Physics Solution Manual for 1100 and 2101

30 m 1 567 108 j s m 2 k 4 77 k 42 k

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: ctivity kss of stainless steel from Table 13.1, we find that the temperature difference ∆Tss between the stainless steel surfaces is ∆Tss ( mLv ) L = ( 0.15 kg ) ( 22.6 ×105 J/kg ) (1.4 ×10−3 m ) = 9.4 C° = kss At ( ) 14 J/ ( s ⋅ m ⋅ C° ) 0.015 m 2 ( 240 s ) 700 THE TRANSFER OF HEAT The temperature at the steel-burner interface is T = 101.2 °C + ∆Tss = 110.6 °C . 18. REASONING If the cylindrical rod were made of solid copper, the amount of heat it would conduct in a time t is, according to Equation 13.1, Qcopper = (kcopper A2 ∆T / L)t . Similarly, the amount of heat conducted by the lead-copper combination is the sum of the heat conducted through the copper portion of the rod and the heat conducted through the lead portion: Qcombination = kcopper ( A2 − A1 ) ∆T / L + klead A1∆T / L t . Since the lead-copper combination conducts one-half the amount of heat than does the solid copper rod, Qcombination = 1 Qcopper , or 2 kcopper ( A2 − A1 ) ∆T L + klead A1∆T L 1 kcopper A2 ∆T = L 2 This expression can be solved for A1 / A2 , the ratio of the cross-sectional areas. Since the cross-sectional area of a cylinder is circular, A = π r 2 . Thus, once the ratio of the areas is known, the ratio of the radii can be determined. SOLUTION Solving for the ratio of the areas, we have kcopper A1 = A2 2 ( kcopper − klead ) The cross-sectional areas are circular so that A1 / A2 = (π r12 ) /(π r22 ) = (r1 / r2 )2 ; therefore, r1 r2 = kcopper 2(kcopper − klead ) = 390 J/(s ⋅ m ⋅ C°) = 0.74 2[390 J/(s ⋅ m ⋅ C°) − 35 J/(s ⋅ m ⋅ C°)] where we have taken the thermal conductivities of copper and lead from Table 13.1. Chapter 13 Problems 701 19. SSM WWW REASONING The rate at which heat is conducted along either rod is given by Equation 13.1, Q / t = ( k A ∆T ) / L . Since both rods conduct the same amount of heat per second, we have ks As ∆T Ls = ki Ai ∆T (1) Li Since the same temperature difference is maintained across both rods, we can algebraically cancel the ∆T terms. Because both rods have the same mass, ms = mi ; in terms of the densities of silver and iron, the statement about the equality of the masses becomes ρs ( Ls As ) = ρi ( Li Ai ) , or As Ai = ρi Li ρs Ls (2) Equations (1) and (2) may be combined to find the ratio of the lengths of the rods. Once the ratio of the lengths is known, Equation (2) can be used to find the ratio of the cross-sectional areas of the rods. If we assume that the rods have circular cross sections, then each has an area of A = π r 2 . Hence, the ratio of the cross-sectional areas can be used to find the ratio of the radii of the rods. SOLUTION a. Solving Equation (1) for the ratio of the lengths and substituting the right hand side of Equation (2) for the ratio of the areas, we have Ls Li = ks As ki Ai = ks ( ρi Li ) ki ( ρs Ls ) 2 or Ls ks ρi = L ki ρs i Solving for the ratio of the lengths, we have Ls Li = ks ρi ki ρs = [420 J/(s ⋅ m ⋅ C°)](7860 kg/m3 ) = 2.0 [79 J/(s ⋅ m ⋅ C°)](10 500 kg/m3 ) b. From Equation (2) we have π rs2 ρL = ii 2 ρs Ls π ri Solving for the ratio of the radii, we have 2 or rs ρi Li = r ρs Ls i 702 THE TRANSFER OF HEAT rs ri = ρi Li 7860 kg/m3 1 = = 0.61 ρs Ls 10 500 kg/m3 2.0 20. REASONING According to Equation 6.10b, power P is the change in energy Q divided by the time t during which the change occurs, or P = Q/t. The power radiated by a filament is given by the Stefan-Boltzmann law as P= Q = eσ T 4 A t (13.2) where e is the emissivity, σ is the Stefan-Boltzmann constant, T is the temperature (in kelvins), and A is the surface area. This expression will be used to find the ratio of the filament areas of the bulbs. SOLUTION Solving Equation (13.2) for the area, we have A= P eσ T 4 Taking the ratio of the areas gives P 1 A1 e1 σ T14 = P2 A2 e2 σ T24 Setting e2 = e1, and P2 = P1, we have that P 1 4 e1 σ T14 A1 T24 ( 2100 K ) = = 4= = 0.37 A2 P T1 ( 2700 K )4 1 e1 σ T24 21. SSM WWW REASONING AND SOLUTION Solving the Stefan-Boltzmann law, Equation 13.2, for the time t, and using the fact that Qblackbody = Qbulb , we have tblackbody = Qblackbody σ T4A = Qbulb σ T4A = Pbulb tbulb σ T4A Chapter 13 Problems 703 where Pbulb is the power rating of the light bulb. Therefore, tblackbody = 5.67 ×10 –8 (100.0 J/s) (3600 s) J/(s ⋅ m ⋅ K ) (303 K) 4 (6 sides)(0.0100 m)2 / side 2 4 1 h 1 d × = 14.5 d 3600 s 24 h 22. REASONING According to the Stefan-Boltzmann law, the radiant power emitted by the Q “radiator” is = e σ T 4 A (Equation 13.2), where Q is the energy radiated in a time t, e is t the emissivity of the surface, σ is the Stefan-Boltzmann constant, T is the temperature in Kelvins, and A is the area of the surface from which the radiant energy is emitted. We will apply this law to the “radiator” before and after it is painted. In either case, the same radiant power is emitted. SOLUTION Applying the Stefan-Boltzmann law, we obtain the following: Q 4 = eafter σ Tafter A t after and Q 4 = ebefore...
View Full Document

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.

Ask a homework question - tutors are online