Understanding Thermal Radiation and Surface Brightness

When objects are heated to a high temperature, they emit electromagnetic radiation known as Thermal Radiation. The amount and spectrum of this radiation depend primarily on the object's temperature and its surface properties. At temperatures around 1400 K, like in this question, the emitted radiation is significant and falls partly within the visible spectrum, making the objects appear luminous or bright.

Surface properties like emissivity and absorptivity play a crucial role in how an object interacts with radiation. Emissivity ($\epsilon$) is a measure of how effectively a surface emits thermal radiation compared to a perfect black body at the same temperature. A perfect black body has an emissivity of 1, while a perfectly reflecting surface has an emissivity of 0.

Kirchhoff's Law Explained

Kirchhoff's Law of Thermal Radiation states that for an object in thermal equilibrium with its surroundings, its emissivity ($\epsilon$) is equal to its absorptivity ($\alpha$). Absorptivity is the fraction of incident radiation that is absorbed by the surface. This law tells us that surfaces that are good absorbers of radiation are also good emitters of thermal radiation, and surfaces that are poor absorbers are poor emitters.

Comparing the Black Spot and Polished Plate

We have two different surfaces on the metal plate: a polished metal surface and a rough, Black Spot. These surfaces have different properties:

• The Polished Plate (metal surface) is typically highly reflective. According to Kirchhoff's Law, a highly reflective surface is a poor absorber ($\alpha$ is low) and therefore also a poor emitter ($\epsilon$ is low).
• The rough, Black Spot is described as black. Black surfaces are excellent absorbers of radiation ($\alpha$ is high), absorbing most of the incident light. By Kirchhoff's Law, a good absorber is also a good emitter ($\epsilon$ is high). A perfectly black surface would have $\epsilon \approx 1$.

So, we can say that the emissivity of the black spot ($\epsilon_{spot}$) is significantly higher than the emissivity of the polished plate ($\epsilon_{polished}$).

Why the Black Spot Appears Brighter

Both the polished metal plate and the rough, Black Spot are heated to the same high temperature (1400 K). When brought into a dark room, we observe the thermal radiation they emit. The amount of thermal radiation emitted by a surface at a given temperature is directly proportional to its emissivity.

Since the Black Spot has a higher emissivity ($\epsilon_{spot} > \epsilon_{polished}$), it emits more thermal radiation per unit area than the Polished Plate at the same temperature. More emitted radiation makes the surface appear brighter to the observer.

Therefore, because the black spot is a much better emitter of Thermal Radiation than the polished plate, it will appear brighter.