Question on blackbody radiance

In summary, the conversation discusses the calculation of the rate of energy absorbed and emitted by a slab of matter, as well as the determination of its equilibrium temperature. The Stefan-Boltzmann law is used to calculate these values, with the assumption that the absorption coefficient is independent of wavelength. The equilibrium temperature is found to be 300K when the rates of energy absorbed and emitted are equal.
  • #1
beasht
3
0
Hi,
just looking for some pointers in how to solve the following question, such as formulas, etc,
any help at all so as i can proceed in the right direction will be much appreciated,
thanks in advance.
not sure how to start it

Consider a slab of matter contained within two infinite planes a distance of s = 1m apart.
absorption coefficent =1m2kg-1
density of slab 1kgm-3
heat capacity 103^3jk-1kg-1
assume the absorption coefficient to be independent of wave length

1.Suppose one face of a slab is illuminated by a source emitting blackbody radiation at temp 300k, compute the rate at which the slab would tend to warm up by absorption of incident radiance.
2. assuming the slab is also at temp of 300k compute the irradiance emitted by the slab. Compute the rate at which the slab would tend to cool by emission.
3. assuming that the incident black body radiation remains unchanged in time, compute the temp at which the slab will come into equilibrium.
 
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  • #2
1. The rate at which the slab warms up can be computed by using the Stefan-Boltzmann law, which states that the rate of energy absorbed by the slab is proportional to the fourth power of the temperature of the incident radiation. Therefore, the rate of energy absorbed by the slab can be calculated as follows:Rate of energy absorbed = σ * A * (T^4 - T0^4) where σ is the Stefan-Boltzmann constant (5.67 x 10^-8 Wm^-2K^-4), A is the area of the slab (1m2), T is the temperature of the incident radiation (300K) and T0 is the initial temperature of the slab (300K). Therefore, the rate of energy absorbed is equal to 0 W.2. The rate at which the slab cools down can be calculated using the Stefan-Boltzmann law, which states that the rate of energy emitted by the slab is proportional to the fourth power of its temperature. Therefore, the rate of energy emitted by the slab can be calculated as follows: Rate of energy emitted = σ * A * (T^4 - T0^4) where σ is the Stefan-Boltzmann constant (5.67 x 10^-8 Wm^-2K^-4), A is the area of the slab (1m2), T is the temperature of the slab (300K) and T0 is the temperature of the surrounding environment (300K). Therefore, the rate of energy emitted is equal to 0 W.3. The equilibrium temperature of the slab can be determined by balancing the rate of energy absorbed and the rate of energy emitted. This can be done by setting the rate of energy absorbed equal to the rate of energy emitted. From the above calculations, we can see that both rates are equal to 0 W. Therefore, the equilibrium temperature of the slab is 300K.
 

1. What is blackbody radiation?

Blackbody radiation is the thermal electromagnetic radiation given off by an idealized object known as a blackbody. It is the spectrum of light emitted by an object at a given temperature, and the distribution of this radiation depends only on the temperature of the object.

2. How is blackbody radiation related to temperature?

The intensity and distribution of blackbody radiation is directly related to the temperature of the emitting object. As the temperature increases, the peak of the radiation shifts to shorter wavelengths and the overall intensity of the radiation increases.

3. What is Planck's law and how does it relate to blackbody radiation?

Planck's law is a mathematical formula that describes the distribution of blackbody radiation at a given temperature. It states that the intensity of the radiation is proportional to the temperature and follows a specific curve known as the Planck curve.

4. How is blackbody radiation important in astronomy?

Blackbody radiation is important in astronomy because it allows us to determine the temperature of celestial objects by studying their emitted radiation. It also provides valuable information about the composition and physical properties of these objects.

5. How does the color of an object relate to its blackbody radiation?

The color of an object is determined by the wavelengths of light it reflects or emits. In the case of blackbody radiation, the color of an object is related to its temperature. Objects with higher temperatures will emit shorter, bluer wavelengths of light while cooler objects will emit longer, redder wavelengths.

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