Physics Solution Manual for 1100 and 2101

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Unformatted text preview: TGlass ∆TLiquid 840 J/ ( kg ⋅ C° ) ( 83.0 C° − 53.0 C° ) = = 2500 J/ ( kg ⋅ C° ) ( 53.0 C° − 43.0 C°) 51. REASONING According to Equation 6.10b, the average power is the change in energy divided by the time. The change in energy in this problem is the heat supplied to the water and the coffee mug to raise their temperature from 15 to 100 ºC, which is the boiling point of water. The time is given as three minutes (180 s). The heat Q that must be added to raise the temperature of a substance of mass m by an amount ∆T is given by Equation 12.4 as Q = cm∆T, where c is the specific heat capacity. This equation will be used for the water and the material of which the mug is made. SOLUTION Using Equation 6.10b, we write the average power P as P= Change in energy QWater + QMug = Time Time The heats QWater and QMug each can be expressed with the aid of Equation 12.4, so that we obtain 656 TEMPERATURE AND HEAT P= QWater + QMug Time = cWater mWater ∆T + cMug mMug ∆T Time 4186 J/ ( kg ⋅ C° ) ( 0.25 kg )(100.0 °C − 15 °C ) = + 920 J/ ( kg ⋅ C° ) ( 0.35 kg )(100.0 °C − 15 °C ) 180 s = 650 W The specific heat of water has been taken from Table 12.2. 52. REASONING Two portions of the same liquid that have the same mass, but different initial temperatures, when mixed together, will yield an equilibrium mixture that has a temperature lying exactly midway between the two initial temperatures. That is, the final temperature is T = 1 (T0A + T0B ) . This occurs only when the mixing occurs without any exchange of heat 2 with the surroundings. Then, all of the heat lost from the warmer portion is gained by the cooler portion. Since each portion is identical except for temperature, the warmer portion cools down by the same number of degrees that the cooler portion warms up, yielding a mixture whose final equilibrium temperature is midway between the two initial temperatures. There are two ways to apply the logic above to the problem at hand. Portions A and B have the same mass m, so they will yield a combined mass of 2m. Thus, we can imagine portions A and B mixed together to yield an equilibrium temperature of 1T 1 94.0 °C + 78.0 °C = 86.0 °C . This mixture has a mass 2m, just like the + T0B ) = 2 ( ) 2 ( 0A mass of portion C, so when it is mixed with portion C, the final equilibrium temperature will be 1 ( 86.0 °C + 34.0 °C ) = 60.0 °C . The other way to apply the logic is to mix half of portion 2 C with portion A and half with portion B. This will produce two mixtures with different temperatures and each having a mass 2m, which can then be combined. The final equilibrium temperature of this combination is again 60.0 ºC. Either way, the guiding principle is that the heat gained by the cooler liquid is equal to the heat lost by the warmer liquid. SOLUTION Since we are assuming negligible heat exchange with the surroundings, the principle of conservation of energy applies in the following form: heat lost equals heat gained. In reaching equilibrium the warmer portions lose heat and the cooler portions gain heat. In either case, the heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount ∆T is given by Q = cm∆T (Equation 12.4), where c is the specific heat capacity. The final temperature is 50.0 ºC. We have, then, cC mC ( 50.0 °C − 34.0 °C ) = cA mA ( 94.0 °C − 50.0 °C ) + cB mB ( 78.0 °C − 50.0 °C ) 4 3 14444244443 1444444444 24444444444 Heat gained Heat lost The specific heat capacities for each portion have the same value c, while mA = mB = m. With these substitutions, we find Chapter 12 Problems 657 cmC ( 50.0 °C − 34.0 °C ) = cm ( 94.0 °C − 50.0 °C ) + cm ( 78.0 °C − 50.0 °C ) mC ( 50.0 °C − 34.0 °C ) = m ( 94.0 °C − 50.0 °C ) + m ( 78.0 °C − 50.0 °C ) Solving this equation for the mass mC of portion C gives mC = m ( 94.0 °C − 50.0 °C ) + m ( 78.0 °C − 50.0 °C ) 50.0 °C − 34.0 °C = 4.50 m 53. REASONING When heat Q is supplied to the silver bar, 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 of silver and m is the mass. Solving this equation for m yields m= Q c ∆T (1) When the temperature of the bar changes by an amount ∆T, the change ∆L in its length is given by Equation 12.2 as ∆L = α L0 ∆T . Solving this equation for ∆T gives ∆T = ∆L α L0 (2) where L0 is the initial length and α is the coefficient of linear expansion for silver. Substituting the expression for ∆T in Equation (2) into Equation (1) gives m= α Q L0 Q Q = = c ∆T c ∆L ∆L c α L0 SOLUTION Taking the values for α and c from Tables 12.1 and 12.2, respectively, the mass of the silver bar is m= α Q L0 c ∆L = 19 × 10−6 ( C° ) −1 ( 4200 J )( 0.15 m ) 235 J/ ( kg ⋅ C° ) ( 4.3 × 10−3 m ) = 1.2 × 10−2 kg ______________________________________________________________________________ 658 TEMPERATURE AND HEAT 54. REASONING Each second, the water heater increases the temperature of the water p...
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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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