Question

A perfectly black body emits a radiation at temperature '$T_1$' K. If it is to radiate at 16 times this power, its temperature '$T_2$' K should be

• A. $8T_1$
• B. $4T_1$
• C. $2T_1$
• D. $16T_1$

Hint:

Because power scales with the fourth power of temperature ($P \propto T^4$), doubling the absolute temperature increases the radiated power exponentially by $2^4 = 16$. This short conceptual fact lets you pick option (C) immediately!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem relates the total radiant power emitted by a perfectly black body to its absolute thermodynamic temperature. We need to find the new temperature required to increase the total radiated power by a factor of 16.

Step 2: Key Formula or Approach:
According to Stefan-Boltzmann Law, the total emissive power ($P$) of a perfectly black body is directly proportional to the fourth power of its absolute temperature ($T$): $$ P = \sigma A T^4 \implies P \propto T^4 $$ where $\sigma$ is Stefan's constant and $A$ is the surface area of the black body.

Step 3: Detailed Explanation:
Let's set up the ratio of the power emissions for the two distinct temperature states: $$ \frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 $$ From the given problem conditions, the final power is 16 times the initial power ($P_2 = 16P_1$): $$ \frac{16P_1}{P_1} = \left(\frac{T_2}{T_1}\right)^4 $$ $$ 16 = \left(\frac{T_2}{T_1}\right)^4 $$ Taking the fourth root on both sides of the equation to isolate the temperature ratio: $$ \frac{T_2}{T_1} = \sqrt[4]{16} = 2 $$ $$ T_2 = 2T_1 $$

Step 4: Final Answer:
The final temperature $T_2$ should be $2T_1$, which corresponds to option (C).