91Ó°ÊÓ

When considering thermal radiation, emissivity is a crucial concept. This property of a material determines how efficiently it emits thermal radiation compared to an ideal black body. A black body would have an emissivity of 1, radiating 100% of the energy it theoretically could at a given temperature. On the other hand, an emissivity of 0 means no thermal radiation is emitted.
In our scenario, the tungsten sphere has an emissivity of 0.35. This means it emits only 35% of the energy it would if it were a perfect black body.
Understanding emissivity is key to solving problems involving heat exchange through radiation, as it fundamentally affects the rate at which energy is radiated by a surface.
This is why we use the emissivity value in the Stefan-Boltzmann formula, allowing us to calculate the sphere's energy emission more accurately.

The Stefan-Boltzmann law is a cornerstone in understanding radiative heat transfer. It relates the power emitted by a body to its temperature and its ability to emit radiation, which is quantified by the emissivity.
The law is expressed mathematically as:

• \[ P = \varepsilon \, \sigma \, A \,(T^4 - T_{\text{surrounding}}^4) \]

Here,

• \( P \) represents the power emitted,
• \( \varepsilon \) is the emissivity,
• \( \sigma \) is the Stefan-Boltzmann constant,
• \( A \) is the surface area of the radiating body,
• \( T \) is the temperature of the body, and
• \( T_{\text{surrounding}} \) is the temperature of the surroundings.

This law is significant because it underscores how the temperature dramatically affects the power emitted, with power output increasing sharply with temperature due to the fourth power relationship.
In this exercise, the calculated power demonstrates how much energy must be supplied to overcome the energy loss due to radiation.

To calculate the energy radiated by a sphere, we must first determine its surface area. The formula for the surface area of a sphere is:
\[ A = 4\pi r^2 \] where \( r \) is the radius of the sphere.
In our scenario, substituting 0.015 m for the radius gives:
\[ A = 4\pi (0.015)^2 \approx 2.827 \times 10^{-3} \text{ m}^2 \] This calculation is essential because the surface area affects how much thermal radiation the sphere can emit.
A larger surface area means more area over which radiation occurs, increasing the total power radiated. Understanding how to find this area allows us to utilize the Stefan-Boltzmann law correctly to find the radiant energy loss or gain for spherical objects.