## 20160622

### Physics quiz question: comparing surface temperatures of Bellatrix and Pollux

Physics 205A Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA

Stars can be modeled as spherical blackbodies that radiate heat out to an environment assumed to have a temperature of 0 K. Pollux is a star that radiates less heat per time, but is larger in size than Bellatrix. Pollux has a surface temperature __________ Bellatrix.
(A) cooler than.
(B) same as.
(C) hotter than.
(D) (Not enough information given.)

The net power radiated by a star with an environment assumed to be at absolute zero is given by:

Power = –e·σ·A·((T)4 – (Tenv)4) = –e·σ·A·((T)4 – (0 K)4).

where the emissitivity e = 1 (for an ideal blackbody), the Stefan-Boltzmann constant is σ = 5.670×10–8 watts/(m2·K4), A is the surface area, and the surface temperature T is unknown, and can be solved for:

(T)4 = –Power/(e·σ·A),

T = (–Power/(e·σ·A))(1/4).

This allows a ratio to be set up to compare the surface temperatures of Bellatrix and Pollux:

TBellatrix = (–PowerBellatrix/(e·σ·ABellatrix))(1/4),

TPollux = (–PowerPollux/(e·σ·APollux))(1/4),

such that:

(TBellatrix/TPollux) = (PowerBellatrix/PowerPollux))·(APollux/ABellatrix)(1/4).

Since Pollux radiates less heat per time than Bellatrix, then the ratio (PowerBellatrix/PowerPollux)) is greater than 1; and since Pollux is larger in size than Bellatrix, the ratio (APollux/ABellatrix)(1/4) is also greater than 1. Thus the ratio (TBellatrix/TPollux) must then be greater than 1, so Pollux has a cooler surface temperature than Bellatrix.

Sections 70854, 70855
Exam code: quiz07zSOL
(A) : 50 students
(B) : 12 students
(C) : 7 students
(D) : 0 students

Success level: 72%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.70