Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e Problem 14.67
The bottom of the Space Shuttle is covered with an area of 250 m2 of High-temperature Reusable Surface Insulation (HRSI) silica tiles (emissivity 0.90) to withstand temperatures of 1,920 K during reentry into the atmosphere.[*][**] Assume that the environment is 0 K. During reentry the rate of heat per time radiated by the bottom of the Space Shuttle is:
(A) 4.3×105 watts.
(B) 6.9×105 watts.
(C) 1.7×108 watts.
(D) 2.1×108 watts.
[**] Andreas Ulovec, "Space Shuttle Landing," math2earth.oriw.eu/publications/04_Space%20shuttle%20landing.pdf.
Correct answer (highlight to unhide): (C)
The power (rate of heat per time) radiated is given by:
Power = –e·σ·A·((Tobj)4 – (Tenv)4),
where a negative value for power corresponds to a net amount of heat being removed (radiated) from the object, while a positive value corresponds to a net amount of heat being put into (absorbed) by the object (in order to be consistent with the ±Q convention for removing heat from (–) or putting heat into (+) a thermodynamic system).
Then the power radiated by the bottom of the Space Shuttle is:
Power = –(0.90)·(5.670×10–8 watts/(m2·K4))·(250 m2)·((1,920 K)4 – (0 K)4),
Power = –1.736861983×108 watts,
or to two significant figures, the magnitude of the heat radiated per time is 1.7×108 watts.
(Response (A) is e·A·Tobj; response (B) is e·σ·A·(Tobj)4; response (D) is σ·A·(Tobj)4/e.)
Sections 70854, 70855, 73320
Exam code: quiz07cO4t
(A) : 7 students
(B) : 4 students
(C) : 51 students
(D) : 1 student
Success level: 80%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.75