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Western University
Physics 1301A/B
Martin Zinke- Allmang

Selection of Even–Numbered Problem Solutions Chapter 11 P–11.2 Fourier's law is given in Eq. [11.1]: (1) We rewrite Fourier’s law to separate the unknown temperature: (2) Next we substitute the given values, including (Q/t)/A = 55 mJ/(s m²): (3) The temperature at the base of Earth's crust is about 1000C. This clearly excludes science fiction ideas like those in "Voyage au centre de la terre" by Jules Verne, published in 1864. Indeed, we couldn't even travel 0.5% of the distance to the centre of Earth and expect to live to tell about it! P–11.4 Solution to part (a): We need to convert to SI–units for the comparison: (4) This value is consistent with the value given for quartz glass in textbook Table 11.1. Solution to part (b): Using the definition of thermal resistance R for a sheet of thickness l: (5) we find: (6) P–11.6 We use Fourier’s law with the unknown thermal conductivity as the dependent variable: (7) During a time interval of ∆t = 100 s, in which the upper plate temperature drops 2.5 K, the amount of heat released by the upper plate is calculated with Joule’s definition, divided by ∆t: (8) The mass of the upper plate is needed inEq. [8]. It follows from the given data,d =10.03 m for the iron plate thickness and its area A = 0.15 m A 0.15 m = 0.0225 m²: (9) Substituting this value in Eq. [8], we find for the heat loss in time ∆t: (10) where we used c =Fe50 J/(kg K). The result of Eq. [10] is now substituted into Eq. [7] for Q/t: (11) in which we used d = 2.004 m for th e thickness of the unknown material. Note that we neglected the temperature change across the unknown material in this last formula and did not include heat loss to other components in the environment. P–11.8 We quantified the interface temperature T in Problem P–11.7, which you find in the Student Solutions x Manual. Look at Eq. [2] on p. 79 of the SSM: (12) We divide both numerator and denominator of Eq. [12] by λ λ and obtai1:2 (13) Using Eq. [5] for the blocks in the forms R = l /1 an1 R1= l /λ 2 we 2ew2ite Eq. [13]: (14) This equation is the result we sought. P–11.10 We start with Fourier's law, which is applied to cases (a) and (b) in the attached figure: (15) Dividing the two formulas in Eq. [15] leads to the ratio of the two heat flow rates: (16) To find the values to substitute into Eq. [16] we study the geometric properties of the arrangements in the given figure. We find: (17) Substituting Eq. [17] in Eq. [15] gives: (18) With Q = 1.0 J for both cases, we find from Eq. [18] that t /t =b4a0, and therefore with t = 1.0amin we get tb= 4.0 min. P–11.12 To raise its temperature from T initialT finala sphere of radius R has to acquire an amount of heat: (19) based on Joule’s definition of heat and withc the specific heat capacity of the material of the sphere. The tem- perature gradient has to point radially outof the sphere and has to have a value of (T final T initial. Based on Fourier’s law this drives a heat flow per
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