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: Understand the change.
A black surface emits energy $E$ per second at $27^\circ$C. Its length and breadth each shrink to one third, and its temperature rises to $327^\circ$C. We find the new emission rate.

Step 2: Recall Stefan's law.
A black body radiates at a rate $E=\sigma A T^4$, so $E$ depends on the area $A$ and the fourth power of the absolute temperature $T$.

Step 3: Change the temperatures to kelvin.
Always use absolute temperature. \[ T_1=27+273=300\ \text{K},\qquad T_2=327+273=600\ \text{K}. \]

Step 4: Find the new area.
Area is length times breadth. Each shrinks to one third, so \[ A_2=\frac{L}{3}\times\frac{B}{3}=\frac{LB}{9}=\frac{A}{9}. \]

Step 5: Set up the ratio.
\[ \frac{E_2}{E_1}=\frac{A_2}{A_1}\left(\frac{T_2}{T_1}\right)^4. \]

Step 6: Put in the numbers.
\[ \frac{E_2}{E}=\frac{1}{9}\times\left(\frac{600}{300}\right)^4=\frac{1}{9}\times2^4=\frac{16}{9}. \]

Step 7: State the result.
\[ E_2=\frac{16E}{9}, \] which is option (4).
\[ \boxed{\dfrac{16E}{9}} \]