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

To find the temperature \( T_2 \) of the second black body, we use the Stefan-Boltzmann Law, which states that the power \( P \) radiated by a black body is given by \( P = \sigma A T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the absolute temperature.

Given two black bodies radiating the same power:

\[P_1 = \sigma \times 4\pi R_1^2 \times T_1^4\]

\[P_2 = \sigma \times 4\pi R_2^2 \times T_2^4\]

Since \( P_1 = P_2 \):

\[\sigma \times 4\pi R_1^2 \times T_1^4 = \sigma \times 4\pi R_2^2 \times T_2^4\]

Canceling out the common terms, we have:

\[R_1^2 \times T_1^4 = R_2^2 \times T_2^4\]

Substitute the given values \( R_1 = 2 \), \( T_1 = 400 \), \( R_2 = 4 \):

\[(2)^2 \times (400)^4 = (4)^2 \times T_2^4\]

\[4 \times 256 \times 10^6 = 16 \times T_2^4\]

Simplifying:

\[256 \times 10^6 = 4 \times T_2^4\]

\[64 \times 10^6 = T_2^4\]

Taking the fourth root:

\[T_2 = 300 \, \text{K}\]

Thus, the temperature of the second black body is \(\boxed{300 \, \text{K}}\).