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), \( E_2 = 16E_1 \).

Answer 2

To solve the problem, we need to calculate the rate of emission of radiation at two different temperatures based on the Stefan-Boltzmann law.

1. Understanding the Stefan-Boltzmann Law:
The rate of emission of radiation (E) of a black body is proportional to the fourth power of its temperature, as per the Stefan-Boltzmann law: \[ E \propto T^4 \] where \( T \) is the absolute temperature of the body in Kelvin. Let the rate of emission at temperature \( T_1 \) be \( E_1 \), and at temperature \( T_2 \) be \( E_2 \).

2. Converting Temperatures to Kelvin:
We are given the temperatures in Celsius: - \( T_1 = 27^\circ C = 27 + 273 = 300 \, \text{K} \) - \( T_2 = 327^\circ C = 327 + 273 = 600 \, \text{K} \)

3. Applying the Stefan-Boltzmann Law:
We know: \[ \frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4 \] Substituting the values of \( T_2 \) and \( T_1 \): \[ \frac{E_2}{E_1} = \left( \frac{600}{300} \right)^4 = 2^4 = 16 \] Thus, \( E_2 = 16 E_1 \).

4. Identifying the Correct Answer:
The rate of emission at 327°C is 16 times the emission at 27°C, which corresponds to Option 3.

Final Answer:
The correct answer is Option C: 16 E1.