Question

A black rectangular surface of area A emits energy E per second at 27°C. If length and breadth is reduced to \( \frac{1}{3} \) of initial value and temperature is raised to 327°C, then energy emitted per second becomes

• A. \( \frac{2E}{9} \)
• B. \( \frac{E}{9} \)
• C. \( \frac{16E}{9} \)
• D. \( \frac{4E}{9} \)

Hint:

In problems involving radiation and temperature, remember that energy radiated by a black body is proportional to the fourth power of the temperature and the area of the surface.

Answer 1

The Correct Option is C

Step 1: Understanding Stefan-Boltzmann Law.
According to Stefan-Boltzmann law, the energy radiated by a black body is given by: \[ E = \sigma A T^4 \] where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the area, and \( T \) is the temperature in Kelvin.
Step 2: Calculating the energy emitted.
The initial temperature is \( 27^\circ \text{C} = 300 \, \text{K} \) and the final temperature is \( 327^\circ \text{C} = 600 \, \text{K} \). The new area is \( \left(\frac{1}{3}\right)^2 \) of the initial area. Thus, the energy radiated is proportional to the temperature raised to the fourth power and the area. Therefore, the new energy is: \[ E' = \frac{A}{9} \times (600)^4 = \frac{16E}{9} \] Step 3: Conclusion.
Thus, the energy emitted per second becomes \( \frac{16E}{9} \), which is option (C).