Physics Solution Manual for 1100 and 2101

Since each portion is identical except for

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Unformatted text preview: the net change in volume for the alcohol thermometer is ∆ Va − ∆ Vg = ( βa − βg )V0 ∆T In each case, this volume change is related to a movement of the liquid into a cylindrical region of the thermometer with volume π r 2 h , where r is the radius of the region and h is the height of the region. For the mercury thermometer, therefore, hm = ( β m − βg )V0 ∆T πr2 Similarly, for the alcohol thermometer ha = (β a − βg )V0∆ T πr2 These two expressions can be combined to give the ratio of the heights, ha /hm . SOLUTION Taking the values for the coefficients of volumetric expansion for methyl alcohol, Pyrex glass, and mercury from Table 12.1, we divide the two expressions for the heights of the liquids in the thermometers and find that ha hm = βa − βg β m − βg = 1200 × 10 –6 (C° ) −1 − 9.9 × 10 –6 (C ° ) −1 = 6.9 182 × 10 –6 (C° ) −1 − 9.9 × 10 –6 (C ° ) −1 Therefore, the degree marks are 6.9 times further apart on the alcohol thermometer than on the mercury thermometer. ______________________________________________________________________________ 42. REASONING AND SOLUTION When the temperature is 0.0 °C, P = ρ0gh0, and when the temperature is 38.0 °C, P = ρgh. Equating and solving for h gives h = (ρ0/ρ)h0. Now ρ0/ρ = V/V0, since the mass of mercury in the tube remains constant. Thus, we have that h = (V/V0)h0. Now, ∆V = V − V0 = βV0∆T or V/V0 = 1 + β ∆T Therefore, h = (1 + β ∆T)h0 = {1 + [182 × 10−6 (C°)−1](38.0 °C − 0.0 °C)}(0.760 m) = 0.765 m where we have taken the value for the coefficient of volumetric expansion β for mercury from Table 12.1. ______________________________________________________________________________ Chapter 12 Problems 43. 651 SSM REASONING Since there is no heat lost or gained by the system, the heat lost by the water in cooling down must be equal to the heat gained by the thermometer in warming 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, m is the mass, and ∆T is the change in temperature. Thus, we have that cH O mH O ∆TH O = ctherm mtherm ∆Ttherm 2 2 2 144 2444 4 3 1442443 Heat lost by water Heat gained by thermometer We can use this equation to find the temperature of the water before the insertion of the thermometer. SOLUTION Solving the equation above for ∆TH2O , and using the value of cH O from Table 2 12.2, we have ∆TH O = 2 ctherm mtherm ∆Ttherm cH O mH O 2 2 815 J/ ( kg ⋅ C° ) ( 31.0 g )( 41.5 °C − 12.0 °C ) = = 1.50 C° 4186 J/ ( kg ⋅ C° ) (119 g ) The temperature of the water before the insertion of the thermometer was T = 41.5 °C + 1.50 C° = 43.0 °C ______________________________________________________________________________ 44. REASONING According to Equation 12.4, the heat required to warm the pool can be calculated from Q = cm∆T . The specific heat capacity c of water is given in Table 12.2. In order to use Equation 12.4, we must first determine the mass of the water in the pool. Equation 11.1 indicates that the mass can be calculated from m = ρ V , where ρ is the density of water and V is the volume of water in the pool. SOLUTION Combining these two expressions, we have Q = cρV∆T , or Q = [4186 J/(kg ⋅ C °) ](1.00 × 10 3 kg/m 3 )(12.0 m × 9.00 m × 1.5 m) (27 ° C − 15 °C) = 8.14 × 10 9 J Using the fact that 1 kWh = 3.6 × 106 J , the cost of using electrical energy to heat the water in the pool at a cost of $0.10 per kWh is $0.10 = $230. 3.6 × 106 J ______________________________________________________________________________ 9 (8.14 × 10 J) 652 TEMPERATURE AND HEAT 45. REASONING We assume that no heat is lost through the chest to the outside. Then, energy conservation dictates that the heat gained by the soda is equal to the heat lost by the watermelon in reaching the final temperature Tf . Each quantity of heat is given by Equation 12.4, Q = cm∆T , where we write the change in temperature ∆ T as the higher temperature minus the lower temperature. SOLUTION Starting with the statement of energy conservation, we have Heat gained by soda = Heat lost by watermelon (cm∆T )soda = (cm∆ T )watermelon Since the watermelon is being treated like water, we take the specific heat capacity of water from Table 12.2. Thus, the above equation becomes [3800 J/(kg ⋅ C°)](12 × 0.35 kg)( Tf – 5.0 ° C) = [4186 J/(kg ⋅ C° )](6.5 kg)(27 °C – Tf ) Suppressing units for convenience and algebraically simplifying, we have 4 4 5 4 1.6 × 10 Tf – 8.0 × 10 = 7.3 × 10 – 2.7 × 10 Tf Solving for Tf , we obtain Tf = 8.1 × 10 5 4 = 19 °C 4.3 × 10 ______________________________________________________________________________ 46. REASONING The volume V of a mass m of water is given by V = m ρ (Equation 11.1), where ρ is the mass density of water (see Table 11.1). In order to warm a mass m of ice water to body temperature, the body must provide an amount Q of hea...
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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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