Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 14.65
NASA/JPL-Caltech
nasa.gov/mission_pages/voyager/voyager20120117.html
The power generated by NASA's Voyager 1 spacecraft was 420 watts in 1977, but today only 285 watts are generated.[*] Assume that all of this power is radiated into space (taken to be absolute zero). Neglect changes in the spacecraft's dimensions due to temperature changes, and assume that it is an ideal blackbody. If the surface temperature of Voyager 1 now is 194 K[**], the temperature of Voyager 1 in 1977 was:
(A) 210 K.
(B) 290 K.
(C) 470 K.
(D) 780 K.
[*] voyager.jpl.nasa.gov/spacecraft/spacecraftlife.html.
[**] "The spectrometer is likely operating at a temperature somewhat lower than –79° C," nasa.gov/mission_pages/voyager/voyager20120117.html.
Correct answer: (A)
The power output of Voyager 1 in 1977 is set equal to the net power radiated:
Power1977 = –e·σ·A·((T1977)4 – (0 K)4) = –e·σ·A·(T1977)4,
where the heat radiated per time was 420 watts, e = 1 (assuming an ideal blackbody), A is the surface area, and the surface temperature in 1977 is unknown. Similarly in 2012:
Power2012 = –e·σ·A·((T2012)4 – (0 K)4) = –e·σ·A·(T2012)4,
where the heat radiated per time is 285 watts, and the surface temperature is 194 K. Since e, σ, and A are the same for both 1977 and 2012 equations, then we can set:
–e·σ·A = –e·σ·A,
Power2012/(T2012)4 = Power1977/(T1977)4,
and solving for the temperature in 1977:
(T1977/T2012)4 = Power1977/Power2012,
T1977/T2012 = (Power1977/Power2012)(1/4),
T1977 = T2012·(Power1977/Power2012)(1/4),
T1977 = (194 K)·(420 watts)/(285 watts)(1/4) = 213.748382773 K,
or to two significant figures, 210 K.
(Response (B) is T2012·(Power1977/Power2012); response (C) is T2012 + 273 K; response (D) is 4·T2012.)
Sections 70854, 70855
Exam code: quiz07Di5k
(A) : 10 students
(B) : 33 students
(C) : 7 students
(D) : 0 students
Success level: 20%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.47
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