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. `0.55^(@)C//sec`
B. `0.66^(@)C//sec`
C. `0.44^(@)C//sec`
D. `0.03^(@)C//sec`

Explanation

To find the maximum rate at which the temperature of the metallic sphere will fall, we can use the Stefan-Boltzmann law for radiation. The formula for the rate of cooling is given by:

\[ \frac{dT}{dt} = -\frac{\sigma \cdot e \cdot A \cdot (T^4 - T_w^4)}{m \cdot s} \]

Where:
- \(\sigma\) is the Stefan-Boltzmann constant,
- \(e\) is the emissivity (given as 1),
- \(A\) is the surface area of the sphere,
- \(T\) is the initial temperature of the sphere in Kelvin,
- \(T_w\) is the temperature of the walls in Kelvin,
- \(m\) is the mass of the sphere,
- \(s\) is the specific heat of the metal.

### Step 1: Convert temperatures to Kelvin
- Initial temperature of the sphere:
\[ T = 227^\circ C = 227 + 273 = 500 \, K \]
- Temperature of the walls:
\[ T_w = 27^\circ C = 27 + 273 = 300 \, K \]

### Step 2: Calculate the surface area of the sphere
The surface area \(A\) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]
Where \(r\) is the radius of the sphere. Given \(r = 0.08 \, m\):
\[ A = 4\pi (0.08)^2 = 4\pi (0.0064) \approx 0.0804 \, m^2 \]

### Step 3: Substitute values into the cooling rate formula
Using the values:
- \(\sigma = 5.8 \times 10^{-8} \, W/m^2K^4\)
- \(e = 1\)
- \(A \approx 0.0804 \, m^2\)
- \(T = 500 \, K\)
- \(T_w = 300 \, K\)
- \(m = 10 \, kg\)
- \(s = 90 \, cal/kg \cdot K = 90 \times 4.2 \, J/kg \cdot K = 378 \, J/kg \cdot K\)

Now substituting these values into the cooling rate formula:
\[ \frac{dT}{dt} = -\frac{(5.8 \times 10^{-8}) \cdot (1) \cdot (0.0804) \cdot ((500)^4 - (300)^4)}{10 \cdot 378} \]

### Step 4: Calculate \( (500)^4 - (300)^4 \)
Calculating \( (500)^4 \) and \( (300)^4 \):
\[ (500)^4 = 62500000000 \]
\[ (300)^4 = 8100000000 \]
Thus,
\[ (500)^4 - (300)^4 = 62500000000 - 8100000000 = 54400000000 \]

### Step 5: Substitute back into the formula
Now substituting this back into the formula:
\[ \frac{dT}{dt} = -\frac{(5.8 \times 10^{-8}) \cdot (0.0804) \cdot (54400000000)}{10 \cdot 378} \]

### Step 6: Calculate the numerator and denominator
Calculating the numerator:
\[ 5.8 \times 10^{-8} \times 0.0804 \times 54400000000 \approx 0.252 \]
Calculating the denominator:
\[ 10 \cdot 378 = 3780 \]

### Step 7: Final calculation
Now, substituting these values:
\[ \frac{dT}{dt} = -\frac{0.252}{3780} \approx -0.0000667 \, K/s \]

### Step 8: Convert to Celsius per second
Since the change in temperature in Kelvin is equivalent to Celsius, we have:
\[ \frac{dT}{dt} \approx -0.0667 \, ^\circ C/s \]

### Final Answer
The maximum rate at which the temperature of the metallic sphere will fall is approximately:
\[ \frac{dT}{dt} \approx -0.0667 \, ^\circ C/s \]