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

Explanation

To find the thermal conductivity of the slab, we can use the formula for heat transfer through a material and the relationship between the heat transferred and the melting of ice.

### Step-by-Step Solution:

1. **Identify the given values:**
- Area of the slab, \( A = 0.36 \, m^2 \)
- Thickness of the slab, \( X = 0.1 \, m \)
- Temperature of steam, \( T_1 = 100^\circ C \)
- Temperature of ice, \( T_2 = 0^\circ C \)
- Mass of ice melted, \( m = 4.8 \, kg \)
- Latent heat of fusion of ice, \( L = 3.63 \times 10^5 \, J/kg \)
- Time, \( \Delta t = 1 \, hour = 3600 \, seconds \)

2. **Calculate the total heat absorbed by the ice:**
The heat absorbed by the ice to melt it can be calculated using the formula:
\[ Q = mL \]
Substituting the values:
\[ Q = 4.8 \, kg \times 3.63 \times 10^5 \, J/kg = 1.73424 \times 10^6 \, J \]

3. **Use the heat transfer formula:**
The heat transfer through the slab can be expressed as:
\[ Q = \frac{KA(T_1 - T_2)}{X} \Delta t \]
Rearranging this gives us:
\[ K = \frac{QX}{A(T_1 - T_2) \Delta t} \]

4. **Substitute the known values into the equation for K:**
\[ K = \frac{(1.73424 \times 10^6 \, J)(0.1 \, m)}{(0.36 \, m^2)(100^\circ C - 0^\circ C)(3600 \, s)} \]
Simplifying the temperature difference:
\[ K = \frac{(1.73424 \times 10^6)(0.1)}{(0.36)(100)(3600)} \]

5. **Calculate the denominator:**
\[ 0.36 \times 100 \times 3600 = 129600 \]

6. **Calculate K:**
\[ K = \frac{173424}{129600} \approx 1.34 \, W/m \cdot K \]

### Final Answer:
The thermal conductivity of the slab is approximately \( K \approx 1.34 \, W/m \cdot K \).

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