
The absorptivity of Platinum black is
(A) 0.91
(B) 0.98
(C) 1.00
(D) 0.99
Answer
235.5k+ views
Hint: We should know that in science, absorptivity may refer to: Molar absorptivity, in chemistry, a measurement of how strongly a chemical species absorbs light at a given wavelength. Absorptance, in physics, the fraction of radiation absorbed at a given wavelength. For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity. Absorptivity ($\alpha$) is a measure of how much of the radiation is absorbed by the body. Reflectivity ($\rho$) is a measure of how much is reflected, and transmissivity ($\tau$) is a measure of how much passes through the object. Emissivity ($\varepsilon$) is a measure of how much thermal radiation a body emits to its environment.
Complete step-by step answer:
We should know that the primary law governing blackbody radiation is the Planck Radiation Law, which governs the intensity of radiation emitted by unit surface area into a fixed direction from the blackbody as a function of wavelength for a fixed temperature. The Planck law is represented by the shape of curves in the figure to the right.
Thus, we can say that, the algebraic sum of ALL the currents entering and leaving a node must be equal to zero, ${{I}_{(exiting)}}+\text{ }{{I}_{(entering)}}=\text{ }0$. This idea by Kirchhoff is commonly known as the Conservation of Charge.
The emissivity of Platinum black is 0.98. According to Kirchhoff’s law of radiation, absorptivity is equal to emissivity at a given temperature.
Hence, the correct answer is Option B.
Note: We should know that emissivity changes with temperature because of energy that is tied up in the behaviour of the molecules that form the surface. As the material gets to a higher temperature, the molecules move more and more, this means they will usually emit more energy. Large variations in temperature can cause a variation in emissivity. The emissivity of metals tends to increase as the temperature increases. The emissivity of non-metals tends to decrease as the temperature increases.
It should be known to us the solar absorptance of a surface is the fraction of the sun's radiation that the surface absorbs. Solar absorptance is a factor in the calculation of the PV array temperature. If an object has some wavelengths at which radiation is scattered or reflected, then the object will have an emissivity less than 1 at the wavelength, and the fraction that is absorbed will be equal to the emissivity at each wavelength.
Complete step-by step answer:
We should know that the primary law governing blackbody radiation is the Planck Radiation Law, which governs the intensity of radiation emitted by unit surface area into a fixed direction from the blackbody as a function of wavelength for a fixed temperature. The Planck law is represented by the shape of curves in the figure to the right.
Thus, we can say that, the algebraic sum of ALL the currents entering and leaving a node must be equal to zero, ${{I}_{(exiting)}}+\text{ }{{I}_{(entering)}}=\text{ }0$. This idea by Kirchhoff is commonly known as the Conservation of Charge.
The emissivity of Platinum black is 0.98. According to Kirchhoff’s law of radiation, absorptivity is equal to emissivity at a given temperature.
Hence, the correct answer is Option B.
Note: We should know that emissivity changes with temperature because of energy that is tied up in the behaviour of the molecules that form the surface. As the material gets to a higher temperature, the molecules move more and more, this means they will usually emit more energy. Large variations in temperature can cause a variation in emissivity. The emissivity of metals tends to increase as the temperature increases. The emissivity of non-metals tends to decrease as the temperature increases.
It should be known to us the solar absorptance of a surface is the fraction of the sun's radiation that the surface absorbs. Solar absorptance is a factor in the calculation of the PV array temperature. If an object has some wavelengths at which radiation is scattered or reflected, then the object will have an emissivity less than 1 at the wavelength, and the fraction that is absorbed will be equal to the emissivity at each wavelength.
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