Question:medium

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?

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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.
Updated On: Nov 26, 2025
  • \( 200 \, \text{K} \)
  • \( 300 \, \text{K} \)
  • \( 250 \, \text{K} \)
  • \( 400 \, \text{K} \)
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The Correct Option is B

Solution and Explanation

The temperature \( T_2 \) of the second black body is determined using the Stefan-Boltzmann Law, \( P = \sigma A T^4 \), which relates radiated power \( P \) to surface area \( A \) and absolute temperature \( T \). \( \sigma \) represents the Stefan-Boltzmann constant.
Given two black bodies with equal radiated 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 \( P_1 = P_2 \):
\[\sigma \times 4\pi R_1^2 \times T_1^4 = \sigma \times 4\pi R_2^2 \times T_2^4\]
Simplifying by canceling common terms yields:
\[R_1^2 \times T_1^4 = R_2^2 \times T_2^4\]
Substituting the provided values \( R_1 = 2 \), \( T_1 = 400 \), and \( R_2 = 4 \):
\[(2)^2 \times (400)^4 = (4)^2 \times T_2^4\]
\[4 \times 256 \times 10^6 = 16 \times T_2^4\]
Further simplification results in:
\[256 \times 10^6 = 4 \times T_2^4\]
\[64 \times 10^6 = T_2^4\]
Taking the fourth root to solve for \( T_2 \):
\[T_2 = 300 \, \text{K}\]
Therefore, the temperature of the second black body is \(\boxed{300 \, \text{K}}\).
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