   Chapter 11.5, Problem 11.5QQ

Chapter
Section
Textbook Problem

Stars A and B have the same temperature, but star A has twice the radius of star B. (a) What is the ratio of star A’s power output to star B’s output due to electromagnetic radiation? The emissivity of both stars can be assumed to be 1. (b) Repeat the question if the stars have the same radius, but star A has twice the absolute temperature of star B. (c) What’s the ratio if star A has both twice the radius and twice the absolute temperature of star B?

(a)

To determine
The ratio of star A’s power output to star B’s output.

Explanation

Given Info: Star A and B are of same temperature but star A has twice the radius of star B.

Formula to calculate the power radiated by Star A is,

PA=σAAeTA4

• PA is the power radiated by star A
• σ is the Stefan-Boltzmann constant.
• AA area of star A.
• eA is the emissivity of the star A.
• TA is the temperature of the star A.

Formula to calculate the surface area of the star A is,

AA=4πrA2

• rA is the radius of the sphere.

Substitute 4πrA2 for AA to rewrite PA .

PA=σ(4πrA2)eATA4=4πσeArA2TA4

Similarly formula to calculate the power radiated by the star B is,

PB=4πσeBrB2TB4

• rB is the radius of the star B.
• TB is the temperature of the star B.
• eB is the emissivity of star B

(b)

To determine
Ratio of star A’s power output to star B’s power output.

(c)

To determine
Ratio of star A’s power output to star B’s power output when star A has both twice the radius and twice the absolute temperature of star B.

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