Physics Solution Manual for 1100 and 2101

Table 122 gives the specific heat capacity c of steel

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: erature of T = 123 °C . ______________________________________________________________________________ 85. REASONING The change in length ∆L of the pipe is proportional to the coefficient of linear expansion α for steel, the original length L0 of the pipe, and the change in temperature ∆T . The coefficient of linear expansion for steel can be found in Table 12.1. SOLUTION The change in length of the pipe is −1 ∆L = α L0 ∆T = 1.2 × 10−5 ( C° ) ( 65 m ) 18 °C − ( −45 °C ) = 4.9 × 10−2 m (12.2) Chapter 12 Problems 677 86. REASONING AND SOLUTION The heat required to evaporate the water is Q = mLv, and to lower the temperature of the jogger we have Q = mjc∆T. Equating these two expressions and solving for the mass m of the water, we have m= m= m jc ∆T Lv ( 75 kg ) [3500 J/(kg ⋅ C°)] (1.5 C° ) = 0.16 kg 2.42 ×106 J/kg ______________________________________________________________________________ 87. SSM REASONING From the principle of conservation of energy, the heat lost by the coin must be equal to the heat gained by the liquid nitrogen. The heat lost by the silver coin is, from Equation 12.4, Q = ccoin m coin ∆Tcoin (see Table 12.2 for the specific heat capacity of silver). If the liquid nitrogen is at its boiling point, –195.8 °C, then the heat gained by the nitrogen will cause it to change phase from a liquid to a vapor. The heat gained by the liquid nitrogen is Q = mnitrogen Lv , where mnitrogen is the mass of liquid nitrogen that vaporizes, and Lv is the latent heat of vaporization for nitrogen (see Table 12.3). SOLUTION Qlost by = Qgained by coin nitrogen c coin mcoin ∆ Tcoin = mnitrogen Lv Solving for the mass of the nitrogen that vaporizes m nitrogen = = c coin m coin ∆ Tcoin Lv [235 J/(kg ⋅ C °)](1.5 × 10 –2 kg)[25 °C – (–195.8 °C)] = 3.9 × 10 −3 kg 2.00 × 10 J/kg ______________________________________________________________________________ 5 88. REASONING Since the container is being ignored and since we are assuming negligible heat exchange with the environment, the principle of conservation of energy applies in the following form: heat gained equals heat lost. In reaching equilibrium the colder aluminum gains heat in warming to 0.0 ºC, and the warmer water loses heat in cooling to 0.0 ºC. 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 Equation 12.4 as Q = cm∆T, where c is the specific heat capacity. In using this equation as we apply the energy-conservation principle, we must remember to express the change in temperature ∆T as the higher minus the lower temperature. The water that freezes into ice also loses heat. The heat Q lost when a mass m 678 TEMPERATURE AND HEAT of water freezes is given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion. By including this amount of lost heat in the energy-conservation equation, we will be able to calculate the mass of water that is frozen. SOLUTION Using the energy-conservation principle and Equations 12.4 and 12.5 gives cAluminum mAluminum ∆TAluminum = cWater mWater ∆TWater + 144444 244444 3 144 2444 4 3 Heat gained by aluminum Heat lost by water mIce L 1444 f, Water 3 2444 Heat lost by water that freezes Solving for mIce, taking values for the specific heat capacities from Table 12.2, and taking the latent heat for water from Table 12.3, we find that mIce = cAluminum mAluminum ∆TAluminum − cWater mWater ∆TWater Lf, Water 9.00 × 102 J/ ( kg ⋅ C° ) ( 0.200 kg ) 0.0 °C − ( −155 °C ) = 4 33.5 ×10 J/kg 4186J/ ( kg ⋅ C° ) (1.5 kg )( 3.0 °C − 0.0 °C ) − = 0.027 kg 33.5 ×104 J/kg 89. REASONING AND SOLUTION The value for the coefficient of thermal expansion of steel is given in Table 12.1. The relation, ∆L = α L0∆T, written in terms of the diameter d of the rod, is ∆d 0.0026 cm ∆T = = = 110 C° (12.2) α d0 12 × 10−6 ( C° )−1 ( 2.0026 cm ) ______________________________________________________________________________ 90. REASONING To determine the fractional decrease in length ∆L , we need the L0,Silver + L0, Gold decrease ∆L in the rod’s length. It is the sum of the decreases in the silver part and the gold part of the rod, or ∆L = ∆LSilver + ∆LGold. Each of the decreases can be expressed in terms of the coefficient of linear expansion α, the initial length L0, and the change in temperature ∆T, according to Equation 12.2. Chapter 12 Problems 679 SOLUTION Using Equation 12.2 to express the decrease in length of each part of the rod, we find the total decrease in the rod’s length to be ∆L = αSilver L0, Silver ∆T + α Gold L0, Gold ∆T 1442443 144 4 4 244 3 ∆LSilver ∆LGold The fractional decrease in the rod’s length is, then, αSilver L0, Silver ∆T + α Gold L0, Gold ∆T ∆L = L0,Silver + L0, Gold L0,Silver + L0, Gold L0, Silver L0, Gold = αSilver ∆T + α Gold ∆T L L + L0, Gold + L0, Gold 0,Silver2444 0,Silver2444 144 4 3 144 4 3 Si...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online