Physics Solution Manual for 1100 and 2101

# Solution according to the principle of energy

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Unformatted text preview: elt or liquefy the aluminum is Q2 = mLf , where Lf is the latent heat of fusion of aluminum. Therefore, the total amount of heat which must be added to the aluminum in its solid phase to liquefy it is Qtotal = Q1 + Q2 = m(c∆T + Lf ) SOLUTION Substituting values, we obtain { } Qtotal = (0.45 kg) [9.00 ×102 J/(kg ⋅ C°)](660 °C –130 °C) + 4.0 ×105 J/kg = 3.9 ×105 J Chapter 12 Problems 661 ______________________________________________________________________________ 58. REASONING a. When water changes from the liquid to the ice phase at 0 °C, the amount of heat released is given by Q = mLf (Equation 12.5), where m is the mass of the water and Lf is its latent heat of fusion. b. When heat Q is supplied to the tree, its temperature changes by an amount ∆T. The relation between Q and ∆T is given by Equation 12.4 as Q = c m ∆T , where c is the specific heat capacity and m is the mass. This equation can be used to find the change in the tree’s temperature. SOLUTION 4 a. Taking the latent heat of fusion for water as Lf = 33.5 × 10 J/kg (see Table 12.3), we find that the heat released by the water when it freezes is Q = mLf = ( 7.2 kg ) ( 33.5 ×104 J/kg ) = 2.4 ×106 J b. Solving Equation 12.4 for the change in temperature, we have Q 2.4 × 106 J = = 5.3 C° c m 2.5 × 103 J/ ( kg ⋅ C° ) (180 kg ) ______________________________________________________________________________ ∆T = 59. REASONING AND SOLUTION a. The latent heat of vaporization Lv of water is given in Table 12.3. To change water at 100.0 °C to steam we have from Equation 12.5 that Q = mLv = (2.00 kg)(22.6 × 105 J/kg) = 4.52 × 106 J b. For liquid water at 0.0 °C we must include the heat needed to raise the temperature to the boiling point. This heat is depends on the specific heat c of water, which is given in Table 12.2. Using Equations 12.4 and 12.5, we have Q = (cm∆T)water + mLv 6 Q = [4186 J/(kg⋅C°)](2.00 kg)(100.0 C°) + 4.52 × 10 J = 5.36 × 10 J 6 ______________________________________________________________________________ 662 TEMPERATURE AND HEAT 60. REASONING a. The amount of heat Q required to melt a mass m of a substance is directly proportional to the latent heat of fusion Lf, according to Q = mLf (Equation 12.5). Since each object has the same mass and more heat is required to melt B, the substance from which B is made has the larger latent heat of fusion. We will use Equation 12.5 to determine the latent heat. b. The amount of heat required to melt a substance is also directly proportional to its mass, according to Equation 12.5. If the mass is doubled, the heat required to melt the substance also doubles. SOLUTION a. According to Equation 12.5, the latent heats of fusion for A and B are Lf,A = QA 3.0 × 104 J 4 = = 1.0 × 10 J/kg m 3.0 kg Lf,B = QB 9.0 × 104 J 4 = = 3.0 × 10 J/kg m 3.0 kg b. The amount of heat required to melt object A when its mass is 6.0 kg is ( ) Q = mA Lf ,A = ( 6.0 kg ) 1.0 × 10 J/kg = 6.0 × 10 J 4 4 (12.5) ______________________________________________________________________________ 61. SSM REASONING From the conservation of energy, the heat lost by the mercury is equal to the heat gained by the water. As the mercury loses heat, its temperature decreases; as the water gains heat, its temperature rises to its boiling point. Any remaining heat gained by the water will then be used to vaporize the water. According to Equation 12.4, the heat lost by the mercury is Qmercury = (cm∆T )mercury . The heat required to vaporize the water is, from Equation 12.5, Qvap = (mvap Lv )water . Thus, the total amount of heat gained by the water is Qwater = (cm∆T )water + ( mvap Lv ) water . SOLUTION Qlost by = Qgained by mercury water ( cm∆T ) mercury = ( cm∆T ) water + ( mvap Lv ) water where ∆ Tmercury = (205 °C − 100.0 °C) and ∆ Twater = (100.0 °C − 80.0 °C) . The specific heats of mercury and water are given in Table 12.2, and the latent heat of vaporization of water is given in Table 12.3. Solving for the mass of the water that vaporizes gives Chapter 12 Problems m vap = = 663 c mercury mmercury ∆Tmercury − c water mwater ∆ Twater ( Lv ) water [139 J/(kg ⋅ C ° )](2.10 kg)(105 C ° ) − [4186 J/(kg ⋅ C °)](0.110 kg)(20.0 C ° ) 22.6 × 10 5 J/kg = 9.49 × 10 –3 kg ______________________________________________________________________________ 62. REASONING The minimum required mass mw of water is the mass that just reaches equilibrium with the condensed benzene at the boiling point of benzene (80.1 °C; see Table 12.3 in the text). If less water is used, not all of the benzene will condense, and if more water is used, then the benzene will be cooled below its boiling point before reaching equilibrium. Therefore, the final temperature of the water is 80.1 °C, the boiling point of benzene. The mass mw of the water needed to condense the benzene vapor depends upon the amount of heat Q that the water must absorb from the benzene. The benzene vapor is at its boiling point,...
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