heat Q required to vaporize this mass of liquid is given by Equation 12.5 as Q = m Lv, where Lv is the latent heat of vaporization for water. Thus, we have Q Lv The heat required to vaporize the water comes from the heat that is removed from the water at 0 °C when it changes phase from the liquid state to ice. This heat is also given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion for water. Thus, the remaining mass of liquid water can be written as mremaining = m − mvaporized = m − mremaining = m − mLf L Q =m− = m 1 − f Lv Lv Lv SOLUTION Using the values of Lf and Lv from Table 12.3, we find that the mass of liquid water that remains at 100 °C is Lf 3.35 × 105 J/kg mremaining = m 1 − = ( 2.00 g ) 1 − = 1.70 g 6 Lv 2.26 × 10 J/kg ______________________________________________________________________________ 73. REASONING AND SOLUTION The steel band must be heated so that it can expand to fit the wheel. The diameter of the band must increase in length by an amount ∆L = 6.00 ×10 –4 m . Also, ∆L = α L0∆T, where the coefficient of thermal expansion α of steel is given in Table 12.1, so that ∆T = ∆L 6.00 ×10−4 m = = 5.0 ×101 C° −6 −1 ( α L0 (12 ×10 C° ) 1.00 m ) The heat to expand the steel band comes from the heat released from the steam as it changes to water. Therefore Qstm = Qsb, or mstmLv + (cmstm∆T)water = cmsb∆T Chapter 12 Problems 671 Substituting the values for the latent heat of vaporization of water (see Table 12.3), the specific heat capacity of water (see Table 12.2), and the specific heat capacity of steel (see Table 12.2) gives mstm(22.6 × 105 J/kg) + [4186 J/(kg⋅C°)]mstm(100.0 °C − 70.0 °C) = [452 J/(kg⋅C°)](25.0 kg)(50.0 C°) Solving for the mass of the steam, we obtain mstm = 0.237 kg . ______________________________________________________________________________ 74. REASONING AND SOLUTION From inspection of the graph that accompanies this problem, a pressure of 3.5 × 106 Pa corresponds to a temperature of 0...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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