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The relation \(R_{T}=\sigma T^{4}\) is exact for blackbodies and holds for all temperatures. Why is this relation not used as the basis of a definition of temperature at, for instance, \(100^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The Stefan-Boltzmann law, although universally applicable, is not used as a temperature standard because it requires precise measurement of radiated energy from a perfect black body, which is experimentally challenging, and the temperature measurement with this method is highly sensitive to slight variations in radiated energy hence making it impractical.

Step by step solution

01

Understanding the Stefan-Boltzmann law

The Stefan-Boltzmann law is an empirical law, derived from observational data, which states that the total energy radiated per unit surface area of a black body in unit time, also known as the black body radiant emittance (\(R_{T}\)), is directly proportional to the fourth power of the black body's thermodynamic temperature (T). The constant of proportionality (\(\sigma\)) is the Stefan-Boltzmann constant. So, the relation \(R_{T}=\sigma T^{4}\) is a mathematical representation of this law.
02

Understanding the concept of temperature standard

A temperature standard is a universally agreed-upon reference point for the measurement of temperature. It must be easily reproducible and applicable to various types of temperature measurements in different fields. The triple point of water (the unique condition where water can simultaneously exist as a solid, liquid, and gas) is currently used as the standard, serving as the defining point for the Kelvin scale.
03

Explaining why the Stefan-Boltzmann law is not used as temperature standard

Although the Stefan-Boltzmann relationship is universally applicable, there are a few reasons why it is not used as a temperature standard. Firstly, it requires precise measurement of the radiated energy from a perfect black body, which is experimentally challenging as true black bodies don't exist in nature. Secondly, the Stefan-Boltzmann law describes an idealized situation and doesn't account for factors such as emissivity of a real object which affects the radiated energy. Thirdly, the temperature measurement using this relationship would be extremely sensitive to slight variations in radiated energy, making it impractical for use as a standard temperature measurement method.

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Key Concepts

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

Black Body Radiation
Imagine an object that absorbs all the radiation that falls on it, no matter what wavelength or angle the radiation arrives from. This theoretical object is known as a 'black body', and it is a perfect emitter as well as a perfect absorber of radiation. Black body radiation refers to the electromagnetic radiation (light, microwaves, etc.) that a black body emits when it is at a certain temperature.

The spectrum of black body radiation is characterized by its continuous, smooth nature and it peaks at different wavelengths depending on the temperature of the black body. An important aspect of black body radiation is that it is solely determined by the temperature, and not by the body’s shape or composition. This is what the Stefan-Boltzmann law, represented by the equation \( R_{T}=\sigma T^{4} \), mathematically expresses: the radiant energy emitted per unit area of a black body is proportional to the fourth power of its absolute temperature.

However, in practical applications, such exact black bodies are idealizations; real-world objects always reflect some radiation and are not perfect emitters, which is one reason the Stefan-Boltzmann law provides challenges when we try to use it for precise temperature measurements.
Thermodynamic Temperature
Thermodynamic temperature is one of the most crucial concepts in the study of heat and thermodynamics. It is an absolute measure of the kinetic energy contained in the particles of an object. Unlike the temperature scales we might be familiar with, such as Celsius or Fahrenheit, which are relative (they have arbitrarily defined points like the freezing and boiling points of water), thermodynamic temperature is measured on a scale that starts at absolute zero. This is the point where no more thermal energy can be removed from a system, theoretically the coldest possible temperature.

The Kelvin scale is used to measure thermodynamic temperature. One Kelvin (K) is defined as \(\frac{1}{273.16}\) of the thermodynamic temperature of the triple point of water. This means that the Kelvin scale provides a basis for temperature measurements that is informed by physical constants, rather than the more arbitrary designations used in other scales. The Stefan-Boltzmann law relates to thermodynamic temperature because it describes how the energy emitted by a black body scales with the temperature measured on an absolute scale, such as Kelvin.
Temperature Standards
Temperature standards serve as vital reference points for the calibration and measurement of temperature. A reliable standard must be stable, reproducible, and universally agreed upon. Currently, the triple point of water is one such standard. It is the singular condition under which water can coexist in all three phases – solid, liquid, and gas – and it occurs at a precise and reproducible temperature of 0.01 degrees Celsius, or 273.16 Kelvin.

These standards are essential because they provide the foundation for temperature measurements across various scientific and industrial fields. One might wonder why the Stefan-Boltzmann law, which ties in so neatly with the concept of thermodynamic temperature, is not used as a temperature standard. As outlined in the exercise solution, this is due to practicality. It's exceedingly difficult to create a perfect black body and accurately measure the tiny amounts of radiation it would emit, especially for standard temperature measurements which require extraordinary precision. The triple point of water, on the other hand, offers a much more accessible and reproducible point of reference for establishing a temperature scale. Moreover, it is less susceptible to the sorts of errors that can arise from trying to measure radiated energy, as is necessary for using the Stefan-Boltzmann law as a temperature standard.

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Most popular questions from this chapter

If we look into a cavity whose walls are kept at a constant temperature no details of the interior are visible. Explain.

The average rate of solar radiation incident per unit area on the earth is \(0.485 \mathrm{cal} / \mathrm{cm}^{2}\) min (or \(338 \mathrm{~W} / \mathrm{m}^{2}\) ). (a) Explain the consistency of this number with the solar constant (the solar energy falling per unit time at normal incidence on a unit area) whose value is \(1.94 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}\) (or \(1353 \mathrm{~W} / \mathrm{m}^{2}\) ). (b) Consider the earth to be a blackbody radiating energy into space at this same rate. What surface temperature would the earth have under these circumstances?

Consider two cavities of arbitrary shape and material, each at the same temperature \(T\), connected by a narrow tube in which can be placed color filters (assumed ideal) which will allow only radiation of a specified frequency \(v\) to pass through. (a) Suppose at a certain frequency \(v^{\prime}, \rho_{T}\left(v^{\prime}\right) d v\) for cavity 1 was greater than \(\rho_{T}\left(v^{\prime}\right) d v\) for cavity 2. A color filter which passes only the frequency \(v^{\prime}\) is placed in the connecting tube. Discuss what will happen in terms of energy flow. (b) What will happen to their respective temperatures? (c) Show that this would violate the second law of thermodynamics; hence prove that all blackbodies at the same temperature must emit thermal radiation with the same spectrum independent of the details of their composition.

Use the relation \(R_{T}(v) d v=(c / 4) \rho_{T}(v) d v\) between spectral radiancy and energy density, together with Planck's radiation law, to derive Stefan's law. That is, show that $$ R_{T}=\int_{0}^{\infty} \frac{2 \pi h}{c^{2}} e^{v^{3} d v}{e^{h v / k T}-1}=\sigma T^{4} $$ where \(\sigma=2 \pi^{5} k^{4} / 15 c^{2} h^{3}\). $$ \left(\text { Hint: } \int_{0}^{\infty} \frac{q^{3} d q}{e^{q}-1}=\frac{\pi^{4}}{15}\right) $$

Does Planck's theory suggest quantized atomic energy states?

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