Questions 5 - 6 are considered challenge problems and often involve synthesis of
ID: 1528735 • Letter: Q
Question
Questions 5 - 6 are considered challenge problems and often involve synthesis of the material. They are at or above the level of an advanced exam question. The hope is that if you can handle these problems then you should be well prepared for the exams. Two square wires are composed differently, see the figure of their cross-sections. Wire 1 is a square wire of material A but with a circular center of material C. The diameter of the circular inside is equal to the length of side of the square. The second square wire has the same cross-sectional area as the first but is entirely comprised of material A. Both wires are the same length and setup to have the same temperature difference maintain between their ends. (a) Find an expression, in terms of their thermal conductivities k, for the ratio of the heat transfer through wire 1 over that through wire 2. b) If material A is Aluminum and material C is copper, calculate the ratio of their heat transfers. At our distance from the Sun, the intensity of solar radiation is 1370 W/m^2. The temperature of the Earth is affected by the greenhouse effect of the atmosphere. This phenomenon describes the effect of absorption of infrared light emitted by the surface so as to make the surface temperature of the Earth higher than if it were airless. For comparison, consider a spherical object of radius r with no atmosphere at the same distance from the Sun as the Earth. Assume its emissivity is the same for all kinds of electromagnetic radiation and its temperature is uniform over its surface. (a) Explain why the projected area over which it absorbs sunlight is pi r^2 and the surface area over which it radiates is 4 pi r^2. (b) Compute its steady-state temperature. Is it chilly? (c) Why would an increase in greenhouse gasses such as water vapor, carbon dioxide, and methane be cause for alarm?Explanation / Answer
We know that,
Q/t = k A (Th - Tc)/d
Where, Q is the heat transfer in time t ; k is the thermak conductivity, A is the Area, Th is the hot temp and Tc is the cold temp, d is the thickness.
Q1 = k1 A1 (Th - Tc)/d1
Q2 = k2 A2 (Th - Tc)/d2
As stated in the question, all the parameter except the thermal conducitives are same, so when we take ratio :
Q1/Q2 = k1/k2
b)Q1/Q2 = k1/k2 = 205/385 = 0.532
Hence, Q1/Q2 = 0.532
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