Question

The emissivity of a perfect black body is increased to 16 times by increasing its temperature. If the initial temperature is \( T \), then final temperature of that black body is:

• A. \( 4T \)
• B. \( 8T \)
• C. \( 2T \)
• D. \( 16T \)

Hint:

When a physical property like emissivity changes by a factor, temperature changes by the fourth root of that factor.

Answer 1

The Correct Option is C

The emissivity \( \epsilon \) of a perfect black body is related to the Stefan-Boltzmann law, which states that the energy radiated by the black body is proportional to the fourth power of its temperature: \[ \epsilon \propto T^4 \] If the emissivity increases by 16 times, then the temperature must increase by the square root of 16 (because the temperature is raised to the fourth power). Hence, \[ \frac{\epsilon_2}{\epsilon_1} = \left( \frac{T_2}{T_1} \right)^4 = 16 \] \[ \frac{T_2}{T_1} = \sqrt{16} = 4 \] Thus, the final temperature is: \[ T_2 = 4T \]

Answer 2

Step 1: Understand emissivity and black body radiation
A perfect black body has an emissivity \( \epsilon = 1 \). The power radiated by a black body is given by Stefan-Boltzmann law:
\[ P = \epsilon \sigma T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant and \( T \) is the absolute temperature.

Step 2: Given data
- Initial emissivity corresponds to power \( P_1 = \sigma T^4 \) (since \( \epsilon = 1 \))
- Final emissivity is 16 times initial emissivity, so power radiated \( P_2 = 16 P_1 \)

Step 3: Find final temperature \( T_f \)
Since emissivity of a perfect black body cannot change, the increase in radiation must be due to temperature change:
\[ P_2 = \sigma T_f^4 = 16 \sigma T^4 \]
Dividing both sides by \( \sigma \):
\[ T_f^4 = 16 T^4 \]
Taking fourth root on both sides:
\[ T_f = 16^{1/4} T = 2 T \]

Step 4: Conclusion
The final temperature of the black body is \( 2T \).