91Ó°ÊÓ

The Stefan-Boltzmann law plays a crucial role in understanding radiation heat transfer. It relates the power radiated by a black body to its temperature. The formula is:

\[ P = \text{emissivity} \times \text{Stefan-Boltzmann constant} \times \text{surface area} \times \text{temperature}^4 \]
In symbols, this equation is written as:

\[ P = \theta \times \boldsymbol{\text{σ}} \times A \times T^4 \]
Where:

• \text{\(P\)} is the radiated power
• \text{\(\theta\)} is the emissivity
• \text{\(\boldsymbol{\text{σ}}\)} is the Stefan-Boltzmann constant \(( 5.67 \times 10^{-8} \text{ W} \text{ m}^{-2} \text{ K}^{-4})\)
• \text{\(A\)} is the surface area
• \text{\(T\)} is the temperature in Kelvin

This law helps us understand how objects emit energy in the form of thermal radiation. By rearranging this equation, we can solve for the temperature if the other quantities are known.

To use the Stefan-Boltzmann law effectively, we need the surface area of the body in question. For a sphere, the surface area \(A\) is calculated using:

\[ A = 4 \times \text{Ï€} \times \text{R}^2 \]
First, determine the radius \(R\):

\[ R = \frac{D}{2} \]
For a diameter \(D\) of 0.5 meters, \(R\) becomes:

\[ R = 0.25 \text{ m} \]
Next, plug \(R\) into the surface area formula:

\[ A = 4 \times \text{Ï€} \times (0.25 \text{ m})^2 = \text{Ï€} \text{ m}^2 \]
So, the surface area \(A\) of the probe is \(\text{Ï€} \text{ m}^2\). Accurate surface area calculation is essential for solving radiation heat transfer problems.

Thermal emissivity \(\varepsilon\) is a measure of a material's ability to emit thermal radiation compared to a perfect black body. The value ranges from 0 to 1, where 1 represents a perfect emitter.
For our probe, the emissivity is given as 0.8. This means the probe emits 80% of the radiation that a perfect black body would. Emissivity is critical in the Stefan-Boltzmann law because it directly impacts the amount of radiated power:

\[ P = \theta \times \boldsymbol{\text{σ}} \times A \times T^4 \]
Materials with different emissivities will radiate heat at different rates, even when their temperatures are the same.
Therefore, understanding and knowing the emissivity of the object in question is vital for accurate calculation of temperature or radiated power.

The concept of steady state is important in thermal problems. When a system is in a steady state, its properties do not change over time. For the given problem, the probe transfers energy by radiation at a constant rate, meaning it is in thermal equilibrium.
At steady state, the energy being radiated away from the probe exactly equals the energy being absorbed. In our scenario, the probe does not receive additional radiation from the sun or deep space. Thus, the only thermal energy exchange is through its radiation.
The steady state simplifies the problem because we do not have to account for any changes in temperature over time. We can focus solely on the Stefan-Boltzmann law to find the surface temperature.