Physics Solution Manual for 1100 and 2101

3 this expression can be solved for v the minimum

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Unformatted text preview: so the heat Q it must give up is only the amount of heat associated with the phase change from vapor to liquid. This amount of heat is found from Q = mb Lv (Equation 12.5), where mb is the mass of the benzene vapor and Lv is the heat of vaporization for benzene (see Table 12.3 in the text). As the water absorbs heat from the condensing benzene, its temperature will rise by an amount ∆T. We will use Q = cmw ∆T (Equation 12.4), where c is the specific heat of water and mw is the mass of water mixed with the benzene vapor, to determine the amount of heat Q absorbed by the water. Assuming that heat is exchanged only between the water and the benzene, this is the same as the amount of heat given up by the benzene. SOLUTION Solving Q = cmw ∆T (Equation 12.4) for the required mass mw of water, we obtain Q (1) mw = c ∆T Substituting Q = mb Lv (Equation 12.5) into Equation (1) yields the required mass of water: mw = mb Lv c∆T = ( 0.054 kg ) ( 3.94 × 105 J/kg ) ( )( 4186 J/ kg ⋅ Co 80.1o C − 41o C ) = 0.13 kg 664 TEMPERATURE AND HEAT 63. REASONING As the body perspires, heat Q must be added to change the water from the liquid to the gaseous state. The amount of heat depends on the mass m of the water and the latent heat of vaporization Lv, according to Q = mLv (Equation 12.5). SOLUTION The mass of water lost to perspiration is 4186 J 1 Calorie ( 240 Calories ) Q = = 0.42 kg Lv 2.42 × 106 J/kg ______________________________________________________________________________ m= 64. REASONING AND SOLUTION The heat required is Q = mLf + cm∆T, where m = ρV. See Table 12.2 for the specific heat c and Table 12.3 for the latent heat Lf. Thus, Q = ρVLf + cρV∆T 3 Q = (917 kg/m )(4.50 × 10 –4 2 4 m)(1.25 m ){33.5 × 10 J/kg + [2.00 × 103 J/(kg⋅C°)](12.0 C°)} = 1.85 × 10 5 J ______________________________________________________________________________ 65. REASONING Since there is no heat lost or gained by the system, the heat lost by the coffee in cooling down must be equal to the heat gained by the ice as it melts plus the heat gained by the melted water as it subsequently heats up. The heat Q lost or gained by a substance is given by Equation 12.4 as Q = cm∆T, where c is the specific heat capacity (see Table 12.2), m is the mass, and ∆T is the change in temperature. The heat that is required to change ice at 0 °C into liquid water at 0 °C is given by Equation 12.5 as Q = miceLf, where mice is the mass of ice and Lf is the latent heat of fusion for water (see Table 12.3). Thus, we have that mcoffee c ∆T = mice Lf + mice cwater ∆Twater 144coffee coffee 1444 24444 4 2444 3 4 3 Heat lost by coffee Heat gained by ice and liquid water The mass of the coffee can be expressed in terms of its density as mcoffee= ρcoffeeVcoffee (Equation 11.1). The change in temperature of the coffee is ∆Tcoffee= 85 °C – T, where T is the final temperature of the coffee. The change in temperature of the water is ∆Twater = T – 0 °C. With these substitutions, the equation above becomes ρcoffeeVcoffee ccoffee ( 85 °C − T ) = mice Lf + micecwater (T − 0 °C ) Solving this equation for the final temperature gives Chapter 12 Problems T= 665 −mice Lf + ρ coffeeVcoffee ccoffee ( 85 °C ) ρ coffeeVcoffee ccoffee + mice cwater − ( 2 ) (11 × 10−3 kg )( 3.35 × 105 J/kg ) + (1.0 × 103 kg/m3 )(150 × 10−6 m3 ) 4186 J/ ( kg ⋅ C° ) ( 85 °C ) = 3 3 −6 3 ( 2 ) (11 × 10−3 kg ) 4186 J/ ( kg ⋅ C° ) (1.0 × 10 kg/m )(150 × 10 m ) 4186 J/ ( kg ⋅ C°) + = 64 °C ______________________________________________________________________________ 66. REASONING The droplets of water in the air give up an amount of heat Qw while cooling to the freezing point, and an amount Qf while freezing into snow. The snow loses a further amount of heat Qs as it cools from the freezing point to the air temperature. The total amount of heat transferred to the air as the lake water is transformed into a blanket of snow is the sum of the heat transferred during all three processes: Qtotal = Qw + Qf + Qs (1) The heat lost by the cooling water droplets is found from Qw = cw m∆Tw (Equation 12.4), where m is the total mass of water sprayed into the air in a minute, cw is the specific heat capacity of water, and ∆Tw = 12.0 C° is the difference between the temperature of the lake water (12.0 °C) and the freezing point (0 °C). For the freezing process, the heat loss is given by Qf = mLf (Equation 12.5), where Lf is the latent heat of fusion of water (see Table 12.3 in the text). Lastly, the heat lost by the snow as it cools to the air temperature is given by Qs = cs m∆Ts (Equation 12.4), where ∆Ts = 7.0 C° is the higher initial temperature of the snow (0.0 °C, the freezing point) minus the lower final temperature (−7.0 °C, air temperature), and cs is the specific heat capacity of snow. SOLUTION Substituting Qw = cw m∆Tw (Equation 12.4), Qf = mLf (Equation 12.5), and Qs = cs m∆Ts (Equation 12.4) into Equation (1) yields Qtotal = Qw + Qf + Qs = cw m∆Tw + mLf +...
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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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