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Be sure to show all calculations clearly and state your final answers in complete sentences. Taking the Sun's Temperature. The Sun radiates a total power of about \(4 \times 10^{26}\) watts into space. The Sun's radius is about \(7 \times 10^{8}\) meters. a. Calculate the average power radiated by each square meter of the Sun's surface. (Hint: The formula for the surface area of a sphere is \(A=4 \pi r^{2} .\) ) b. Using your answer from part a and the Stefan-Boltzmann law, calculate the average surface temperature of the Sun. (Note: The temperature calculated this way is called the Sun's effective temperature.)

Short Answer

Expert verified
The average power radiated per square meter is approximately \(6.49 \times 10^7\) watts/m². The Sun's effective temperature is approximately 5778 K.

Step by step solution

01

Calculate the Surface Area

First, we calculate the surface area of the Sun using the formula for the surface area of a sphere, which is \( A = 4 \pi r^2 \). The radius \( r \) of the Sun is given as \( 7 \times 10^8 \) meters. Substitute this into the formula:\[ A = 4 \pi (7 \times 10^8)^2 \]This simplifies to:\[ A \approx 4 \times 3.14159 \times (49 \times 10^{16}) \approx 6.16 \times 10^{18} \text{ m}^2 \].
02

Calculate the Average Power Radiated per Square Meter

Using the total power radiated by the Sun, \( 4 \times 10^{26} \) watts, and the surface area calculated in Step 1, we can find the average power per square meter by dividing total power by surface area:\[ \text{Power per square meter} = \frac{4 \times 10^{26} \text{ watts}}{6.16 \times 10^{18} \text{ m}^2} \]This calculation gives:\[ \text{Power per square meter} \approx 6.49 \times 10^7 \text{ watts/m}^2 \].
03

Use Stefan-Boltzmann Law to Find Temperature

The Stefan-Boltzmann Law states that the power radiated per unit area \( P \) is given by\[ P = \sigma T^4 \]where \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \) is the Stefan-Boltzmann constant and \( T \) is the temperature. We have found \( P = 6.49 \times 10^7 \text{ watts/m}^2 \). Solve for \( T \):\[ T^4 = \frac{P}{\sigma} = \frac{6.49 \times 10^7}{5.67 \times 10^{-8}} \]This simplifies to:\[ T^4 \approx 1.145 \times 10^{15} \]Solving for \( T \) gives\[ T \approx 5778 \text{ K} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power of the Sun
The Sun is an astonishingly powerful source of energy, radiating about \(4 \times 10^{26}\) watts into space. This enormous amount of power is an unfathomable figure for most of us, but it is a crucial aspect of how the Sun impacts our solar system. As a star, the Sun produces this energy through nuclear fusion processes in its core. These processes transform hydrogen into helium, releasing a tremendous amount of energy in the form of light and heat. This total power is dispersed evenly in all directions over the surface of the Sun. It gives us the light and warmth we experience on Earth, and it is what we calculate when trying to determine the Sun's effective temperature.
Stefan-Boltzmann Law
The radiation emitted by the Sun can be better understood through the Stefan-Boltzmann Law. This fundamental principle in physics relates the power radiated per unit area of a black body to the fourth power of its temperature. The law is expressed as \( P = \sigma T^4 \), where \( P \) is the power per unit area, \( T \) is the temperature in Kelvin, and \( \sigma \) is the Stefan-Boltzmann constant approximately equal to \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \).

By using the calculated average power per square meter from the Sun, we can rearrange the Stefan-Boltzmann equation to solve for the temperature:
  • Rearrange the formula: \( T^4 = \frac{P}{\sigma} \)
  • Substitute the given values to find \( T \)
The result gives us an estimate of the Sun's effective temperature, which is a key parameter in understanding stellar characteristics.
Surface Area Calculation
To find out how much power each square meter of the Sun's surface radiates, we first need to calculate its total surface area. The Sun can be considered a sphere, and the formula for the surface area of a sphere is \( A = 4 \pi r^2 \). Here, \( r \), the radius of the Sun, is about \(7 \times 10^8\) meters.

By substituting this radius into the formula, we calculate:
  • \( A = 4 \pi (7 \times 10^8)^2 \)
  • This results in a surface area of approximately \( 6.16 \times 10^{18} \text{ m}^2 \).
This extensive surface area is what allows the Sun to distribute its immense power across the cosmos. Once this value is known, we can calculate the power radiated per square meter, which is essential for subsequent calculations like determining the Sun's effective temperature.
Sun's Effective Temperature
From the calculations, we find that the Sun's effective temperature is around 5778 Kelvin. This temperature is not the surface temperature in terms of layers we can see. Instead, it is the temperature of a black body that would emit the same total amount of electromagnetic energy. It's based on the energy radiated per square meter, as determined using the Stefan-Boltzmann Law:
  • Calculate the power per square meter: \( 6.49 \times 10^7 \text{ watts/m}^2 \)
  • Use the Stefan-Boltzmann Law to find \( T \)
  • The result is \( T \approx 5778 \text{ K} \)
This effective temperature is a valuable concept as it allows astronomers to compare different stars and understand their properties based simply on observable characteristics such as their luminosity.

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Most popular questions from this chapter

Be sure to show all calculations clearly and state your final answers in complete sentences. Understanding Light Bulbs. A traditional incandescent light bulb uses a hot tungsten coil to produce a thermal radiation spectrum. The temperature of this coil is typically about \(3000 \mathrm{K}\) a. What is the wavelength of maximum intensity for this light bulb? Compare to the 500 -nm wavelength of maximum intensity for the Sun. b. Overall, do you expect the light from this bulb to be the same as, redder than, or bluer than light from the Sun? Why? Use your answer to explain why professional photographers use a different type of film for indoor photography than for outdoor photography. c. Do incandescent light bulbs emit all their energy as visible light? Use your answer to explain why these light bulbs are usually hot to touch. d. Fluorescent light bulbs primarily produce emission line spectra rather than thermal radiation spectra. Explain why, if the emission lines are in the visible part of the spectrum, a fluorescent bulb can emit more visible light than a standard bulb of the same wattage. e. Compact fluorescent light bulbs are designed to produce so many emission lines in the visible part of the spectrum that their light looks very similar to the light of incandescent bulbs. However, they are much more energy efficient: \(A\) 15-watt compact fluorescent bulb typically emits as much visible light as a traditional incandescent 75 -watt bulb. Although compact fluorescent bulbs generally cost more than incandescent bulbs, is it possible that they could save you money? Besides initial cost and energy efficiency, what other factors must be considered?

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