Question

The initial and the final temperatures of a black body are $27^\circ C$ and $177^\circ C$ respectively. The increase in the amount of radiation emitted per second is:

• A. 506.25%
• B. 150.25%
• C. 225.75%
• D. 406.25%

Hint:

Remember, the Stefan-Boltzmann law relates the power radiated to the fourth power of the temperature. This relationship is key for solving radiation problems.

Answer 1

The Correct Option is D

The radiation emitted by a black body is proportional to the fourth power of its temperature. Using the Stefan-Boltzmann law: \[ \frac{I_2}{I_1} = \left( \frac{T_2}{T_1} \right)^4 \] Where $T_1$ is the initial temperature and $T_2$ is the final temperature. Converting the temperatures to Kelvin: \[ T_1 = 27^\circ C + 273 = 300 \, K, \quad T_2 = 177^\circ C + 273 = 450 \, K \] Then, \[ \frac{I_2}{I_1} = \left( \frac{450}{300} \right)^4 = 1.5^4 = 5.0625 \] Thus, the increase in radiation emitted is: \[ (5.0625 - 1) \times 100 = 406.25\% \]