Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 14.56
A Harvey Building Products vinyl casement window [*] has a thermal resistance of 0.33 K/watt. The temperature difference between inside and outside is 30° C. The SymphonyShades™ Virtuoso® window shade[**] is also claimed to have a thermal resistance of approximately 0.33 K/watt. If this window shade completely covers the window, the rate of heat conduction per time will be __________ of the original (uncovered) value.
(E) (No change in rate of heat conduction.)
[*] R-factor = 0.33, ENERGY STAR rated, 19 square feet, harveybp.com/upload/products/literature/Harvey_AccessoryWindows_Brochure.pdf
[**] R-factor = 0.34, single-cellular light-filtering fabric, 19 square feet, symphonyshades.com/single_cell_shades.html.
Correct answer: (B)
The rate at which heat is conducted per time to the environment through the window given by:
Powerwindow = ((heat flow)/time)window = ∆T/Rwindow,
where Rwindow is the thermal resistance of the window:
Rwindow = dwindow/(κwindow·Awindow),
which is already given as 0.33 K/watt, such that the heat flow per time conducted through the uncovered window is:
Powerwindow = (30° C)/(0.33 K/watt) = 90.909090 watts.
Thermal resistances are additive for materials that are stacked in series, so the heat flow per time through the window and shade will be:
Powerwindow & shade = ((heat flow)/time)window & shade = ∆T/Rwindow & shade,
Powerwindow & shade = ∆T/(Rwindow + Rshade),
Powerwindow & shade = (30° C)/(0.33 K/watt + 0.33 K/watt) = 45.45454545 watts.
Thus increasing the overall thermal resistance by a factor of two will reduce the heat flow per time to one-half of its original (uncovered) value.
Sections 70854, 70855
Exam code: quiz07h3A7
(A) : 9 students
(B) : 27 students
(C) : 8 students
(D) : 7 students
(E) : 1 student
Success level: 52%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.67