Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 13.101
National Institute of Standards and Technology
The international prototype meter[*][**] was set by the distance between two markings on a platinum-iridium alloy bar (α = 8.7×10–6 K–1). What temperature increase would correspond to the distance between these markings expanding by 0.01 mm?
(A) 0.001° C.
(B) 0.01° C.
(D) 1° C.
[*] "Differences...lie within 0.01 millimetre," bipm.org/en/CGPM/db/1/1/.
[**] "Pt90/Ir10 coefficient of thermal expansion @20-100° C," goodfellow.com/E/Platinum-Iridium.html.
Correct answer: (D)
The relation between the change in length ∆L due to a temperature change ∆T is given by:
α·∆T = ∆L/L,
∆T = ∆L/(α·L) = (0.01×10–3 m)/((8.7×10–6 K–1)·(1 m)) = 1.14942528736 K,
or to one significant figure, the change in temperature is 1 K or 1° C.
Sections 70854, 70855
Exam code: quiz07Di5k
(A) : 5 students
(B) : 6 students
(C) : 4 students
(D) : 35 students
Success level: 70%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.09