Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e Problem 14.55, Comprehensive Problem 14.85
A 0.91 m2 area section of CertainTeed SMARTBATT™ fiberglass insulation with a thickness of 8.9×10–2 m has a thermal conductivity of 4.2×10–3 watts/(m·K)[*]. The thermal resistance of this section of insulation is:
(A) 3.4×10–4 K/watt.
(B) 23 K/watt.
(C) 2.4×102 K/watt.
(D) 2.9×103 K/watt.
Correct answer (highlight to unhide): (B)
The thermal resistance R of an object can be related to its thermal conductivity κ by:
R = d/(κ·A),
where d is the thickness of the object that heat must conduct through, and A is the cross-sectional area, such that:
R = (8.9×10–2 m)/((4.2×10–3 watts/(m·K))·(0.91 m2)) = 23.286237572 K/watt,
or to two significant figures, the thermal conductivity of the wall is 23 K/watt.
(Response (A) is κ·A·d; response (C) is 1/κ; response (D) is 1/(κ·A·d.)
Sections 70854, 70855, 73320
Exam code: quiz07cO4t
(A) : 1 student
(B) : 60 students
(C) : 3 students
(D) : 0 students
Success level: 94%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.21