A copper sphere is suspended in an evacuated chamber...

A copper sphere is suspended in an evacuated chamber maintained at $300 \mathrm{~K}$. The sphere is maintained at constant temperature of $900 \mathrm{~K}$ by heating electrically. A total of $300 \mathrm{~W}$ electric power is needed to do this. When half of the surface of the copper sphere is completely blackened, $600 \mathrm{~W}$ is needed to maintain the same temperature of sphere. The emissivity of copper is (A) $1 / 4$ (B) $1 / 3$ (C) $1 / 2$ (D) 1

Key Concepts

Emissivity

Emissivity is a measure of how effectively a surface emits thermal radiation compared to a perfect black body. It is a dimensionless quantity ranging from 0 to 1, indicating the efficiency of radiation from a material's surface. A higher emissivity means the object radiates energy more effectively.

Stefan-Boltzmann Law

The Stefan-Boltzmann law relates the power radiated from a surface to its absolute temperature. It states that the radiated power is proportional to the emissivity, the surface area, and the fourth power of the temperature (P = ??AT^4). This principle is fundamental in analyzing thermal radiation in various systems.

Radiative Heat Transfer

Radiative heat transfer is the process by which thermal energy is exchanged between bodies through electromagnetic radiation. It is particularly important in vacuum environments where conduction and convection are negligible. The efficiency of this transfer is highly dependent on the emissivity and temperature of the surfaces involved.

Surface Modification for Thermal Emission

Modifying a surface, such as by blackening, alters its emissive properties and, consequently, its thermal radiation characteristics. Changing the surface finish can affect the overall energy balance by increasing or decreasing the rate at which energy is emitted, which is critical in controlling the thermal environment of the object.

Transcript

00:01 In the given question we have to find out the emissivity of copper and we've been given an entire situation for that.

00:07 So the temperature of the chamber, which is t chamber, has been given to us as 300 kelvin.

00:18 The temperature of the copper sphere has been given to us as 900 kelvin.

00:25 And the other values which is the par at the first half of the question is given to us as 300 watts whereas the par in the second part of the question when the area is blackened is given to us as 600 watts now we'll be using the stephen bolzman law to calculate to find out the answer so this is the statement the equation of the law which is par is equal to e which is the emissivity sigma which is the surface density into area and here comes the temperature the temperature difference and both of which is raised to the par 4.

01:10 So this is the basic equation that we'll be using to solve the question.

01:14 Now applying this law we will apply this in the first situation when p1 was the par which is in this case it will be e sigma a so t4 will be the temperature of the sphere which is raised to the power 4 minus the temperature of the chamber raised to the power 4.

01:39 This can further be written as 300 kelvin which is equal to e the value of the emissivity of copper into sigma into a we can leave all the values as it is and here we'll write down 900 raised to the power 4 minus 300.

01:58 Raised to the power pole.

02:00 So this is our first equation that we've obtained using the bolzman steepen bolzman law.

02:06 Now for the second equation after the surface is blackened after half area of sphere is blackened, the formula gets changed in the following manner...

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