Question

Two black bodies emit the same amount of radiation per second. The radius of the first is \( R_1 = 2 \, \text{m} \) and its temperature is \( T_1 = 400 \, \text{K} \). If the second body has a radius \( R_2 = 4 \, \text{m} \), what is its temperature \( T_2 \) in Kelvin?

• A. \( 200 \, \text{K} \)
• B. \( 300 \, \text{K} \)
• C. \( 250 \, \text{K} \)
• D. \( 400 \, \text{K} \)

Hint:

The Stefan-Boltzmann law relates the power emitted by a body to its temperature and surface area. For bodies emitting the same power, the temperature ratio depends on the radius ratio raised to the fourth power.

Answer 1

The Correct Option is B

The power emitted by a black body is given by the Stefan-Boltzmann law: \[ P = \sigma A T^4 \] Where: - \( P \) is the power emitted, - \( \sigma \) is the Stefan-Boltzmann constant, - \( A \) is the surface area of the body, - \( T \) is the temperature. The surface area \( A \) of a sphere is given by \( A = 4 \pi R^2 \), so: \[ P_1 = \sigma \times 4 \pi R_1^2 \times T_1^4 \] \[ P_2 = \sigma \times 4 \pi R_2^2 \times T_2^4 \] Since the two bodies emit the same power: \[ P_1 = P_2 \] Substituting the expressions for \( P_1 \) and \( P_2 \): \[ \sigma \times 4 \pi R_1^2 \times T_1^4 = \sigma \times 4 \pi R_2^2 \times T_2^4 \] Canceling out the common terms: \[ R_1^2 \times T_1^4 = R_2^2 \times T_2^4 \] Substituting the given values: \[ (2)^2 \times (400)^4 = (4)^2 \times T_2^4 \] Simplifying: \[ 4 \times 400^4 = 16 \times T_2^4 \] \[ T_2^4 = \frac{400^4}{4} \] \[ T_2^4 = 300^4 \] Taking the fourth root of both sides: \[ T_2 = 300 \, \text{K} \] Thus, the temperature of the second body is \( 300 \, \text{K} \).