Cuesta College, San Luis Obispo, CA
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(A) smaller, e = 0.1.
(B) larger, e = 0.9.
(C) (There is a tie.)
(D) (Not enough information given.)
Correct answer (highlight to unhide): (B)
The power radiated by an object (where the negative sign indicates that heat is continuously leaving its surface) is given by:
Power = –e·σ·A·T4,
where the Stefan-Boltzmann constant σ = 5.670×10–8 watts/(m2·K4) and the surface temperatures T = 600 K are the same for both objects.
For the smaller, lower emissitivity (e = 0.1) object, the power radiated is:
Powersmaller = –(0.1)·σ·Asmaller·(600 K)4,
and for the larger, higher emissivity (e = 0.9) object, the power radiated is:
Powerlarger = –(0.9)·σ·Alarger·(600 K)4,
and from inspection, since 0.1 < 0.9 and Asmaller < Alarger, then the larger, higher emissivity object will radiate heat at a higher power rate than the smaller, lower emissivity object.
Sections 70854, 70855
Exam code: quiz07PeA7
(A) : 16 students
(B) : 32 students
(C) : 2 students
(D) : 0 students
Success level: 64%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.37
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