Question

Consider an isolated system of two concentric spherical black bodies. The inner sphere of radius \( R \) is at temperature \( T \), and the outer sphere of radius \( 4R \) is at temperature \( 2T \). The rate of absorption of radiant energy by the outer sphere is: 

• A. \( \mathbf{4 \sigma \pi R^2 T^4} \)
• B. \( 8 \sigma \pi R^2 T^4 \)
• C. \( 16 \sigma \pi R^2 T^4 \)
• D. \( 64 \sigma \pi R^2 T^4 \)

Hint:

- The Stefan-Boltzmann Law states that radiated power is proportional to \( A T^4 \). - A concentric outer sphere absorbs all radiation from the inner sphere. - The rate of absorption equals the rate of radiation from the inner sphere.

Answer 1

The Correct Option is A

Step 1: Understanding Black Body Radiation The power radiated by a black body follows Stefan-Boltzmann law: \[ P = \sigma A T^4 \] where: - \( P \) is the radiated power, - \( \sigma \) is the Stefan-Boltzmann constant, - \( A \) is the surface area of the radiating body, - \( T \) is the temperature in Kelvin.

 Step 2: Calculating the Power Radiated by the Inner Sphere The inner sphere has radius \( R \), so its surface area is: \[ A_{\text{inner}} = 4 \pi R^2 \] The power radiated by the inner sphere: \[ P_{\text{inner}} = \sigma A_{\text{inner}} T^4 \] \[ P_{\text{inner}} = \sigma (4 \pi R^2) T^4 \] \[ P_{\text{inner}} = 4 \sigma \pi R^2 T^4 \] 

Step 3: Absorption by the Outer Sphere The outer sphere completely surrounds the inner sphere and absorbs all the radiation. Thus, the rate of absorption of radiant energy by the outer sphere is equal to the power radiated by the inner sphere: \[ P_{\text{absorbed}} = P_{\text{inner}} = 4 \sigma \pi R^2 T^4 \] 

Step 4: Verifying the Correct Option Comparing with the given options, the correct answer is: \[ \mathbf{4 \sigma \pi R^2 T^4} \]