To determine how much energy emitted by the inner surface of the hollow spherical shell B falls back onto itself, we can use the principles of blackbody radiation and the geometry of the spheres involved. Both the inner surface of B and the surface of A are considered blackbodies, meaning they absorb and emit radiation perfectly. Let's break this down step by step.
Understanding Blackbody Radiation
A blackbody is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. According to Planck's law, the amount of energy emitted by a blackbody is dependent on its temperature. The Stefan-Boltzmann Law gives us a formula to calculate the total energy radiated per unit area of a blackbody in terms of its temperature:
j* = σT^4
where:
- j* is the total energy emitted per unit area (W/m²)
- σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/m²K^4)
- T is the absolute temperature in Kelvin (K)
Calculating the Energy Emitted
Given that both spheres are at a temperature of 300 K, we can calculate the energy emitted by the inner surface of shell B. First, we need to find the area of the inner surface of B:
The surface area of B is given as 80 cm², which we convert to m² for consistency in units:
Area of B = 80 cm² = 80 x 10^-4 m² = 0.008 m²
Now, we can calculate the energy emitted by the inner surface of B:
j* = σT^4 = (5.67 x 10^-8 W/m²K^4)(300 K)^4
Calculating this gives:
j* ≈ 5.67 x 10^-8 x 8.1 x 10^6 ≈ 459 W/m²
Next, we find the total power emitted by the inner surface of B:
Power emitted by B = j* × Area of B = 459 W/m² × 0.008 m² ≈ 3.672 W
Energy Falling Back on the Inner Surface
Since A is at the center of B and both are blackbodies, the energy emitted by the inner surface of B will be uniformly distributed in all directions. The energy that falls back onto the inner surface of B can be calculated based on the geometry of the situation.
The area of the spherical surface A is:
Area of A = 20 cm² = 20 x 10^-4 m² = 0.002 m²
Now, the fraction of the energy emitted by B that falls back onto itself can be calculated as the ratio of the area of A to the area of B:
Fraction = Area of A / Area of B = 0.002 m² / 0.008 m² = 0.25
Thus, the energy falling back onto the inner surface of B is:
Energy falling back = Total power emitted by B × Fraction = 3.672 W × 0.25 ≈ 0.918 W
Final Result
In summary, the amount of energy emitted by the inner surface of the hollow spherical shell B that falls back onto itself is approximately 0.918 watts. This demonstrates how the geometry of the system and the properties of blackbody radiation interact to determine energy transfer in thermal systems.
