Question

Three objects coloured black, grey and white can withstand hostile conditions at $2800^{\circ}C$. These objects are thrown into a furnace where each of them attains a temperature of $2000^{\circ}C$. Which object will glow brightest?
A. The white object
B. The black object.
C. All glow with equal brightness
D. Grey object.

Answer 1

Hint: We know that Kirchhoff's law of thermal radiation gives the relation of a material to the radiations emitted or absorbed, at any temperature. We can use this to find the individual relationships of the material and then compare them.

Complete step by step answer:
We know that according to Kirchhoff’s law of thermal radiation, for any body of material which is in thermodynamic equilibrium, the absorption and emission of radiations, the function of temperature and wavelength gives the ratio of the power emitted to the coefficient of absorption. Where emissivity is, the ratio of power of the emitted radiation to the power of radiation emitted by a black body. And absorptivity or coefficient of absorption is the fraction of light that is absorbed when incident on the material.
According to this law, all the wavelength of the electromagnetic radiation is absorbed at any temperature. Also, a perfect black body can absorb or emit all the radiations at any temperature.
Here, clearly, none of the given bodies disintegrate at $2000^{\circ}C$. Then all the objects will absorb the heat but the black object will absorb maximum heat. Hence clearly black object will absorb all radiations and glow brightly due to its has high power of absorptivity and emissivity
Hence, the correct answer is option B.

Note:
In simple words, the law says that only if the body is in thermodynamic equilibrium, then the radiation emitted or absorbed given by the emissivity and the absorptivity of the material are equal. Also, black body emits and absorbs maximum radiations.