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 0K. Then the time required to decrease the temperature of the sphere from `T_(0)` to `(T_(0))/(2)` kelvin is (Assume sphere behaves like a black body)

Options

A. `(rhoSR)/(9sigmaT_(0)^(3))`
B. `(5rhoSR)/(9sigmaT_(0)^(3))`
C. `(7rhoSR)/(9sigmaT_(0)^(3))`
D. `(8rhoSR)/(9sigmaT_(0)^(3))`

Explanation

To solve the problem of determining the time required for a solid sphere to cool from an initial temperature \( T_0 \) to \( \frac{T_0}{2} \) in a surrounding at 0 K, we can use the principles of heat transfer and the Stefan-Boltzmann law. Here is the step-by-step solution:

### Step 1: Understand the Heat Loss Mechanism
The sphere loses heat due to radiation, and the rate of heat loss can be expressed using the Stefan-Boltzmann law:
\[ \frac{dQ}{dt} = \sigma A T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area of the sphere, and \( T \) is the temperature of the sphere.

### Step 2: Calculate the Mass of the Sphere
The mass \( m \) of the sphere can be calculated using its density \( \rho \) and volume \( V \):
\[ m = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]

### Step 3: Relate Heat Loss to Temperature Change
The heat lost by the sphere can also be expressed in terms of its specific heat \( S \) and the change in temperature:
\[ dQ = m S dT \]
Substituting the expression for mass:
\[ dQ = \rho \left(\frac{4}{3} \pi R^3\right) S dT \]

### Step 4: Set Up the Differential Equation
From the two expressions for \( dQ \), we can equate them:
\[ \rho \left(\frac{4}{3} \pi R^3\right) S dT = -\sigma A T^4 dt \]
The surface area \( A \) of the sphere is given by:
\[ A = 4 \pi R^2 \]
Substituting this into the equation gives:
\[ \rho \left(\frac{4}{3} \pi R^3\right) S dT = -\sigma (4 \pi R^2) T^4 dt \]

### Step 5: Simplify the Equation
Dividing both sides by \( 4 \pi R^2 \) leads to:
\[ \frac{\rho R S}{3} dT = -\sigma T^4 dt \]
Rearranging gives:
\[ \frac{dT}{T^4} = -\frac{3 \sigma}{\rho R S} dt \]

### Step 6: Integrate Both Sides
Integrate both sides from \( T_0 \) to \( \frac{T_0}{2} \) on the left and from \( 0 \) to \( t \) on the right:
\[ \int_{T_0}^{\frac{T_0}{2}} \frac{dT}{T^4} = -\frac{3 \sigma}{\rho R S} \int_0^t dt \]

### Step 7: Solve the Integrals
The left side integral evaluates to:
\[ \left[-\frac{1}{3 T^3}\right]_{T_0}^{\frac{T_0}{2}} = -\frac{1}{3} \left(\frac{1}{\left(\frac{T_0}{2}\right)^3} - \frac{1}{T_0^3}\right) = -\frac{1}{3} \left(\frac{8}{T_0^3} - \frac{1}{T_0^3}\right) = -\frac{7}{3 T_0^3} \]
The right side evaluates to:
\[ -\frac{3 \sigma}{\rho R S} t \]

### Step 8: Equate and Solve for Time \( t \)
Setting the two sides equal gives:
\[ -\frac{7}{3 T_0^3} = -\frac{3 \sigma}{\rho R S} t \]
Solving for \( t \):
\[ t = \frac{7 \rho R S}{9 \sigma T_0^3} \]

### Final Answer
The time required to decrease the temperature of the sphere from \( T_0 \) to \( \frac{T_0}{2} \) is:
\[ t = \frac{7 \rho R S}{9 \sigma T_0^3} \]