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Chapter 14: Problem 132

Astronomy Find the total power radiated by our Sun. Assume it is a perfect emitter of radiation \((z=1)\) with a radius of \(6.96 \times 10^{3} \mathrm{~m}\) and a temperature of 5800 \(\mathrm{K}\).

Short Answer

Expert verified
The total power radiated by the Sun is approximately \(3.8 \times 10^{26}\) watts.

Step by step solution

01

- Calculate the Surface Area of the Sun

Given that the Sun is spherical, the surface area \(A\) is calculated with the formula \(A=4\pi r^{2}\), where \(r\) is the radius of the Sun. Substituting \(r=6.96 \times 10^{3} \mathrm{~km}\) or \(6.96 \times 10^{8} \mathrm{~m}\) into the formula, we get \(A=4\pi (6.96 \times 10^{8})^{2}\).
02

- Substitute into Stefan-Boltzmann Law

We can now substitute the calculated area and given temperature into the Stefan-Boltzmann Law. This gives us \(P = \sigma \cdot A \cdot T^{4} = 5.67 \times 10^{-8} \mathrm{W} \mathrm{m}^{-2} \mathrm{K}^{-4} \cdot 4\pi (6.96 \times 10^{8})^{2} \mathrm{m}^{2} \cdot (5800 \mathrm{K})^{4}\).
03

- Calculate the Result

Performing the calculation will give us the total power radiated by the Sun.

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Key Concepts

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

Surface Area of a Sphere
The surface area of a sphere is a critical concept for understanding many physical sciences. This formula is not limited to Earth-bound objects but can extend to celestial bodies like stars and planets. The sphere is defined mathematically by the formula for its surface area:
  • The formula is \[ A = 4\pi r^2 \]where:
    • \(A\) is the surface area,
    • \(r\) is the radius of the sphere,
    • \(\pi\) is approximately 3.14159.
  • In the example of the Sun, the given radius is converted from kilometers to meters.
  • The radius of the Sun is 6.96 x 10^8 meters.
  • By substituting in this radius, the formula helps calculate the Sun鈥檚 total surface area.
The large surface area is why stars like the Sun have such immense capabilities for radiating energy.
Blackbody Radiation
In the realm of physics and astronomy, blackbody radiation is a fundamental concept. A perfect blackbody is an idealized object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.
  • Crucially, the Sun can be approximated as a perfect blackbody emitter.
  • This means its radiation can be described by the Stefan-Boltzmann Law:
This law states that the power (\(P\)) radiated by a blackbody is related to its temperature (\(T\)) and surface area (\(A\)). Specifically:
  • The formula is:\[ P = \sigma \cdot A \cdot T^4 \]where:
    • \(\sigma\) is the Stefan-Boltzmann constant, approximately 5.67 x 10^{-8} W m^{-2} K^{-4},
    • \(T\) is the absolute temperature of the body in Kelvin.
  • For the Sun, with its surface temperature of about 5800 K, the blackbody approximation provides a powerful insight into its radiated energy output.
Solar Constant
The solar constant is an important concept in understanding how solar energy is distributed in space and on Earth.
  • It is defined as the amount of solar energy received per unit area at a distance of one astronomical unit (AU) from the Sun.
  • The average value of the solar constant is about 1361 W/m^2.
  • This energy measure is not "constant" in the strictest sense because it can vary slightly due to changes in the Earth-Sun distance and solar activity.
Understanding the solar constant is key in many applications:
  • It helps in climate modeling, determining energy supply for photovoltaic systems, and understanding solar influence on weather patterns.
  • The solar constant also illustrates the immense power radiated by the Sun, as only a fraction of the total power actually reaches the outer limits of Earth's atmosphere.
These variations are important in the study of Earth's climate system and the efficiency of solar power utilization.

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Most popular questions from this chapter

If you add heat to water at \(0^{\circ} \mathrm{C}\), the water will decrease in volume until it reaches A) \(1^{\circ} \mathrm{C}\) B) \(2^{\circ} \mathrm{C}\) C) \(3^{\circ} \mathrm{C}\) D) \(4^{\circ} \mathrm{C}\) E) \(100^{\circ} \mathrm{C}\)

When a substance goes directly from a solid state to a gaseous form, the process is known as A. vaporization. B. fusion. C. melting. D. condensation. E. sublimation.

Suppose a person who lives in a house next to a busy urban freeway attempts to "harness" the sound energy from the nonstop traffic to heat the water in his home. He places a transducer on his roof, "carches" the sound waves, and converts the sound waves into an clectrical signal that warms a cistern of water. However, after running the system for 7 days, the \(5 \mathrm{~kg}\) of water increases in temperature by only \(0.01{ }^{\circ} \mathrm{C}\) ! Assuming \(100 \%\) transfer efficiency, calculate the acoustic power "caught" by the transducer. SSM

Helium condenses at \(-268.93^{\circ} \mathrm{C}\) and has a latent heat of vaporization of \(21,000 \mathrm{~J} / \mathrm{kg}\). If you start with \(5 \mathrm{~g}\) of helium gas at \(30^{\circ} \mathrm{C}\), calculate the amount of heat required to change the sample to liquid helium. The specific heat of helium is \(5193 \mathrm{~J} /\left(\mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\) at \(300 \mathrm{~K}\). SSM

Astronomy A distant star radiates 1000 times more energy than our own Sun even though the temperature of the star is only \(70 \%\) of the Sun's. If both stars are perfect emitters, estimate the radius of the distant star. Recall, the radius of the Sun is \(6.96 \times 10^{8} \mathrm{~m}\). \(5 \mathrm{SM}\)
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