Problem Set 3 Solutions - Thermal Physics | PHYS 224, Assignments of Thermal Physics

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224 problem set 3 solutions

224 problem set 3 solutions Problem 1,60. To make a rough estimate, imagine that the handle is divided into two halves: a half nearer the burner at 200°C, and the half that { wish to grab, initially at 25°C. This half will definitely be too hot to grab when its temperature reaches 75°C, an increase of about 50° above room temperature, The heat required to accormplish this increase is Q=mcAT = pVcAT = pA tc AT, where p is the density of iron, ¢ is the specific heat, A is the handie’s cross-sectional area, and £ is the tength of the half-handle. On the other hand, by the Fourier heat conduction law, the heat that flows past the handle’s mid-point is a= hawt at, dz where dT and dz can be approximated by taking differences between the mid-points of the two half-handles: dT’ = 150°C on average, and dx = £. Equating the two expressions for Q and solving for A¢ then gives Are PCAPAT _ pot AT _ (7900 kg/m'}(450 J/ka°C)(0.8 m)7(50°C) k,ACaT dz) k, dP (80 W/m.°G}{150°C) or about two and a half minutes. Seems about right, in my experience. = 150 s, Problem 1.68. Suppose that the perfume has already spread over half of the room, densely enough to barely be smelled. How long will it take to then spread over the entire room? If the length of the room is Az and its cross-sectional area is A, then we can write, as in equation 1.71, N N/V N —- = D——— = D——— AAt Ac A(Az)?’ with the same N on each side to indicate that we want a substantial fraction of the molecules to cross from one side of the room to the other. Canceling the N’s and A’s and solving for At then gives (Az)? (4m)? f2 "D108 m/s where I’ve assumed that the room is a few meters across and used a typical D value for diffusion through air. Obviously this is much longer than you would actually have to wait to smell the perfume, so convection must be the dominant transport mechanism in this situation. Ai s 10°s = 2 weeks, Problem 1.57. (R values and heat loss through a window.) (a) For plate glass, Ar 32x 104m = 0.0040 Kem R= +> = oa W/n K W For a 1-mm layer of air, 2 10° m og Kem = 0.038 Wo ~ 0.026 W/m - K nearly ten times as much (b) Because we're dealing only with temperature differences, the conversion for tempera- ture units is 1°F = 8 K. To figure out what a Btu is, I need to compute the energy needed to raise the temperature of a pound (453.6 g) of water by 1°F: 1Btu = (453.6 g)(4.186 J/g -K)(2.K) = 1055 J.