A solid sphere of radius 'R' density `'rho'` , and specific heat 'S' initially at temperature `T_0` Kelvin is suspended in a surrounding at temperature 0 K. Then the time required to decrease the temperature of the sphere from `T_0` to `T_0/2` Kelvin is `alpha xx (rhoSR)/(sigmaT_0^3)` . Find `alpha`. (Assume sphere behaves like a black body )

Answer

To solve the problem, we need to determine the time required for a solid sphere to cool from an initial temperature \( T_0 \) to \( \frac{T_0}{2} \) when suspended in an environment at 0 K. We will use the principles of heat transfer and radiation.

### Step-by-Step Solution:

1. **Identify the Heat Loss**:
The heat lost by the sphere can be expressed as:
\[ \text{Heat lost} = m S dT \]
where \( m \) is the mass of the sphere, \( S \) is the specific heat, and \( dT \) is the change in temperature.

2. **Express Mass**:
The mass \( m \) of the sphere can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ m = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]

3. **Heat Loss in Terms of Temperature Change**:
The heat loss can be rewritten as:
\[ \text{Heat lost} = \rho \left(\frac{4}{3} \pi R^3\right) S dT \]

4. **Radiative Heat Loss**:
The heat lost due to radiation can be expressed using the Stefan-Boltzmann law:
\[ \text{Radiative Heat Loss} = \sigma A T^4 dt \]
where \( A \) is the surface area of the sphere, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the temperature of the sphere. The surface area \( A \) of the sphere is given by:
\[ A = 4 \pi R^2 \]

5. **Setting Up the Equation**:
Equating the heat lost to the radiative heat loss gives:
\[ -\rho \left(\frac{4}{3} \pi R^3\right) S dT = \sigma (4 \pi R^2) T^4 dt \]

6. **Rearranging the Equation**:
Rearranging this equation, we have:
\[ \frac{dT}{T^4} = -\frac{3 \sigma R}{\rho S} dt \]

7. **Integrating Both Sides**:
Integrate both sides from the initial temperature \( T_0 \) to \( \frac{T_0}{2} \) and from \( 0 \) to \( t \):
\[ \int_{T_0}^{\frac{T_0}{2}} dT \cdot \frac{1}{T^4} = -\frac{3 \sigma R}{\rho S} \int_{0}^{t} dt \]

8. **Calculating the Integral**:
The left side integrates to:
\[ \left[-\frac{1}{3 T^3}\right]_{T_0}^{\frac{T_0}{2}} = -\frac{1}{3 \left(\frac{T_0}{2}\right)^3} + \frac{1}{3 T_0^3} = -\frac{8}{3 T_0^3} + \frac{1}{3 T_0^3} = -\frac{7}{3 T_0^3} \]
Thus, we have:
\[ -\frac{7}{3 T_0^3} = -\frac{3 \sigma R}{\rho S} t \]

9. **Solving for Time \( t \)**:
Rearranging gives:
\[ t = \frac{7 \rho S R}{9 \sigma T_0^3} \]

10. **Identifying \( \alpha \)**:
From the expression for time, we can identify \( \alpha \):
\[ \alpha = \frac{7}{9} \]

### Final Answer:
Thus, the value of \( \alpha \) is:
\[ \alpha = \frac{7}{9} \]