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' ( ) + , . 4 > ? @ L V W X Y c d g m n r | }            e m p u v y | }                      & 2 B N   y XX XX N XXhXX  FX M=MM5PHYsics 1150 Howmework Solutions Chapter 13, Heat Transfer Ch 13: 5, 17, 21, 24, 26, 30 13.5 A goose-down sleeping bag has a surface area of 2.25 m2 and is filled iwth a layer of down 5 cm thick. What is the heat-transfer rate (Hcd) through it from a person with skin temperature of 35C to the outside air at 5C. How does this rate compare with the bodys minimal metabolism rate, about 100 W? From Table 13,1, p 457, we find the thermal conductivity of goose down to be K = 0.023 W/m C We can now apply the conduction heat-transfer equation, Eq 13.1, Hcd = [KA(T2 T1] / L From the problem, we know the other pieces of this equation, DT = T2 T1 = 35C ( 5C) = 40 C A = 2.25 m2 L = 5 cm = 0.05 m Hcd = [KA(T2 T1] / L Hcd = [(0.023W/m C)(2.25 m2)(40 C)]/0.05 m = Hcd = 41 W This is about half of the bodys normal metabolism so this sleeping bag should keep our camper a happy camper (or, at least, a warm camper). 13.17 A lamp bulb inside an ornamental spherical glass shell 20 cm in radius and 0.50 cm thick radiats 100 W of thermal power. What is the difference in temperature between the inner and outer surfaces of the glass? (Hmmm, . . . go ahead and use e = 1.00, even in this case).  To find the temperature, we can use the Radiation equation, Eq 13.4, Remit = e s A T4 We will use e 1.0, even for this. s is the Stefan-Boltzman constant, s = 5.67 x 108 W/m2 K4 The surface area of a sphere is A = (4/3) r2 For the inner surface, Ainner = (4/3) (0.200 m)2 = 0.1676 m2 For the outer surface, Aout = (4/3) (0.205 m)2 = 0.1760 m2 The temperatures T must be absolute temperatures, measured on the Kelvin scale. Now we are ready to solve for the temperaturs T4 = Remit / e s A Tinner4 = 100 W/[(1.0)(5.67 x 108 W/m2 K4)(0.1676 m2)] = 1.0523 x 1010 K4 Tinner = 320.3 K Touter4 = 100 W/[(1.0)(5.67 x 108 W/m2 K4)(0.1760 m2)] = 1.0078 x 1010 K4 Touter = 316.8 K Our use of e = 1.0 makes our actual numbers rather iffy. But the point of this problem was to show that the inner surface will be at a higher temperature than the outer surface. That is certainly true for our calculation. Changing the value of the emissivity e may change our actual numbers but we will still find Tinner>Touter.13. 21 Victoria, Canada, is located at 48.5 north latitude. How high above the horizon is the Sun on the first day of summer and the first day of winter?  N and S mark the poles, the axis of rotation. Z is the zenith, the point directly overhead, 48.5 from the equator (48.5 north latitude). Summer:  q is the angle from directly overhead (the zenith) to the Sun. Therefore, the angle between the horizon and the Sun will be 90 q = 90 25 = 65. The Sun will be 65 above the horizon at the summer solstice.Winter: q is again the angle between directly overhead (the zenith) and the Sun. Therefore, the Sun is 90 72 = 18 above the horizon at the winter solstice. 13.24 A solar home is to be built in Minneapolis, 45 north latitude. What is the elevation of the Sun at midday on the first day of summer and on the first day of winter? An overhaning eave 85 cm wide is planned. How far below this eave will sunlight fall on the first day of summer and the first day of winter. The first part of this questions requires exactly the same reasoning, using exactly the same kind of diagrams, as question 13.21.  N and S mark the poles, the axis of rotation. Z is the zenith, the point directly overhead, 45 from the equator (45 north latitude). Summer:  If the Sun is 21.5 from the zenith (the position directly overhead) then it must be 9021.5=68.5 above the horizon. Winter:  If the Sun is 68.5 from the zenith, then it must be 9068.5 = 21.5 above the horizon. Now we can look at the shadow cast by the eave on these two days: Summer:  The length of the shadow, S in the diagram, is the opposite side of a 68.5 right triangle (85 cm is the adjacent side). tan q = opp/adj tan 68.5 = S / 85 cm S = (85 cm)(tan 68.5) S = (85 cm)(2.54) = 216 cm = 2.16 cm Winter:  The length of the shadow, S in the diagram, is the opposite side of a 21.5 right triangle (85 cm is the adjacent side). tan q = opp/adj tan 21.5 = S / 85 cm S = (85 cm)(tan 21.5) S = (85 cm)(0.393) = 33.5 cm 13.26 Calculate the R value of an insulated wall like that in Figure 13.30 if 2 x 6s are used for wall studs instead of 2 x 4s. The larger board allow 5.5 in of fiberglass insulation instead of 3.5 in. By what percentage does this increase in thickness change the R value? How does the increased thickness change the heat loss calculated in Example 13.14? The R-values for the components of the wall in Figure 13.30 are given in Table 13.4, on page 481. By increasing the fiberglass insulation from 3.5 inches to 5.5 inches, the R-value for this insulation increases from 10.90 to 18.80, according to Table 13.3, on page 480. This increases the total R-value from 14.33 to 22.23. This is an increase of 7.9/14.33 = 0.55 = 55% for the R-value of the wall. In terms of R-values, the conduction heMat transfer rate is given by equation 13.9, Hcd = A(T2 T1) / R = (1/R) [A (T2 T1)] In Example 13.14, we found the conduction heat transfer rate to be Hcd(un) = 3.8 kW for the uninsulated wall, Hcd(3.5) = 1.2 kW for the insulated wall with 2x4s and 3.5 in of insulation Now, with 2x6s and 5.5 in of insulation, Hcd(5.5) = (1/22.23)[(960)(60)] = 2591 Btu/h = 0.76 kW This is a reduction of heat flow by DH = 1.18 kW 0.76 kW = 0.42 kW DH / H = 0.42 kW / 1.18 kW = 0.36 = 36% 13.30 A roasted turkey cools from 85C to 80C in 10 min when sitting in a 25C room. How long does it require to cool from 85C to 55C? This requires an application of Newtons Law of Cooling, equation 13.3, T(t) = Tsur + DT et/t DT is the initial temperature difference of the turkey and its surroundings; DT=85C25C=60C Knowing that iMt cools from 85C to 80C allows us to solve for the time constant t in this equation; T(t) = Tsur + DT et/t T(10 min) = 80C = 25C + (60C) e(10 min)/t 55C = (60C) e(10 min)/t 55C/60C = e(10 min)/t 55/60 = e(10 min)/t 0.9167 = e(10 min)/t e(10 min)/t = 0.9167 ln[e(10 min)/t ] = ln[0.9167] (10 min)/t = 0.0870 10 min/t = 0.0870 t = (10 min)/0.0870 t = 115 min Now we know the time constant t and we can use Newtons Law of Cooling to go back and solve for t, the time, when T, the temperature, is 55C. 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