Physics Solution Manual for 1100 and 2101

71wg b the number n of water molecules is the product

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Unformatted text preview: at a rate that is given by (Equation 13.1), where the bottom of t L a pot has a thermal conductivity k, a cross-sectional area A, and a thickness L. The thermal conductivities of copper and aluminum (kCu, kAl) are different (see Table 13.1), but the two pot bottoms are identical in every other respect. Further, because both pots are boiling away Q at the same rate, the flow of heat must occur at the same rate through both bottoms. We t will use Equation 13.1 and the temperature TAl of the heating element under the aluminum- 716 THE TRANSFER OF HEAT bottomed pot to determine the temperature TCu of the heating element under the copperbottomed pot. Q of heat flow, as well as equal crosst Q kA ∆T = sectional areas A and thicknesses L. Therefore, (Equation 13.1) yields t L SOLUTION Both pot bottoms have identical rates Q kCu A ( ∆T )Cu kAl A ( ∆T )Al = = t L L kCu ( ∆T )Cu = kAl ( ∆T )Al or (1) Solving Equation (1) for ( ∆T )Cu yields ( ∆T )Cu = kAl ( ∆T )Al kCu (2) Substituting ( ∆T )Cu = TCu − T0 into Equation (2) and solving for TCu, we obtain TCu − T0 = kAl ( ∆T )Al kCu or TCu = kAl ( ∆T )Al + T0 kCu Therefore, the temperature of the heating element underneath the copper-bottomed pot is TCu ( )( ( ) 240 J/ s ⋅ m ⋅ Co 155.0 oC − 100.0 oC = + 100.0 oC = 134 oC o 390 J/ s ⋅ m ⋅ C ) 43. SSM WWW REASONING Heat flows along the rods via conduction, so that Equation ( k A∆T ) t , where Q is the amount of heat that flows in a time t, k is the 13.1 applies: Q = L thermal conductivity of the material from which a rod is made, A is the cross-sectional area of the rod, and ∆T is the difference in temperature between the ends of a rod. In arrangement a, this expression applies to each rod and ∆T has the same value of ∆T = TW − TC . The total heat Q ′ is the sum of the heats through each rod. In arrangement b, the situation is more complicated. We will use the fact that the same heat flows through each rod to determine the temperature at the interface between the rods and then use this temperature to determine ∆T and the heat flow through either rod. Chapter 13 Problems 717 SOLUTION For arrangement a, we apply Equation 13.1 to each rod and obtain for the total heat that k A ( TW − TC ) t k2 A ( TW − TC ) t ( k1 + k2 ) A ( TW − TC ) t Q′ = Q1 + Q2 = 1 + = (1) L L L For arrangement b, we use T to denote the temperature at the interface between the rods and note that the same heat flows through each rod. Thus, using Equation 13.1 to express the heat flowing in each rod, we have k1 A ( TW − T ) t 144L 244 3 = k2 A ( T − TC ) t 144L 244 3 Heat flowing through rod 1 or k1 ( TW − T ) = k2 ( T − TC ) Heat flowing through rod 2 Solving this expression for the temperature T gives T= k1TW + k2TC k1 + k2 (2) Applying Equation 13.1 to either rod in arrangement b and using Equation (2) for the interface temperature, we can determine the heat Q that is flowing. Choosing rod 2, we find that kT +k T k2 A 1 W 2 C − TC t k +k k A ( T − TC ) t 1 2 Q= 2 = L L k T −k T k2 A 1 W 1 C t k +k 1 2 = k2 Ak1 ( TW − TC ) t = L L ( k1 + k2 ) Using Equations (1) and (3), we obtain for the desired ratio that Q′ = Q ( k1 + k2 ) A ( TW − TC ) t L k2 Ak1 ( TW − TC ) t L ( k1 + k2 ) ( k1 + k2 ) A ( TW − TC ) t L ( k1 + k2 ) ( k1 + k2 )2 = = k2 k1 L k2 A k1 ( TW − TC ) t Using the fact that k2 = 2k1 , we obtain (3) 718 THE TRANSFER OF HEAT ( k + 2k1 ) = 4.5 Q′ ( k1 + k2 ) = =1 Q k2 k1 2k1 k1 2 2 44. REASONING The drawing shows a crosssectional view of the small sphere inside the larger spherical asbestos shell. The small sphere produces a net radiant energy, because its temperature (800.0 °C) is greater than that of its environment (600.0 °C). This energy is then conducted through the thin asbestos shell (thickness = L). By setting the net radiant energy produced by the small sphere equal to the energy conducted through the asbestos shell, we will be able to obtain the temperature T2 of the outer surface of the shell. T2 600.0 °C L r2 r1 800.0 °C SOLUTION The heat Q conducted during a time through the thin asbestos shell is given A ∆T ) t (k by Equation 13.1 as Q = asbestos 2 , where kasbestos is the thermal conductivity of L 2 asbestos (see Table 13.1), A2 is the surface area of the spherical shell A2 = 4π r2 , ∆T is the ( ) temperature difference between the inner and outer surfaces of the shell (∆T = 600.0 °C − T2), and L is the thickness of the shell. Solving this equation for the T2 yields QL T2 = 600.0 °C − kasbestos 4π r22 t ( ) The heat Q is produced by the net radiant energy generated by the small sphere inside the asbestos shell. According to Equation 13.3, the net radiant energy is ( ) Q = Pnet t = eσ A1 T 4 − T04 t , where e is the emissivity, σ is the Stefan-Boltzmann constant, A1 is the surface area of the sphere ( A1 = 4π r12 ) , T is the temperature of the sphere (T = 800.0 °C = 1073.2 K) and T0 is the temperature of the environment that surrounds the sphere (T0 = 600.0 °C = 873.2 K). Substit...
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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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