Question

The rate of emission of radiation of a black body at 27°C is \(E_1\). If the temperature increased to 327°C the emission is \(E_2\), then \(E_2\) is:

• A. \(4 E_1\)
• B. \(8 E_1\)
• C. \(16 E_1\)
• D. \(24 E_1\)

Hint:

Remember, the Stefan-Boltzmann law demonstrates how dramatically the radiative power of a black body changes with temperature.

Answer 1

The Correct Option is C

The rate of emission of radiation from a black body is given by Stefan's Law, which states that the power emitted per unit area is proportional to the fourth power of the absolute temperature:

\[ E = \sigma T^4 \]

where:

• \( E \) is the rate of emission of radiation,
• \( \sigma \) is the Stefan-Boltzmann constant,
• \( T \) is the absolute temperature in Kelvin.

Let the initial temperature be \( T_1 = 27^\circ \text{C} = 273 + 27 = 300 \, \text{K} \) and the final temperature be \( T_2 = 327^\circ \text{C} = 273 + 327 = 600 \, \text{K} \).

According to Stefan's Law, the ratio of the emissions is given by:

\[ \frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4 \]

Substituting the values of \( T_1 \) and \( T_2 \):

\[ \frac{E_2}{E_1} = \left( \frac{600}{300} \right)^4 = 2^4 = 16 \]

Thus, \( E_2 = 16E_1 \).

So, the correct answer is Option (3), \( E_2 = 16E_1 \).