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The amount of radiant power produced by the sun is approximately \(3.9 \times 10^{26} \mathrm{~W}\). Assuming the sun to be a perfect blackbody sphere with a radius of \(6.96 \times 10^{8} \mathrm{~m},\) find its surface temperature (in kelvins).

Short Answer

Expert verified
The surface temperature of the sun is approximately 5778 K.

Step by step solution

01

Understanding the Blackbody Concept

A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It also emits radiation at a maximum given its temperature. The sun can be approximated as a perfect blackbody sphere.
02

Apply the Stefan-Boltzmann Law

The Stefan-Boltzmann Law relates the radiant heat energy emitted by a blackbody in terms of its temperature. It states that the power emitted by a blackbody is proportional to the fourth power of its temperature: \[ P = \sigma A T^4 \] where \( P \) is the power radiated, \( \sigma \) is the Stefan-Boltzmann constant \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \), \( A \) is the surface area of the blackbody, and \( T \) is its temperature in kelvins.
03

Calculate the Surface Area of the Sun

Calculate the surface area \( A \) of the sun, treated as a sphere, using the formula for the surface area of a sphere: \[ A = 4 \pi r^2 \] where \( r = 6.96 \times 10^8 \, \text{m} \). So:\[ A = 4 \pi (6.96 \times 10^8 \, \text{m})^2 \approx 6.09 \times 10^{18} \, \text{m}^2 \]
04

Rearrange for Temperature

Rearrange the Stefan-Boltzmann Equation to solve for temperature \( T \): \[ T = \left(\frac{P}{\sigma A}\right)^{1/4} \]
05

Substitute Values into the Temperature Equation

Substitute \( P = 3.9 \times 10^{26} \, \text{W} \), \( A = 6.09 \times 10^{18} \, \text{m}^2 \), and \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \) into the equation:\[ T = \left(\frac{3.9 \times 10^{26}}{5.67 \times 10^{-8} \times 6.09 \times 10^{18}}\right)^{1/4} \]
06

Calculate the Temperature

Perform the calculation:1. Calculate the divisor: \( 5.67 \times 10^{-8} \times 6.09 \times 10^{18} = 3.45603 \times 10^{11} \).2. Divide: \( \frac{3.9 \times 10^{26}}{3.45603 \times 10^{11}} \approx 1.1281 \times 10^{15} \).3. Take the fourth root to find the temperature: \( T \approx (1.1281 \times 10^{15})^{1/4} \approx 5778 \, \text{K} \).

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Key Concepts

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

Blackbody Radiation
Blackbody radiation is an essential concept in physics, referring to the way an idealized object, called a blackbody, absorbs and emits electromagnetic radiation. A blackbody is considered perfect because it absorbs all incoming light without reflecting any, regardless of the light's color or the angle at which it strikes the surface.
When heated to a certain temperature, a blackbody emits radiation in a characteristic fashion depending on its temperature. This emitted radiation covers a continuous range of wavelengths. The sun, as a massive ball of hot plasma, is often approximated to a perfect blackbody, as it absorbs nearly all incoming light and emits radiation over a broad spectrum.
  • A blackbody is an ideal absorber and emitter.
  • It emits radiation based on its temperature.
  • The sun is an example of a body that closely approximates a blackbody.
Understanding blackbody radiation helps in comprehending how stars, including our sun, radiate energy effectively.
Surface Temperature
Surface temperature in the context of celestial bodies like the sun refers to the temperature at the outermost layer of the object from which radiation is emitted. This temperature is crucial for understanding the intensity and characteristics of the radiation emitted by the body. Given the sun's massive surface area and intense internal processes, its surface temperature determines how much energy it emits into space.
We use physical laws like the Stefan-Boltzmann Law to calculate this temperature based on the emitted power and the body's surface area, assuming it behaves like a perfect blackbody.
  • Surface temperature determines radiative characteristics.
  • It can be calculated using known physical laws.
  • The sun’s surface temperature is about 5778 K.
This calculation is essential in astrophysics and helps astronomers understand stars' energy outputs and lifecycles.
Radiant Power
Radiant power, or radiative power, is the total amount of energy radiated by an object per unit of time. In the case of the sun, this power is immense, approximately \(3.9 \times 10^{26} \, \text{W}\). It is a fundamental concept since it quantifies the energy output of the sun, which is crucial for understanding its influence on the solar system.
This amount of power affects everything from the climate on Earth to the habitability of planets within our solar system. Radiant power calculation involves understanding how much energy is emitted, which is connected to the surface area and temperature of the body.
  • Radiant power measures energy output per second.
  • For the sun, it is about \(3.9 \times 10^{26} \, \text{W}\).
  • It affects astrological and climate-related phenomena.
Grasping radiant power is crucial for both astrophysical studies and practical applications on Earth.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant \(\sigma\) is vital for converting a blackbody's temperature into total emitted energy. This constant appears in the Stefan-Boltzmann Law, which is mathematically expressed as \( P = \sigma A T^4 \), where \(P\) stands for the emitted power, \(A\) is the surface area, and \(T\) is the temperature in kelvins.
This law shows the relationship between an object's temperature and its energy emission: specifically, it dictates how energy emitted scales dramatically with temperature, raised to the fourth power. The value of the Stefan-Boltzmann constant is \(5.67 \times 10^{-8} \text{W/m}^2\text{K}^4\), acting as a bridge between the abstract concept of temperature and measurable power output.
  • Highly significant for astrophysical calculations.
  • Makes temperature-energy conversion possible.
  • Its value is \(5.67 \times 10^{-8} \text{W/m}^2\text{K}^4\).
The constant is essential for predicting the radiation properties of stellar bodies and understanding the effects of heat radiation in various scientific fields.

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Most popular questions from this chapter

Multiple-Concept Example 8 discusses the ideas on which this problem depends. Interactive LearningWare \(13.2\) at reviews the concepts that are involved in this problem. Suppose the skin temperature of a naked person is \(34{ }^{\circ} \mathrm{C}\) when the person is standing inside a room whose temperature is \(25^{\circ} \mathrm{C}\). The skin area of the individual is \(1.5\) \(\mathrm{m}^{2}\). (a) Assuming the emissivity is \(0.80\), find the net loss of radiant power from the body. (b) Determine the number of food Calories of energy (1 food Calorie \(=4186 \mathrm{~J}\) ) that are lost in one hour due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.

Two pots are identical except that the flat bottom of one is aluminum, whereas that of the other is copper. Water in these pots is boiling away at \(100.0^{\circ} \mathrm{C}\) at the same rate. The temperature of the heating element on which the aluminum bottom is sitting is \(155.0^{\circ} \mathrm{C}\). Assume that heat enters the water only through the bottoms of the pots and find the temperature of the heating element on which the copper bottom rests.

A solid cylinder is radiating power. It has a length that is ten times its radius. It is cut into a number of smaller cylinders, each of which has the same length. Each small cylinder has the same temperature as the original cylinder. The total radiant power emitted by the pieces is twice that emitted by the original cylinder. How many smaller cylinders are there?

Liquid helium is stored at its boiling-point temperature of \(4.2 \mathrm{~K}\) in a spherical container \((r=0.30 \mathrm{~m})\). The container is a perfect blackbody radiator. The container is surrounded by a spherical shield whose temperature is \(77 \mathrm{~K}\). A vacuum exists in the space between the container and the shield. The latent heat of vaporization for helium is \(2.1 \times 10^{4} \mathrm{~J} / \mathrm{kg} .\) What mass of liquid helium boils away through a venting valve in one hour?

A car parked in the sun absorbs energy at a rate of 560 watts per square meter of surface area. The car reaches a temperature at which it radiates energy at this same rate. Treating the car as a perfect radiator \((e=1)\), find the temperature.

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