Physics Solution Manual for 1100 and 2101

In reality the mass of the block decreases as the

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Unformatted text preview: on 12.4 as Q = cm∆T, where c is the specific heat capacity. The amount ∆T by which the temperature changes is the boiling temperature minus the initial temperature of 100.0 ºC. The boiling temperature is the temperature at which the vapor pressure of the water equals the external pressure of 3.0 × 105 Pa and can be read from the vapor pressure curve for water given in Figure 12.32. SOLUTION Using Equation 12.4, with TBP being the boiling temperature and T0 being the initial temperature, we have Q = cm∆T = cm (TBP − T0 ) According to Figure 12.32, an external pressure of 3.0 × 105 Pa corresponds to a boiling point temperature of TBP = 134 ºC. Using this value in Equation 12.4 and taking the specific heat capacity for water from Table 12.2, we determine the heat to be Q = cm (TBP − T0 ) = 4186 J/ ( kg ⋅ C° ) ( 2.0 kg )(134 °C − 100.0 °C ) = 2.8 × 105 J 80. REASONING AND SOLUTION The pressure inside the container is due to the weight on the piston in addition to the pressure of the atmosphere. The 120-kg mass produces a pressure of P= F mg (120 kg ) ( 9.80 m/s 2 ) = = = 1.0 × 105 Pa 2 A π r2 π ( 0.061 m ) If we include atmospheric pressure (1.01 × 105 Pa), the total pressure inside the container is Ptotal = P + Patm = 2.0 × 105 Pa Examination of the vaporization curve for water in Figure 12.32 shows that the temperature corresponding to this pressure at equilibrium is T = 120 °C . ______________________________________________________________________________ Chapter 12 Problems 675 81. REASONING AND SOLUTION At 10 °C the equilibrium vapor pressure is 1250 Pa. At 25 °C, the equilibrium vapor pressure is 3200 Pa. To get the smallest possible value for the relative humidity assume that the air at 10 °C is saturated. That is, take the partial pressure to be 1250 Pa (see the vapor pressure curve for water that accompanies Problem 75). Then we have, 1250 Pa Partial pressure Percent relative humidity = × 100 = × 100 = 39% Equilibrium vapor pressure 3200 Pa ______________________________________________________________________________ 82. REASONING AND SOLUTION Equation 12.6 defines the relative humidity as Partial pressure of water vapor Percent relative humidity = × 100 Equilibrium vapor pressure of water at the existing temeprature According to the vapor pressure curve that accompanies Problem 75, the equilibrium vapor pressure of water at 36°C is 5.8 x 103 Pa. Since the water condenses on the coils when the temperature of the coils is 30 °C, the relative humidity at 30 °C is 100 percent. From Equation 12.6 this implies that the partial pressure of water vapor in the air must be equal to the equilibrium vapor pressure of water at 30 °C. From the vapor pressure curve, this pressure is 4.4 x 103 Pa. Thus, the relative humidity in the room is 4.4 × 10 3 Pa × 100 = 76% 5.8 × 10 3 Pa ______________________________________________________________________________ Percent relative humidity = 83. SSM REASONING We must first find the equilibrium temperature Teq of the iced tea. Once this is known, we can use the vapor pressure curve that accompanies Problem 75 to find the partial pressure of water vapor at that temperature and then estimate the relative humidity using Equation 12.6. According to the principle of energy conservation, when the ice is mixed with the tea, the heat lost by the tea is gained by the ice, or Qtea = Qice . The heat gained by the ice is used to melt the ice at 0.0 °C; the remainder of the heat is used to bring the water at 0.0 °C up to the final equilibrium temperature Teq . SOLUTION Qtea = Qice c water m tea (30.0 °C − Teq ) = m ice Lf + cwater mice ( Teq – 0.00 °C ) 676 TEMPERATURE AND HEAT The specific heat capacity of water is given in Table 12.2, and the latent heat of fusion Lf of water is given in Table 12.3. Solving for Teq , we have Teq = = c water m tea ( 30.0 °C) − m ice Lf c water ( m tea + mice ) [4186 J/(kg ⋅ C °) ] (0.300 kg)(30.0 °C) − (0.0670 kg)(33.5 × 10 4 [4186 J/(kg ⋅ C ° )] (0.300 kg + 0.0670 kg) J/kg) = 9.91 °C According to the vapor pressure curve that accompanies Problem 75, at a temperature of 9.91 °C, the equilibrium vapor pressure is approximately 1250 Pa. At 30 °C, the equilibrium vapor pressure is approximately 4400 Pa. Therefore, according to Equation 12.6, the percent relative humidity is approximately 1250 Pa Percent relative humidity = × 100 = 28% 4400 Pa ______________________________________________________________________________ 84. REASONING AND SOLUTION The water will boil if the vapor pressure of the water is equal to the ambient pressure. The pressure at a depth h in the water can be determined from Equation 11.4: P2 = P + ρ gh . When h = 10.3 m, 1 P2 = (1.01×105 Pa ) + (1.000 ×103 kg/m3 ) ( 9.80 m/s 2 ) (10.3 m ) = 2.02 ×105 Pa The vapor pressure curve in Figure 12.32 shows that the vapor pressure of water is equal to 2.02 ×105 Pa at a temperature of 123 °C. Thus, the water at that depth has a temp...
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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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