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The Solar Constant The surface of the Sun has a temperature of \(5500^{\circ} \mathrm{C}\). (a) Treating the Sun as a perfect blackbody, with an emissivity of \(1.0,\) find the power that it radiates into space. The radius of the Sun is \(7.0 \times 10^{8} \mathrm{m},\) and the temperature of space can be taken to be \(3.0 \mathrm{K}\). (b) The solar constant is the number of watts of sunlight power falling on a square meter of the Earth's upper atmosphere. Use your result from part (a) to calculate the solar constant, given that the distance from the Sun to the Earth is \(1.5 \times 10^{11} \mathrm{m}\).

Short Answer

Expert verified
The power radiated by the Sun is approximately \(3.846 \times 10^{26}\,W\). The solar constant is about \(1361\,W/m^2\).

Step by step solution

01

Convert Temperature to Kelvin

Convert the surface temperature of the Sun from Celsius to Kelvin by adding 273.15. \[ T = 5500 + 273.15 = 5773.15\, \text{K} \]
02

Calculate the Power Radiated by the Sun

Use the Stefan-Boltzmann Law: \[ P = \sigma A T^4 \] where \( \sigma \) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \text{W/m}^2\text{K}^4\), \( A \) is the surface area of the Sun \(4 \pi R^2\), and \( T \) is the temperature in Kelvin. Calculate the surface area: \[ A = 4 \pi (7.0 \times 10^8)^2 \]. Then, calculate the power: \[ P = 5.67 \times 10^{-8} \times 4 \pi (7.0 \times 10^8)^2 \times (5773.15)^4 \].
03

Find the Solar Constant

The solar constant is the power per unit area at the distance Earth is from the Sun. Use the inverse square law: \[ S = \frac{P}{4 \pi d^2} \] where \( P \) is the power calculated in the previous step and \( d \) is the distance from the Sun to the Earth. Substitute \( d = 1.5 \times 10^{11} \) m into the equation to find \( S \).

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in physics that describes how much power a blackbody radiates per unit area based on its temperature. This law is essential in astrophysics and climate science.

According to the Stefan-Boltzmann Law, the power radiated by a blackbody is directly proportional to the fourth power of its temperature in Kelvin. Mathematically, it's stated as:
  • \( P = \sigma A T^4 \)
Here, \( P \) is the total 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, and \( T \) is the temperature.

For the Sun, considering it as a perfect blackbody, this calculation helps us understand the amount of energy it emits into space. This model is simplified, but it gives us a good approximation of the Sun’s radiant energy output.
blackbody radiation
Blackbody radiation refers to the theoretical concept of a perfect emitter and absorber of energy. In physics, a blackbody is an idealized object that absorbs all incident electromagnetic radiation and re-emits it in a spectrum that is solely determined by its temperature.

No actual physical body is a perfect blackbody, but many objects approximate this behavior very closely. The Sun, for instance, is often modeled as a blackbody with an emissivity of 1, meaning it is considered an ideal emitter of radiation.

The radiation emitted follows Planck’s law and depends solely on the temperature of the blackbody. As the temperature increases, the peak wavelength of the emitted radiation shifts to shorter wavelengths—a principle known as Wien’s Law. This behavior is crucial for understanding the thermal emission characteristics of stars and other astronomical objects.
inverse square law
The inverse square law describes how a physical quantity diminishes as the distance from its source increases. It states that the intensity of the effect (such as light, sound, or radiation) is inversely proportional to the square of the distance from the source.

In the context of calculating the solar constant, this principle is applied to estimate the sunlight power reaching Earth. It implies that as the distance from the Sun increases, the energy per unit area decreases according to the formula:
  • \( S = \frac{P}{4\pi d^2} \)
Here, \( S \) is the solar constant, \( P \) is the total power output of the Sun, and \( d \) is the distance from the Sun to Earth.

This law helps in understanding how sunlight spreads out in all directions and how much of it reaches a specific point, such as our planet, influencing everything from weather to ecosystems.
temperature conversion
Temperature conversion is a crucial step in many scientific calculations, especially when working with thermodynamics and blackbody radiation. Converting temperatures from Celsius to Kelvin is fundamental because most scientific equations, like the Stefan-Boltzmann Law, use temperature in Kelvin.

To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. This transformation is necessary because the Kelvin scale starts at absolute zero, which is the point where all molecular motion stops. This makes Kelvin a natural choice for scientific work.

For example, converting the Sun's surface temperature from 5500°C to Kelvin, you get:
  • \( T = 5500 + 273.15 = 5773.15 \, \text{K} \)
This conversion allows for accurate calculations of emitted power and energy according to the physical laws governing blackbody radiation.

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Most popular questions from this chapter

A pendulum consists of a large weight suspended by a steel wire that is \(0.9500 \mathrm{m}\) long. (a) If the temperature increases, does the period of the pendulum increase, decrease, or stay the same? Explain. (b) Calculate the change in length of the pendulum if the temperature increase is \(150.0 \mathrm{C}^{\circ}\) (c) Calculate the period of the pendulum before and after the temperature increase. (Assume that the coefficient of linear expansion for the wire is \(12.00 \times 10^{-6} \mathrm{K}^{-1}\) and that \(g=9.810 \mathrm{m} / \mathrm{s}^{2}\) at the location of the pendulum.)

An exercise machine indicates that you have worked off 2.5 Calories in a minute-and-a-half of running in place. What was your power output during this time? Give your answer in both watts and horsepower.

The specific heat of alcohol is about half that of water. Suppose you have \(0.5 \mathrm{kg}\) of alcohol at the temperature \(20^{\circ} \mathrm{C}\) in one container, and \(0.5 \mathrm{kg}\) of water at the temperature \(30^{\circ} \mathrm{C}\) in a second container. When these fluids are poured into the same container and allowed to come to thermal equilibrium, (a) is the final temperature greater than, less than, or equal to \(25^{\circ} \mathrm{C} ?\) (b) Choose the best explanation from among the following: I. The low specific heat of alcohol pulls in more heat, giving a final temperature that is less than \(25^{\circ}\). II. More heat is required to change the temperature of water than to change the temperature of alcohol. Therefore, the final temperature will be greater than \(25^{\circ}\). III. Equal masses are mixed together; therefore, the final temperature will be \(25^{\circ},\) the average of the two initial temperatures.

If heat is transferred to 150 g of water at a constant rate for \(2.5 \mathrm{min},\) its temperature increases by \(13 \mathrm{C}^{\circ} .\) When heat is transferred at the same rate for the same amount of time to a \(150-g\) object of unknown material, its temperature increases by \(61 \mathrm{C}^{\circ}\). (a) From what material is the object made? (b) What is the heating rate?

Longest Suspension Bridge The world's longest suspension bridge is the Akashi Kaikyo Bridge in Japan. The bridge is \(3910 \mathrm{m}\) long and is constructed of steel. How much longer is the bridge on a warm summer day \(\left(30.0^{\circ} \mathrm{C}\right)\) than on a cold winter day \(\left(-5.00^{\circ} \mathrm{C}\right) ?\)

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