Question

The temperature of the black body increases from \(T\) to \(2T\). The factor by which the rate of emission will increase is

• A. 4
• B. 2
• C. 16
• D. 8

Hint:

Stefan-Boltzmann constant \(\sigma = 5.67 \times 10^{-8}\) W m\(^{-2}\) K\(^{-4}\).

Answer 1

The Correct Option is C

To determine the factor by which the rate of emission of a black body increases when its temperature increases from \( T \) to \( 2T \), we use the Stefan-Boltzmann Law. This law states that the power radiated per unit area of a black body is directly proportional to the fourth power of the temperature. Mathematically, it is expressed as:

\(P = \sigma \cdot A \cdot T^4\)

where:

• \(P\) is the power emitted by the black body.
• \(\sigma\) is the Stefan-Boltzmann constant.
• \(A\) is the surface area of the black body.
• \(T\) is the absolute temperature.

For the initial temperature \( T \), the power emitted is:

\(P_1 = \sigma \cdot A \cdot T^4\)

When the temperature is increased to \( 2T \), the power emitted becomes:

\(P_2 = \sigma \cdot A \cdot (2T)^4 = \sigma \cdot A \cdot 16T^4\)

Hence, the rate of emission increases by a factor:

\(\frac{P_2}{P_1} = \frac{16 \sigma \cdot A \cdot T^4}{\sigma \cdot A \cdot T^4} = 16\)

Therefore, the correct answer is that the rate of emission increases by a factor of 16 when the temperature of the black body is doubled.