/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 How large is the sun? By measuri... [FREE SOLUTION] | 91Ó°ÊÓ

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How large is the sun? By measuring the spectrum of wave- lengths of light from our sun, we know that its surface temperature is 5800 \(\mathrm{K}\) . By measuring the rate at which we receive its energy on earth, we know that it is radiating a total of \(3.92 \times 10^{26} \mathrm{J} / \mathrm{s}\) and behaves nearly like an ideal blackbody. Use this information to calculate the diameter of our sun.

Short Answer

Expert verified
The diameter of the sun is approximately \(1.39 \times 10^9\) meters.

Step by step solution

01

Understand the Blackbody Concept

A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation and re-emits it according to its temperature. The sun can be approximated as a blackbody radiator, meaning it follows the Stefan-Boltzmann Law for radiative energy output.
02

Apply the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the total energy radiated per unit surface area of a blackbody is given by \[ E = \sigma T^4 \]where \( E \) is the radiant energy per unit area, \( \sigma \) (Stefan-Boltzmann constant) is approximately \(5.67 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4}\), and \( T \) is the temperature in Kelvin.
03

Set Up the Equation for Total Power Radiated

For a spherical body like the sun, the total power emitted \( P \) is given by the formula \[ P = 4 \pi R^2 \sigma T^4 \]We know \( P = 3.92 \times 10^{26} \text{ J/s} \) and \( T = 5800 \text{ K} \). We aim to find the radius \( R \) of the sun.
04

Solve for the Radius

Rearrange the formula to solve for \( R \):\[ R^2 = \frac{P}{4 \pi \sigma T^4} \]Substitute the given values:\[ P = 3.92 \times 10^{26} \text{ J/s}, \quad T = 5800 \text{ K}, \quad \sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4} \]\[ R^2 = \frac{3.92 \times 10^{26}}{4 \pi (5.67 \times 10^{-8}) (5800)^4} \]
05

Calculate the Radius

Use a calculator to find \( R \):\[ R ≈ 6.96 \times 10^8 \text{ meters} \]
06

Calculate the Diameter

The diameter of the sun is twice the radius:\[ D = 2R \approx 2 \times 6.96 \times 10^8 \text{ meters} \]\[ D \approx 1.39 \times 10^9 \text{ meters} \]
07

Finalize the Calculation

The diameter of the sun, given all calculations, is approximately \(1.39 \times 10^9\) meters.

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Key Concepts

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

Blackbody Radiation
Blackbody radiation refers to the way an idealized physical body absorbs and emits electromagnetic radiation. A blackbody is an object that perfectly absorbs all wavelengths of light and re-emits energy based on its temperature, without any reflection. The sun, in this context, is treated as a nearly perfect blackbody because it closely fits the model's assumptions.

When we talk about blackbody radiation, we imply that the object's emitted energy spectrum is entirely determined by its temperature, following the principles of the Stefan-Boltzmann Law. The hotter the blackbody, the more energy it emits. This energy emission explains why stars like our sun can emit enormous quantities of radiation visibly as light.

Key characteristics of blackbody radiation include:
  • The energy emitted increases with the temperature of the object.
  • The radiation emitted covers a broad spectrum of electromagnetic waves, including visible and ultraviolet light.
  • The total emitted power is proportional to the fourth power of the temperature, as per the Stefan-Boltzmann Law.
Solar Diameter Calculation
The calculation of the solar diameter is an application of understanding blackbody radiation and the Stefan-Boltzmann Law. Given that the sun can be modeled as a blackbody, we use the Stefan-Boltzmann Law to determine its size based on its energy output and temperature.

According to the Stefan-Boltzmann Law, the energy emitted by the sun per unit area can be expressed as \[ E = \sigma T^4 \] where \( \sigma \) is Stefan-Boltzmann's constant, approximately equal to \( 5.67 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4} \), and \( T \) is the temperature in Kelvin.

For the sun's total power output \( P \), the formula adapts for a spherical body as:
\[ P = 4 \pi R^2 \sigma T^4 \] where \( R \) is the sun's radius, and \( P \) is the power output, \( 3.92 \times 10^{26} \text{ J/s} \) in this case. By rearranging and solving for \( R \), the radius of the sun can be calculated. The diameter, \( D \), is simply twice the radius:
\[ D = 2R \]
This calculation allows us to determine that the sun's diameter is approximately \( 1.39 \times 10^9 \text{ meters} \).
Solar Energy Output
Solar energy output refers to the total energy the sun emits, which reaches Earth and beyond. As the source of nearly all energy in our solar system, the sun's power output is colossal, influencing the climate and supporting life.

For practical calculations, we assume the sun behaves like a blackbody according to the Stefan-Boltzmann Law, emitting energy across its entire surface. The given power output of the sun is \( 3.92 \times 10^{26} \text{ J/s} \), which refers to the total energy radiated into space per second. This massive output is what lightens and warms the Earth, driving processes such as photosynthesis and the water cycle.

Understanding solar energy output is essential for fields like
  • astronomy, where it aids in understanding stellar characteristics,
  • renewable energy research, focusing on harnessing solar power, and
  • climatology, by influencing atmospheric conditions and ocean temperatures.
Recognizing how this energy is distributed and utilized is key to exploring future energy solutions and understanding our environment's dynamics.

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Most popular questions from this chapter

Jogging in the heat of the day. You have probably seen people jogging in extremely hot weather and wondered "Why? As we shall see, there are good reasons not to do this! When jogging strenuously, an average runner of mass 68 \(\mathrm{kg}\) and surface area 1.85 \(\mathrm{m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{W}, 80 \%\) of which is converted to heat. The jogger radiates heat, but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C} .\) (We shall neglect conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is \(40.0^{\circ} \mathrm{C}\) (104 F)? (Remember that he radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\) . (e) How many 750 \(\mathrm{mL}\) bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 \(\mathrm{kg}\) .

A copper calorimeter can with mass 0.100 \(\mathrm{kg}\) contains 0.160 \(\mathrm{kg}\) of water and 0.018 \(\mathrm{kg}\) of ice in thermal equilibrium at atmospheric pressure. If 0.750 \(\mathrm{kg}\) of lead at a temperature of \(255^{\circ} \mathrm{C}\) is dropped into the can, what is the final temperature of the system if no heat is lost to the surroundings?

\(\bullet\) In a physics lab experiment, a student immersed 200 one-cent coins (each having a mass of 3.00 g) in boiling water. After they reached thermal equilibrium, she quickly fished them out and dropped them into 0.240 \(\mathrm{kg}\) of water at \(20.0^{\circ} \mathrm{C}\) in an insulated container of negligible mass. What was the final temperature of the coins? [One-cent coins are made of a metal alloy-mostly zinc-with a specific heat capacity of 390 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) . ]\)

Shivering. You have no doubt noticed that you usually shiver when you get out of the shower. Shivering is the body's way of generating heat to restore its internal temperature to the normal \(37^{\circ} \mathrm{C},\) and it produces approximately 290 \(\mathrm{W}\) of heat power per square meter of body area. \(\mathrm{A} 68 \mathrm{kg}(150 \mathrm{lb}), 1.78 \mathrm{m}\) (5 foot, 10 inch) person has approximately 1.8 \(\mathrm{m}^{2}\) of surface area. How long would this person have to shiver to raise his or her body temperature by \(1.0 \mathrm{C}^{\circ},\) assuming that none of this heat is lost by the body? The specific heat capacity of the body is about 3500 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\) .

\(\cdot\) Evaporative cooling. The evaporation of sweat is an important mechanism for temperature control in some warmblooded animals. (a) What mass of water must evaporate from the skin of a 70.0 \(\mathrm{kg}\) man to cool his body 1.00 \(\mathrm{C}^{\circ}\) . The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) The specific heat capacity of a typical human body is 3480 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) .\) (b) What volume of water must the man drink to replenish the evaporated water? Compare this result with the volume of a soft-drink can, which is 355 \(\mathrm{cm}^{3} .\)

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