Understanding Black Body Energy Emission

This question asks us to determine the rate at which energy is emitted from a black body filament at a specific temperature (3600 K), given the energy emission rate at another temperature (1800 K). This involves understanding the relationship between a black body's temperature and its radiated energy.

Applying the Stefan-Boltzmann Law

The energy emission rate of a black body is governed by the Stefan-Boltzmann Law. This law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time is directly proportional to the fourth power of its absolute temperature. The formula is given by:

$ P = \sigma A T^4 $

Where:

• \( P \) is the energy emission rate (in Watts, W).
• \( \sigma \) is the Stefan-Boltzmann constant (a physical constant).
• \( A \) is the surface area of the black body (in square meters, m²).
• \( T \) is the absolute temperature of the black body (in Kelvin, K).

In this problem, we assume the surface area \( A \) and the Stefan-Boltzmann constant \( \sigma \) remain constant for the filament. Therefore, the energy emission rate \( P \) is directly proportional to the fourth power of the temperature \( T \).

Setting Up the Proportionality

We can express the relationship between the energy emission rates at two different temperatures, \( T_1 \) and \( T_2 \), as follows:

$ \frac{P_2}{P_1} = \frac{\sigma A T_2^4}{\sigma A T_1^4} $

Since \( \sigma \) and \( A \) are constant, they cancel out:

$ \frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 $

Calculating the New Energy Emission Rate

We are given the following values:

• Initial temperature, \( T_1 = 1800 \) K
• Initial energy emission rate, \( P_1 = 16 \) W
• Final temperature, \( T_2 = 3600 \) K

First, let's find the ratio of the temperatures:

$ \frac{T_2}{T_1} = \frac{3600 \text{ K}}{1800 \text{ K}} = 2 $

Now, we can use this ratio in the Stefan-Boltzmann Law equation:

$ \frac{P_2}{P_1} = (2)^4 $

$ \frac{P_2}{P_1} = 16 $

To find the final energy emission rate \( P_2 \), we rearrange the equation:

$ P_2 = P_1 \times 16 $

Substitute the value of \( P_1 \):

$ P_2 = 16 \text{ W} \times 16 $

$ P_2 = 256 \text{ W} $

Conclusion

Based on the Stefan-Boltzmann Law, if the temperature of the black body filament doubles from 1800 K to 3600 K, the energy emission rate increases by a factor of \( 2^4 = 16 \). Therefore, the energy emission rate at 3600 K is \( 16 \text{ W} \times 16 = 256 \) W.