A slab of stone of area of `0.36 m^(2)` and thickness `0.1 m` is exposed on the lower surface to steam at `100^(@)C`. A block of ice at `0^(@)C` rests on the upper surface of the slab. In one hour `4.8 kg` of ice is melted. The thermal conductivity of slab is
(Given latent heat of fusion of ice `= 3.63 xx 10^(5) J kg^(-1)`)

Options

A. `1.24 J//m//s//^(@)C`
B. `1.29 J//m//s//^(@)C`
C. `2.05 J//m//s//^(@)C`
D. `1.02 J//m//s//^(@)C`

Explanation

To find the thermal conductivity of the slab, we can use the formula for the rate of heat transfer through a material:

\[ \frac{Q}{t} = \frac{k \cdot A \cdot (T_1 - T_2)}{x} \]

Where:
- \( Q \) is the heat transferred,
- \( t \) is the time,
- \( k \) is the thermal conductivity,
- \( A \) is the area,
- \( T_1 \) and \( T_2 \) are the temperatures on either side of the slab,
- \( x \) is the thickness of the slab.

### Step 1: Calculate the heat transferred (Q)
The heat transferred \( Q \) can be calculated using the mass of ice melted and the latent heat of fusion:

\[ Q = m \cdot L \]

Where:
- \( m = 4.8 \, \text{kg} \) (mass of ice melted),
- \( L = 3.63 \times 10^5 \, \text{J/kg} \) (latent heat of fusion).

Calculating \( Q \):

\[ Q = 4.8 \, \text{kg} \cdot 3.63 \times 10^5 \, \text{J/kg} = 1.7384 \times 10^6 \, \text{J} \]

### Step 2: Substitute values into the heat transfer equation
We know:
- \( A = 0.36 \, \text{m}^2 \),
- \( T_1 = 100^\circ C \),
- \( T_2 = 0^\circ C \),
- \( x = 0.1 \, \text{m} \),
- \( t = 3600 \, \text{s} \) (1 hour).

Now we can substitute these values into the heat transfer equation:

\[ \frac{Q}{t} = \frac{k \cdot A \cdot (T_1 - T_2)}{x} \]

Substituting the known values:

\[ \frac{1.7384 \times 10^6 \, \text{J}}{3600 \, \text{s}} = \frac{k \cdot 0.36 \, \text{m}^2 \cdot (100 - 0)}{0.1 \, \text{m}} \]

### Step 3: Simplify and solve for k
Calculating the left side:

\[ \frac{1.7384 \times 10^6}{3600} = 483.44 \, \text{W} \]

Now, substituting this into the equation:

\[ 483.44 = \frac{k \cdot 0.36 \cdot 100}{0.1} \]

This simplifies to:

\[ 483.44 = k \cdot 360 \]

Now, solving for \( k \):

\[ k = \frac{483.44}{360} \approx 1.34 \, \text{W/m·K} \]

### Final Answer
The thermal conductivity of the slab is approximately:

\[ k \approx 1.34 \, \text{W/m·K} \]