A black body of mass .0.10 g is kept in a black enclosure or temperature `27^(@)`C. The temperature of the body is `127^(@)`C and the area of the emittimg surface in `10^(-3) m^(2)`. If its specific heat capacity is 420 J `kg^(-1)k^(-1)` find the rate of cooling of the body . `sigma = 5.67xx10^(-8)Wm^(-2)K^(-4)`.

Explanation

To find the rate of cooling of a black body in a black enclosure, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Given Values to SI Units**:
- Mass of the body, \( m = 0.10 \, \text{g} = 0.10 \times 10^{-3} \, \text{kg} = 10^{-4} \, \text{kg} \)
- Temperature of the body, \( T_b = 127 \, ^\circ C = 127 + 273 = 400 \, \text{K} \)
- Temperature of the surroundings, \( T_s = 27 \, ^\circ C = 27 + 273 = 300 \, \text{K} \)
- Area of the emitting surface, \( A = 10^{-3} \, \text{m}^2 \)
- Specific heat capacity, \( C = 420 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1} \)
- Stefan-Boltzmann constant, \( \sigma = 5.67 \times 10^{-8} \, \text{W} \, \text{m}^{-2} \, \text{K}^{-4} \)

2. **Use the Stefan-Boltzmann Law**:
The rate of heat loss by radiation can be expressed as:
\[ \frac{dQ}{dt} = A \cdot \epsilon \cdot \sigma \cdot (T_b^4 - T_s^4) \]
For a black body, emissivity \( \epsilon = 1 \). Thus, the equation simplifies to:
\[ \frac{dQ}{dt} = A \cdot \sigma \cdot (T_b^4 - T_s^4) \]

3. **Calculate \( T_b^4 \) and \( T_s^4 \)**:
- \( T_b^4 = (400 \, \text{K})^4 = 256 \times 10^8 \, \text{K}^4 \)
- \( T_s^4 = (300 \, \text{K})^4 = 81 \times 10^8 \, \text{K}^4 \)

4. **Substitute Values into the Equation**:
\[ \frac{dQ}{dt} = (10^{-3} \, \text{m}^2) \cdot (5.67 \times 10^{-8} \, \text{W} \, \text{m}^{-2} \, \text{K}^{-4}) \cdot (256 \times 10^8 - 81 \times 10^8) \]
\[ = (10^{-3}) \cdot (5.67 \times 10^{-8}) \cdot (175 \times 10^8) \]

5. **Calculate the Rate of Heat Loss**:
\[ \frac{dQ}{dt} = (10^{-3}) \cdot (5.67 \times 175) \, \text{W} \]
\[ = (10^{-3}) \cdot (992.25) \, \text{W} = 0.99225 \, \text{W} \]

6. **Relate Heat Loss to Temperature Change**:
The heat loss can also be expressed in terms of the specific heat capacity:
\[ \frac{dQ}{dt} = m \cdot C \cdot \frac{dT}{dt} \]
Setting the two expressions for \( \frac{dQ}{dt} \) equal gives:
\[ m \cdot C \cdot \frac{dT}{dt} = 0.99225 \]

7. **Solve for the Rate of Change of Temperature**:
\[ \frac{dT}{dt} = \frac{0.99225}{m \cdot C} \]
\[ = \frac{0.99225}{10^{-4} \cdot 420} \]
\[ = \frac{0.99225}{42} \approx 0.0236 \, \text{K/s} \approx 23.65 \, \text{K/s} \]

### Final Answer:
The rate of cooling of the body is approximately \( 23.65 \, \text{K/s} \).

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