A black body, at temperature T K emits radiation at the rate of 81 W/m2. It the temperature falls to t=T/3 K. then the new rate of thermal radiation will be

Options

A. `(81)^2 W/m^2`
B. 27 W/`m^2`
C. 243 W/`m^2`
D. 1 W/`m^2`

Answer

To solve the problem, we will use Stefan's Law, which states that the power radiated per unit area of a black body is directly proportional to the fourth power of its absolute temperature. The formula is given by:

\[ E = \sigma T^4 \]

where:
- \( E \) is the emissive power (rate of thermal radiation) in watts per square meter (W/m²),
- \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \),
- \( T \) is the absolute temperature in Kelvin (K).

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- The initial temperature is \( T \) K.
- The initial rate of thermal radiation is \( E = 81 \, \text{W/m}^2 \).

2. **Apply Stefan's Law for the initial state**:
\[ E = \sigma T^4 \]
Substituting the known value:
\[ 81 = \sigma T^4 \]

3. **Determine the new temperature**:
- The temperature falls to \( t = \frac{T}{3} \) K.

4. **Apply Stefan's Law for the new state**:
\[ E' = \sigma \left(\frac{T}{3}\right)^4 \]

5. **Simplify the expression for \( E' \)**:
\[ E' = \sigma \frac{T^4}{3^4} = \sigma \frac{T^4}{81} \]

6. **Substitute \( \sigma T^4 \) from the initial condition**:
Since \( \sigma T^4 = 81 \):
\[ E' = \frac{81}{81} = 1 \, \text{W/m}^2 \]

7. **Conclusion**:
The new rate of thermal radiation when the temperature falls to \( \frac{T}{3} \) K is \( 1 \, \text{W/m}^2 \).

### Final Answer:
The new rate of thermal radiation will be **1 W/m²**.