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The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50 \(\mathrm{kW} / \mathrm{m}^{2} .\) The distance from the earth to the sun is \(1.50 \times 10^{11} \mathrm{m},\) and the radius of the sun is \(6.96 \times 10^{8} \mathrm{m}\) . (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

Short Answer

Expert verified
(a) Use step calculations to find \( \,E = 6.33 \times 10^7 \, \text{W/m}² \). (b) Sun's surface temp \( T = 5778 \, \text{K} \).

Step by step solution

01

Calculate the Total Power Received by Earth

The intensity of solar radiation at the earth's atmosphere is given as 1.50 kW/m². To find the total power received by the earth, we have to consider that this power is spread over a sphere with a radius equal to the distance from the earth to the sun. The formula to use is the surface area of a sphere, which is \( 4\pi R^2 \) where \( R \) is the distance from the sun to the earth. Plugging the values, we get: \[ \text{Total Power} = 1.50 \times 10^3 \, \mathrm{W/m^2} \times 4\pi (1.50 \times 10^{11} \mathrm{m})^2 \]
02

Find the Sun's Power Output

The power output from the sun is equal to the total power received by the Earth (calculated above in Step 1), as this power originates from the sun and disperses uniformly in all directions. Calculating this gives an expression for the total solar power output. \[ P = 1.50 \times 10^3 \times 4\pi (1.50 \times 10^{11})^2 \, \text{W} \]
03

Calculate Rate of Radiation per Unit Area from the Sun's Surface

The rate of energy radiation per unit area from the sun's surface can be found using the formula for the surface area of a sphere again, now with the sun's radius: \( 4\pi (6.96 \times 10^8)^2 \). Dividing the total power of the sun by the surface area of the sun gives us the radiation per unit area. \[ \text{Rate of Radiation} = \frac{P}{4\pi (6.96 \times 10^8)^2} \, \text{(W/m²)} \]
04

Determine the Surface Temperature of the Sun Assuming Blackbody Radiation

Assuming the sun radiates as an ideal blackbody, we use the Stefan-Boltzmann Law: \[ E = \sigma T^4 \] where \( E \) is the radiation rate per unit area, \( \sigma \) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \, \text{W/m}^2\text{/K}^4\), and \( T \) is the temperature. Rearranging for \( T \), \[ T = \left(\frac{E}{\sigma}\right)^{1/4} \] Use the result from Step 3 for \( E \) to solve for \( T \).
05

Perform Calculations and Interpret Results

Calculate the exact values using a calculator. First, obtain the value of \( P \) from Step 2. Then, find the radiation rate from Step 3, and finally, using the value of \( E \), apply it in Step 4 to find the temperature of the sun's surface. This involves several arithmetic operations and exponentiation.

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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 the thermal electromagnetic radiation emitted by an object that absorbs all incident electromagnetic waves. This means it emits energy based only on its temperature, without reflecting any light from other sources.
The Sun can be treated as an ideal blackbody, making the concept significant for understanding solar radiation. In essence, any object that is considered a perfect blackbody will emit radiation across all wavelengths.
Importantly, it does so with a peak intensity dependent on its temperature. Per Wien's Displacement Law, the wavelength at which this peak occurs decreases as the temperature increases.
  • The higher the temperature, the more energy is emitted.
  • The radiation pattern is continuous across different wavelengths.
  • The color of the radiation shifts from red to blue as temperature rises.
Understanding blackbody radiation helps us calculate the temperature of stars like the Sun by analyzing the spectrum of emitted radiation. It lays the groundwork for applying the Stefan-Boltzmann Law, which is key to solving problems involving solar radiation.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law provides a formula for calculating the energy emitted by a blackbody. It's one of the cornerstones for understanding how stars like the Sun emit energy.
This law states that the power emitted per unit area of a blackbody is directly proportional to the fourth power of its absolute temperature. Mathematically, it can be expressed as:
\[ E = \sigma T^4 \]
Where:
  • \( E \) is the energy emitted per square meter.
  • \( \sigma \) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \, \text{W/m}^2\text{/K}^4\).
  • \( T \) is the absolute temperature of the blackbody in Kelvin.
This means that a slight increase in temperature leads to a far greater increase in emitted energy. The Stefan-Boltzmann Law is crucial for determining the surface temperature of the Sun by using the energy rate calculated from its radiation. It's simple yet powerful, providing a direct link between temperature and emitted radiation.
Sun's Surface Temperature
Determining the Sun's surface temperature involves understanding both blackbody radiation and the Stefan-Boltzmann Law. It's a pivotal measurement because it helps us understand the Sun's overall energy output and affects everything from climate patterns to solar power design on Earth.
After calculating the rate of solar radiation from the Sun's surface, we apply the Stefan-Boltzmann Law to find its temperature. The formula:
\[ T = \left( \frac{E}{\sigma} \right)^{1/4} \]
shows that by measuring the intensity of radiation \( E \) and using the known constant \( \sigma \), we can calculate \( T \).
  • This process yields the Sun's surface temperature being approximately 5,778 Kelvin.
  • It confirms the immense energy output that makes life possible on Earth.
  • Understanding these principles highlights the interplay of light and heat from the Sun and its influence on our planet.
These calculations showcase the elegance of physics in explaining natural phenomena and our capability to measure and comprehend the universe's workings.

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Most popular questions from this chapter

Out of Tune. The B-string of a guitar is made of steel (density \(7800 \mathrm{kg} / \mathrm{m}^{3} ),\) is 63.5 \(\mathrm{cm}\) long, and has diameter 0.406 \(\mathrm{mm} .\) The fundamental frequency is \(f=247.0 \mathrm{Hz}\) . (a) Find the string tension. (b) If the tension \(F\) is changed by a small amount \(\Delta F,\) the frequency \(f\) changes by a small amount \(\Delta f .\) Show that $$\frac{\Delta f}{f}=\frac{\Delta F}{2 F}$$ (c) The string is tuned to a fundamental frequency of 247.0 Hz when its temperature is \(18.5^{\circ} \mathrm{C}\) . Strenuous playing can make the temperature of the string rise, changing its vibration frequency. Find \(\Delta f\) if the temperature of the string rises to \(29.5^{\circ} \mathrm{C}\) . The steel string has a Young's modulus of \(2.00 \times 10^{11} \mathrm{Pa}\) and a coefficient of linear expansion of \(1.20 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1} .\) Assume that the temperature of the body of the guitar remains constant. Will the vibration frequency rise or fall?

(a) If an area measured on the surface of a solid body is \(A_{0}\) at some initial temperature and then changes by \(\Delta A\) when the temperature changes by \(\Delta T,\) show that $$\Delta A=(2 \alpha) A_{0} \Delta T$$ where \(\alpha\) is the coefficient of linear expansion. (b) A circular sheet of aluminum is 55.0 \(\mathrm{cm}\) in diameter at \(15.0^{\circ} \mathrm{C}\) . By how much does the area of one side of the sheet change when the temperature increases to \(27.5^{\circ} \mathrm{C}\) ?

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a 70.0-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 of a typical human body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) (see Exercise 17.31\() .\) (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{cm}^{3}\right)\).

A U.S. penny has a diameter of 1.9000 \(\mathrm{cm}\) at \(20.0^{\circ} \mathrm{C} .\) The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10^{-5} \mathrm{K}^{-1} .\) What would its diameter be on a hot day in Death Valley \(\left(48.0^{\circ} \mathrm{C}\right) ?\) On a cold night in the mountains of Greenland \(\left(-53^{\circ} \mathrm{C}\right) ?\)

You pour 108 \(\mathrm{cm}^{3}\) of ethanol, at a temperature of \(-10.0^{\circ} \mathrm{C},\) into a graduated cylinder initially at \(20.0^{\circ} \mathrm{C},\) filling it to the very top. The cylinder is made of glass with a specific heat of 840 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and a coefficient of volume expansion of \(1.2 \times 10^{-5} \mathrm{K}^{-1} ;\) its mass is 0.110 \(\mathrm{kg} .\) The mass of the ethanol is 0.0873 \(\mathrm{kg} .\) (a) What will be the final temperature of the ethanol, once thermal equilibrium is reached? (b) How much ethanol will overflow the cylinder before thermal equilibrium is reached?

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