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

Short Answer

Expert verified
Rate is 315.6 W/m² at 273 K; 3.156 x 10⁶ W/m² at 2730 K.

Step by step solution

01

Understanding the Exercise

We need to determine the rate at which a blackbody radiates energy per unit area at two different temperatures: 273 K and 2730 K. This requires using the Stefan-Boltzmann law for blackbody radiation.
02

Introduction to the Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the energy radiated per unit area of a blackbody per unit time is proportional to the fourth power of the absolute temperature. The equation is given by:\[ E = \sigma T^4 \]where \( E \) is the energy emitted per unit area, \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \, \text{W/m}^2\cdot \text{K}^4) \), and \( T \) is the absolute temperature in Kelvins.
03

Calculating Radiation at 273 K

Using the Stefan-Boltzmann law, we can substitute \( T = 273 \) K into the formula:\[ E_{273} = 5.67 \times 10^{-8} \times (273)^4 \]Performing the calculation gives:\[ E_{273} \approx 315.6 \, \text{W/m}^2 \]
04

Calculating Radiation at 2730 K

Now using the same law but with \( T = 2730 \) K:\[ E_{2730} = 5.67 \times 10^{-8} \times (2730)^4 \]Perform the calculation to find:\[ E_{2730} \approx 3.156 \times 10^6 \, \text{W/m}^2 \]
05

Conclusion

The rate of energy radiation per unit area of a blackbody at 273 K is approximately 315.6 W/m², and at 2730 K is approximately 3.156 x 10⁶ W/m².

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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 a fundamental concept in physics, referring to the type of electromagnetic radiation emitted by a perfect blackbody. A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It does not reflect or transmit any light. This makes it a perfect emitter of radiation when it is at thermal equilibrium, emitting a characteristic spectrum of light that depends solely on its temperature.

When something behaves like a blackbody, the energy it emits is uniformly distributed across different wavelengths. This property is key in many fields, including astrophysics and climate science, as it helps in understanding how stars emit light and even in gauging Earth's energy balance. The emitted radiation covers a range of wavelengths, and as the temperature changes, these wavelengths shift. This process follows the principles of the Stefan-Boltzmann law, which connects the temperature of a body to the radiation it emits.
Energy Emission
Energy emission refers to the release of energy, often in the form of electromagnetic radiation, from a source. In the context of blackbody radiation, energy emission is the radiation emitted by the blackbody per unit area. It depends on the temperature of the blackbody and is described using the Stefan-Boltzmann law.

The amount of energy emitted per unit area (denoted as \( E \)) is directly related to the fourth power of the absolute temperature \( (T) \) of the body. The Stefan-Boltzmann constant \( (\sigma) \) acts as the constant of proportionality, making the equation:
  • \( E = \sigma T^4 \)
  • Where \( \sigma \) is approximately \( 5.67 \times 10^{-8} \text{ W/m}^2 \cdot \text{K}^4 \)
  • \( T \) is the absolute temperature in Kelvin

This equation helps compute how much energy a blackbody emits at any given temperature, as seen in calculating energy at 273 K and 2730 K in the provided exercise.
Temperature Dependence
The Stefan-Boltzmann law highlights the temperature dependence of energy emission from a blackbody. Temperature dependence means that the amount and characteristics of radiation emitted by the blackbody change as its temperature changes.

This relationship is specifically a fourth power dependence, as quantified by the equation \( E = \sigma T^4 \). It implies that even small changes in a blackbody's temperature can lead to large changes in the energy it emits. For instance, doubling the temperature increases the energy emission by a factor of sixteen \((2^4)\).

Due to this strong dependence, this law becomes crucial in predicting and understanding phenomena in various systems, from everyday objects to celestial bodies. It helps scientists explain why hotter objects (like the Sun) emit more energy than cooler ones (like the Earth).
Absolute Temperature
Absolute temperature is a measurement based on the thermodynamic scale, which starts at absolute zero—the point where particles have minimal thermal activity. Kelvin (K) is the unit for measuring absolute temperature.

The concept of absolute temperature is vital when dealing with thermal radiation calculations, such as those involving the Stefan-Boltzmann law. Since energy emission of a blackbody (as per the Stefan-Boltzmann equation) depends on \( T^4 \), it is crucial to have accurate temperature readings.

Absolute zero, zero Kelvin, is theoretically the lowest limit of temperature where no energy is emitted, as all particle movement ceases. By using Kelvin, we ensure that all temperature calculations are relative to this baseline, avoiding negative values and ensuring a standard reference for physical laws. This consistency is what allows formulas like the Stefan-Boltzmann law to be universally applicable across different systems and conditions.

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Most popular questions from this chapter

You have 750 g of water at 10.0\(^\circ\)C in a large insulated beaker. How much boiling water at 100.0\(^\circ\)C must you add to this beaker so that the final temperature of the mixture will be 75\(^\circ\)C?

If the air temperature is the same as the temperature of your skin (about 30\(^\circ\)C), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 70-kg person's body produces energy at a rate of about 500 W due to metabolism, 80% of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750-mL bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is 1.0 kg.)

An asteroid with a diameter of 10 km and a mass of \(2.60 \times 10{^1}{^5} kg\) impacts the earth at a speed of 32.0 km/s, landing in the Pacific Ocean. If 1.00% of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of 10.0\(^\circ\)C), what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10{^1}{^5} kg\).)

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50 kW/m\(^2\). The distance from the earth to the sun is \(1.50 \times 10{^1}{^1} m\), and the radius of the sun is \(6.96 \times 10{^8} m\). (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?

During your mechanical engineering internship, you are given two uniform metal bars \(A\) and \(B\), which are made from different metals, to determine their thermal conductivities. Measuring the bars, you determine that both have length 40.0 cm and uniform cross-sectional area 2.50 cm\(^2\). You place one end of bar \(A\) in thermal contact with a very large vat of boiling water at 100.0\(^\circ\)C and the other end in thermal contact with an ice-water mixture at 0.0\(^\circ\)C. To prevent heat loss along the bar's sides, you wrap insulation around the bar. You weigh the amount of ice initially and find it to be 300 g. After 45.0 min has elapsed, you weigh the ice again and find that 191 g of ice remains. The ice-water mixture is in an insulated container, so the only heat entering or leaving it is the heat conducted by the metal bar. You are confident that your data will allow you to calculate the thermal conductivity \(k_A\) of bar \(A\). But this measurement was tedious-you don't want to repeat it for bar \(B\). Instead, you glue the bars together end to end, with adhesive that has very large thermal conductivity, to make a composite bar 80.0 m long. You place the free end of A in thermal contact with the boiling water and the free end of \(B\) in thermal contact with the ice-water mixture. As in the first measurement, the composite bar is thermally insulated. You go to lunch; when you return, you notice that ice remains in the ice-water mixture. Measuring the temperature at the junction of the two bars, you find that it is 62.4\(^\circ\)C. After 10 minutes you repeat that measurement and get the same temperature, with ice remaining in the ice-water mixture. From your data, calculate the thermal conductivities of bar \(A\) and of bar \(B\).

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