Question

A black rectangular surface of area $A$ emits energy $E$ per second at $27^\circ \text{C}$. If length and breadth are reduced to one third of initial value and temperature is raised to $327^\circ \text{C}$, then energy emitted per second becomes

• A. $\frac{4E}{9}$
• B. $\frac{7E}{9}$
• C. $\frac{10E}{9}$
• D. $\frac{16E}{9}$

Hint:

Always remember to convert Celsius temperatures to absolute Kelvin temperatures before applying power laws like the Stefan-Boltzmann Law. Failing to add $273$ is the most common pitfall in these problems!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are analyzing the thermal radiation emitted by a black body. We need to find the new rate of energy emission when the surface dimensions are reduced and its temperature is increased.

Step 2: Key Formula or Approach:
According to the Stefan-Boltzmann Law, the energy emitted per second ($E$) by a black body is directly proportional to its surface area ($A$) and the fourth power of its absolute temperature ($T$):
$$E = \sigma A T^4 \implies E \propto A T^4$$

Step 3: Detailed Explanation:
Let the initial area be $A_1 = A$.
The initial absolute temperature is $T_1 = 27^\circ \text{C} + 273 = 300\ \text{K}$.
The new length and breadth are reduced to one-third of their original values. Since Area = length $\times$ breadth, the new area $A_2$ becomes:
$$A_2 = \left(\frac{L}{3}\right) \times \left(\frac{B}{3}\right) = \frac{LB}{9} = \frac{A}{9}$$ The new absolute temperature is $T_2 = 327^\circ \text{C} + 273 = 600\ \text{K}$.
Using the proportionality, we set up a ratio for the new energy $E_2$:
$$\frac{E_2}{E_1} = \left(\frac{A_2}{A_1}\right) \times \left(\frac{T_2}{T_1}\right)^4$$ Substitute the values into the ratio:
$$\frac{E_2}{E} = \left(\frac{A/9}{A}\right) \times \left(\frac{600}{300}\right)^4$$ $$\frac{E_2}{E} = \frac{1}{9} \times (2)^4$$ $$\frac{E_2}{E} = \frac{1}{9} \times 16 = \frac{16}{9}$$ $$E_2 = \frac{16E}{9}$$

Step 4: Final Answer:
The energy emitted per second becomes $\frac{16E}{9}$, matching option (D).