Question

If a black body at 400 K surrounded by atmosphere at 300 K has rate of cooling \( \propto T^4 \), the same body at 900 K, surrounded by same atmosphere, will have rate of cooling nearly

• A. \( 4T_0 \)
• B. \( 16T_0 \)
• C. \( 36T_0 \)
• D. \( 64T_0 \)

Hint:

The rate of heat radiated by a black body is proportional to the fourth power of its temperature. This means that small increases in temperature result in much larger increases in radiated energy.

Answer 1

The Correct Option is D

Step 1: Stefan-Boltzmann law.
The rate of cooling (or heat radiated) from a body due to its temperature is governed by the Stefan-Boltzmann law:
\[ P = \sigma A (T^4 - T_{\text{ambient}}^4), \]
where:
- \( P \) is the rate of heat radiated (rate of cooling),
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the body,
- \( T \) is the temperature of the body,
- \( T_{\text{ambient}} \) is the temperature of the surrounding atmosphere.
The rate of cooling is proportional to \( T^4 \), the fourth power of the body’s temperature, and the difference between the body’s temperature and the temperature of the surroundings.

Step 2: Rate of cooling at 400 K.
At \( T = 400 \, \text{K} \) and the surrounding temperature \( T_{\text{ambient}} = 300 \, \text{K} \), the rate of cooling is:
\[ P_1 = \sigma A (400^4 - 300^4). \]

Step 3: Rate of cooling at 900 K.
When the body is heated to 900 K while the ambient temperature remains the same, the rate of cooling is:
\[ P_2 = \sigma A (900^4 - 300^4). \]

Step 4: Proportionality of rate of cooling.
Since \( P \) is proportional to \( T^4 \), the rate of cooling at 900 K is proportional to \( 900^4 \). Therefore, the ratio of the rate of cooling at 900 K to that at 400 K is:
\[ \frac{P_2}{P_1} = \frac{900^4}{400^4} = \left( \frac{900}{400} \right)^4 = \left( \frac{9}{4} \right)^4 = \frac{6561}{256} \approx 64. \]
Final Answer:
Thus, the rate of cooling at 900 K is:
\[ \boxed{64 \, P_0}. \]