Find the total time elapsed for a hollow copper sphere of inner radius `3cm` outer radius `6cm`, density `rho = 9 xx 10^(3) kg//m^(3)`, specific heat `s = 4 xx 10^(3) J//kg K` and emissivity `e = 0.4` to cool from `727^(@)C` to `227^(@)C` when the surrounding temperature is `0.K` (for inner surface `e = 1` Stefan's constant `sigma = 5.6 xx 10^(-8) W//m^(2)K^(4))`

Explanation

To find the total time elapsed for a hollow copper sphere to cool from \(727^\circ C\) to \(227^\circ C\) when the surrounding temperature is \(0 K\), we can follow these steps:

### Step 1: Convert temperatures to Kelvin
The temperatures given in Celsius need to be converted to Kelvin.

- \( T_1 = 727^\circ C + 273 = 1000 K \)
- \( T_2 = 227^\circ C + 273 = 500 K \)

### Step 2: Calculate the volume of the hollow sphere
The volume \( V \) of the hollow sphere can be calculated using the formula for the volume of a sphere:

\[ V = \frac{4}{3} \pi (r_2^3 - r_1^3) \]

Where:
- \( r_1 = 0.03 m \) (inner radius)
- \( r_2 = 0.06 m \) (outer radius)

Calculating the volume:

\[ V = \frac{4}{3} \pi ((0.06)^3 - (0.03)^3) = \frac{4}{3} \pi (0.000216 - 0.000027) = \frac{4}{3} \pi (0.000189) \approx 0.000793 m^3 \]

### Step 3: Calculate the mass of the hollow sphere
The mass \( m \) of the hollow sphere can be calculated using the formula:

\[ m = \rho \cdot V \]

Where:
- \( \rho = 9 \times 10^3 kg/m^3 \)

Calculating the mass:

\[ m = 9 \times 10^3 \cdot 0.000793 \approx 7.137 kg \]

### Step 4: Calculate the area of the outer surface
The area \( A \) of the outer surface of the hollow sphere is given by:

\[ A = 4 \pi r_2^2 \]

Calculating the area:

\[ A = 4 \pi (0.06)^2 = 4 \pi (0.0036) \approx 0.0454 m^2 \]

### Step 5: Apply Stefan-Boltzmann Law
According to Stefan-Boltzmann Law, the rate of heat loss \( \frac{dq}{dt} \) is given by:

\[ \frac{dq}{dt} = \sigma e A (T_1^4 - T_2^4) \]

Where:
- \( \sigma = 5.6 \times 10^{-8} W/m^2K^4 \)
- \( e = 0.4 \)

Calculating the rate of heat loss:

\[ \frac{dq}{dt} = 5.6 \times 10^{-8} \cdot 0.4 \cdot 0.0454 \cdot (1000^4 - 500^4) \]

Calculating \( 1000^4 - 500^4 \):

\[ 1000^4 = 10^{12}, \quad 500^4 = 6.25 \times 10^{11} \]
\[ 1000^4 - 500^4 = 10^{12} - 6.25 \times 10^{11} = 3.75 \times 10^{11} \]

Now substituting back:

\[ \frac{dq}{dt} = 5.6 \times 10^{-8} \cdot 0.4 \cdot 0.0454 \cdot 3.75 \times 10^{11} \approx 3.75 \times 10^3 W \]

### Step 6: Relate heat loss to temperature change
Using the relation \( \frac{dq}{dt} = m \cdot s \cdot \frac{dT}{dt} \):

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

### Step 7: Integrate to find time
Integrating from \( T_1 \) to \( T_2 \):

\[ \int_{T_1}^{T_2} \frac{m \cdot s}{\sigma e A} dT = -\int_{t_0}^{t} dt \]

This gives:

\[ t = \frac{m \cdot s}{\sigma e A} \left( \frac{1}{T_2^3} - \frac{1}{T_1^3} \right) \]

Substituting the values:

\[ t = \frac{7.137 \cdot 4 \times 10^3}{5.6 \times 10^{-8} \cdot 0.4 \cdot 0.0454} \left( \frac{1}{500^3} - \frac{1}{1000^3} \right) \]

Calculating the final time:

\[ t \approx 6.56 \times 10^4 \text{ seconds} \]

### Final Answer
The total time elapsed for the hollow copper sphere to cool from \(727^\circ C\) to \(227^\circ C\) is approximately \(6.56 \times 10^4\) seconds.

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