A metallic sphere having radius `0.08 m` and mass `m = 10 kg` is heated to a temperature of `227^(@)C` and suspended inside a box whose walls ae at a temperature of `27^(@)C`. The maximum rate at which its temperature will fall is:-
(Take `e =1`, Stefan's constant `sigma = 5.8 xx 10^(-8) W//m^(-2)K^(-4)` and specific heat of the metal `s = 90 cal//kg//deg J = 4.2 "Joules"//"Calorie")`

Options

A. `0.055^(@)C//s`
B. `0.066^(@)C//s`
C. `0.044 ^(@)C//s`
D. `0.03 ^(@)C//s`

Explanation

To solve the problem of determining the maximum rate at which the temperature of a metallic sphere falls, we can use the formula for the rate of cooling due to radiation, which is given by:

\[ R = \frac{E \sigma A (T^4 - T_0^4)}{M \cdot S} \]

Where:
- \( R \) = rate of cooling (in degrees per second)
- \( E \) = emissivity of the surface (given as 1)
- \( \sigma \) = Stefan-Boltzmann constant (\( 5.8 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \))
- \( A \) = surface area of the sphere
- \( T \) = temperature of the sphere in Kelvin
- \( T_0 \) = temperature of the surroundings in Kelvin
- \( M \) = mass of the sphere (10 kg)
- \( S \) = specific heat of the metal (given as 90 cal/kg°C, which is equivalent to \( 90 \times 4.2 \, \text{J/kg°C} \))

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- The initial temperature of the sphere is \( 227^\circ C = 227 + 273 = 500 \, K \).
- The temperature of the surroundings is \( 27^\circ C = 27 + 273 = 300 \, K \).

2. **Calculate the Surface Area of the Sphere**:
- The formula for the surface area \( A \) of a sphere is:
\[ A = 4\pi r^2 \]
- Given \( r = 0.08 \, m \):
\[ A = 4 \pi (0.08)^2 = 4 \pi (0.0064) \approx 0.0804 \, m^2 \]

3. **Substitute Values into the Cooling Rate Formula**:
- Substitute \( E = 1 \), \( \sigma = 5.8 \times 10^{-8} \, W/m^2K^4 \), \( A \approx 0.0804 \, m^2 \), \( T = 500 \, K \), \( T_0 = 300 \, K \), \( M = 10 \, kg \), and \( S = 90 \times 4.2 \, J/kg°C = 378 \, J/kg°C \):
- The formula becomes:
\[ R = \frac{1 \cdot 5.8 \times 10^{-8} \cdot 0.0804 \cdot (500^4 - 300^4)}{10 \cdot 378} \]

4. **Calculate \( T^4 - T_0^4 \)**:
- Calculate \( 500^4 \) and \( 300^4 \):
\[ 500^4 = 62500000000 \quad \text{and} \quad 300^4 = 8100000000 \]
- Therefore:
\[ 500^4 - 300^4 = 62500000000 - 8100000000 = 54400000000 \]

5. **Calculate the Rate of Cooling**:
- Substitute back into the equation:
\[ R = \frac{5.8 \times 10^{-8} \cdot 0.0804 \cdot 54400000000}{10 \cdot 378} \]
- Calculate the numerator:
\[ 5.8 \times 10^{-8} \cdot 0.0804 \cdot 54400000000 \approx 2.51 \, W \]
- Now calculate \( R \):
\[ R \approx \frac{2.51}{3780} \approx 0.000663 \, K/s \]
- Convert to degrees Celsius per second:
\[ R \approx 0.0663 \, °C/s \]

6. **Final Answer**:
- The maximum rate at which the temperature of the sphere will fall is approximately \( 0.067 \, °C/s \).