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P ( T ) = σ A T 4

where A is the surface area of a blackbody, T is its temperature (in kelvins), and σ is the Stefan–Boltzmann constant    , σ = 5.670 × 10 −8 W / ( m 2 · K 4 ) . Stefan’s law enables us to estimate how much energy a star is radiating by remotely measuring its temperature.

Power radiated by stars

A star such as our Sun will eventually evolve to a “red giant” star and then to a “white dwarf” star. A typical white dwarf is approximately the size of Earth, and its surface temperature is about 2.5 × 10 4 K . A typical red giant has a surface temperature of 3.0 × 10 3 K and a radius ~100,000 times larger than that of a white dwarf. What is the average radiated power per unit area and the total power radiated by each of these types of stars? How do they compare?

Strategy

If we treat the star as a blackbody, then according to Stefan’s law, the total power that the star radiates is proportional to the fourth power of its temperature. To find the power radiated per unit area of the surface, we do not need to make any assumptions about the shape of the star because P / A depends only on temperature. However, to compute the total power, we need to make an assumption that the energy radiates through a spherical surface enclosing the star, so that the surface area is A = 4 π R 2 , where R is its radius.

Solution

A simple proportion based on Stefan’s law gives

P dwarf / A dwarf P giant / A giant = σ T dwarf 4 σ T giant 4 = ( T dwarf T giant ) 4 = ( 2.5 × 10 4 3.0 × 10 3 ) 4 = 4820

The power emitted per unit area by a white dwarf is about 5000 times that the power emitted by a red giant. Denoting this ratio by a = 4.8 × 10 3 , [link] gives

P dwarf P giant = a A dwarf A giant = a 4 π R dwarf 2 4 π R giant 2 = a ( R dwarf R giant ) 2 = 4.8 × 10 3 ( R dwarf 10 5 R dwarf ) 2 = 4.8 × 10 −7

We see that the total power emitted by a white dwarf is a tiny fraction of the total power emitted by a red giant. Despite its relatively lower temperature, the overall power radiated by a red giant far exceeds that of the white dwarf because the red giant has a much larger surface area. To estimate the absolute value of the emitted power per unit area, we again use Stefan’s law. For the white dwarf, we obtain

P dwarf A dwarf = σ T dwarf 4 = 5.670 × 10 −8 W m 2 · K 4 ( 2.5 × 10 4 K ) 4 = 2.2 × 10 10 W/m 2

The analogous result for the red giant is obtained by scaling the result for a white dwarf:

P giant A giant = 2.2 × 10 10 4.82 × 10 3 W m 2 = 4.56 × 10 6 W m 2 4.6 × 10 6 W m 2

Significance

To estimate the total power emitted by a white dwarf, in principle, we could use [link] . However, to find its surface area, we need to know the average radius, which is not given in this example. Therefore, the solution stops here. The same is also true for the red giant star.

Check Your Understanding An iron poker is being heated. As its temperature rises, the poker begins to glow—first dull red, then bright red, then orange, and then yellow. Use either the blackbody radiation curve or Wien’s law to explain these changes in the color of the glow.

The wavelength of the radiation maximum decreases with increasing temperature.

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Check Your Understanding Suppose that two stars, α and β , radiate exactly the same total power. If the radius of star α is three times that of star β , what is the ratio of the surface temperatures of these stars? Which one is hotter?

T α / T β = 1 / 3 0.58 , so the star β is hotter.

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Practice Key Terms 9

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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