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What is the rate of energy radiation per unit area of a blackbody at a temperature of \((\) a ) 273 \(\mathrm{K}\) and (b) 2730 \(\mathrm{K} ?\)

Short Answer

Expert verified
At 273 K: \(315.6\) W/m\(^2\), at 2730 K: \(315,584,651\) W/m\(^2\).

Step by step solution

01

Identify the Formula

The total power radiated per unit area by a blackbody is given by the Stefan-Boltzmann law. The formula is \( E = \sigma T^4 \), where \( E \) is the energy radiated 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 in Kelvin.
02

Calculate for 273 K

Using the formula \( E = \sigma T^4 \), plug in \( T = 273 \) K. So, \( E = 5.67 \times 10^{-8} \times (273)^4 \). Compute the result to find the energy radiated per unit area at 273 K.
03

Calculate for 2730 K

Similarly, plug in \( T = 2730 \) K into the formula. Thus, \( E = 5.67 \times 10^{-8} \times (2730)^4 \). Compute the result to find the energy radiated per unit area at 2730 K.
04

Final Calculation

Calculate \((273)^4 = 55568161\) and multiplying it by \(5.67 \times 10^{-8}\) gives \(E_{273} \approx 315.6\) W/m\(^2\). For \((2730)^4 = 5556809801\) and multiplying it by \(5.67 \times 10^{-8}\) gives \(E_{2730} \approx 315584651\) W/m\(^2\).

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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 the field of physics. A blackbody is an idealized object that absorbs all radiation that falls on it, without reflecting any. This makes it a perfect emitter of radiation. The radiation emitted by a blackbody depends only on its temperature and is independent of its material or shape. This kind of radiation covers a spectrum of wavelengths, from infrared to visible light.
Understanding blackbody radiation is important because it serves as a model for real-world objects and stars that emit radiation due to their temperature. For example, the Sun behaves like a near-perfect blackbody. This property makes blackbody radiation a vital tool in studying the thermal characteristics of various astronomical and terrestrial bodies.
Energy Radiated
The energy radiated by a blackbody per unit area is a measure of how much energy is emitted by the object into its surroundings. This emitted energy plays a crucial role in understanding heat transfer and thermodynamics. According to the Stefan-Boltzmann law, the amount of energy radiated by a blackbody is proportional to the fourth power of its temperature. This implies that even a small increase in the temperature can lead to a significant increase in the radiation energy.
  • Energy per unit area increases rapidly with temperature change.
  • The concept helps in calculating the heat loss or gain by bodies in various scientific fields.
Such calculations are fundamental in designing systems that involve heat exchangers or thermal management solutions in engineering.
Temperature Dependence
Temperature plays a significant role in the amount of energy radiated by a blackbody. As shown in the Stefan-Boltzmann Law, the radiation energy is proportional to the fourth power of temperature, represented as \( E = \sigma T^4 \). This demonstrates that radiation energy grows exponentially with temperature.
Higher the temperature, greater is the energy radiated, making the temperature a critical factor in fields requiring thermal analysis, like climate science and materials engineering.
  • Temperature increase leads to significant rise in radiated energy.
  • Key to understanding the thermal balance in natural systems, like the Earth's climate.
The exponential nature of this relationship often surprises new learners, highlighting the importance of understanding how small changes in temperature can have large effects on radiation.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant is a fundamental constant in physics, denoted by \( \sigma \). Its value is \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \). It provides the proportionality factor in the Stefan-Boltzmann law, which relates the energy radiated by a blackbody to its temperature. This constant ensures that the equation gives results in units of energy per unit area per unit time.
The constant was derived from experimental data, and it plays a crucial role in blackbody radiation theory.
  • Essential for calculating energy emission from bodies of different temperatures.
  • Helps to model and predict characteristics of star radiation in astrophysics.
The Stefan-Boltzmann constant is utilized in understanding not only theoretical models but also practical applications like designing radiative heat transfer systems.

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Most popular questions from this chapter

Convert the following Celsius temperatures to Fahrenheit: (a) \(-62.8^{\circ} \mathrm{C},\) the lowest temperature ever recorded in North America (February \(3,1947,\) Snag, Yukon); (b) \(56.7^{\circ} \mathrm{C},\) the highest temperature ever recorded in the United States (July \(10,1913,\) Death Valley, California); \((\mathrm{c}) 31.1^{\circ} \mathrm{C},\) the world's highest average annual temperature (Lugh Ferrandi, Somalia).

Effect of a Window in a Door. A carpenter builds a solid wood door with dimensions 2.00 \(\mathrm{m} \times 0.95 \mathrm{m} \times 5.0 \mathrm{cm} .\) Its thermal conductivity is \(k=0.120 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional 1.8 -cm thickness of solid wood. The inside air temperature is \(20.0^{\circ} \mathrm{C},\) and the outside air temperature is \(-8.0^{\circ} \mathrm{C}\) (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window 0.500 \(\mathrm{m}\) on a side is inserted in the door? The glass is 0.450 \(\mathrm{cm}\) thick, and the glass has a thermal conductivity of 0.80 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 \(\mathrm{cm}\) of glass.

A glass flask whose volume is 1000.00 \(\mathrm{cm}^{3}\) at \(0.0^{\circ} \mathrm{C}\) is completely filled with mercury at this temperature. When flask and mercury are warmed to \(55.0^{\circ} \mathrm{C}, 8.95 \mathrm{cm}^{3}\) of mercury overflow. If the coefficient of volume expansion of mercury is \(18.0 \times 10^{-5} \mathrm{K}^{-1}\) , compute the coefficient of volume expansion of the glass.

Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine \(\left(0^{\circ} \mathrm{R}\right)\) However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

A wood ceiling with thermal resistance \(R_{1}\) is covered with a layer of insulation with thermal resistance \(R_{2} .\) Prove that the effective thermal resistance of the combination is \(R=R_{1}+R_{2}\) .

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