Question

A surface behaves as a perfect black body at 600 K. Another black body is at 300 K. The ratio of emissive powers is:

• A. 2
• B. 4
• C. 8
• D. 16

Hint:

Radiation heat transfer is highly sensitive to temperature because of the fourth-power dependence.
Doubling the absolute temperature of a black body increases its radiation emission by $2^4 = 16$ times.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question asks for the ratio of emissive powers of two perfect black bodies operating at different absolute temperatures.

Step 2: Key Formula or Approach:
According to the Stefan-Boltzmann Law, the total emissive power ($E$) of a perfect black body is directly proportional to the fourth power of its absolute temperature ($T$):
\[ E = \sigma \cdot T^4 \]
where $\sigma$ is the Stefan-Boltzmann constant.

Step 3: Detailed Explanation:

• Let $E_1$ be the emissive power of the first black body at temperature $T_1 = 600 \text{ K}$.

• Let $E_2$ be the emissive power of the second black body at temperature $T_2 = 300 \text{ K}$.

• The ratio of their emissive powers is given by:
\[ \frac{E_1}{E_2} = \frac{\sigma T_1^4}{\sigma T_2^4} = \left(\frac{T_1}{T_2}\right)^4 \]

• Substitute the temperature values into the ratio:
\[ \frac{E_1}{E_2} = \left(\frac{600}{300}\right)^4 \]
\[ \frac{E_1}{E_2} = (2)^4 = 16 \]

Step 4: Final Answer:
The ratio of the emissive powers of the two black bodies is 16.