   Chapter 11, Problem 16CQ

Chapter
Section
Textbook Problem

Star A has twice the radius and twice the absolute temperature of star B. What is the ratio of the power output of star A to that of star B? The emissivity of both stars can be assumed to be 1. (a) 4 (b) 8 (c) 16 (d) 32 (e) 64

To determine
The ratio PA/PB of the radiated power.

Explanation

Given Info: Star A has twice the radius and twice the absolute temperature of star B.

Formula to calculate the radiated power from star A is,

PA=σAAeATA4 (I)

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

Formula to calculate the radiated power from star B is,

PB=σAeBTB4 (II)

• PB is the power radiated by the star B
• AB is the area of the star B.
• eB is the emissivity of the star B
• TB is the temperature of the star B

Divide (I) by (II).

PAPB=σAAeATA4σABeBTB4=eAAATA4eBABTB4

Use 4πrA2 for AA and 4πrB2 for AB to rewrite the above equation

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