A slab of stone of area `0.34 m ^(2)`and thickness 10 cm is exposed on the lower face to steam at `100^(@)C`.A block of ice at `0^(@)C` rests on the upper face of the slab. In one hour, 3.6 kg of ice is melted. Assume that the heat loss from the sides is negligible. The latent heat of fusion of ice is `3.4 xx 10^(4) J kg^(-1)`. What is the thermal conductivity of the stone in units of `Js^(-1) m^(-1)``""^(@)C^(-1)` ?

Options

A. `0.1`
B. `0.15`
C. `0.2`
D. `0.25`

Explanation

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

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

Where:
- \( Q \) is the heat transferred (in Joules),
- \( K \) is the thermal conductivity (in \( \text{Js}^{-1} \text{m}^{-1} \degree C^{-1} \)),
- \( A \) is the area of the slab (in \( m^2 \)),
- \( T_1 \) and \( T_2 \) are the temperatures on either side of the slab (in °C),
- \( d \) is the thickness of the slab (in meters),
- \( t \) is the time (in seconds).

### Step 1: Calculate the heat transferred (Q)
The heat transferred is equal to the mass of ice melted multiplied by the latent heat of fusion.

Given:
- Mass of ice, \( m = 3.6 \, \text{kg} \)
- Latent heat of fusion, \( L = 3.4 \times 10^4 \, \text{J/kg} \)

\[ Q = m \cdot L = 3.6 \, \text{kg} \cdot 3.4 \times 10^4 \, \text{J/kg} = 122400 \, \text{J} \]

### Step 2: Convert time to seconds
Given that the time is 1 hour, we convert it to seconds:

\[ t = 1 \, \text{hour} = 3600 \, \text{seconds} \]

### Step 3: Identify the temperatures
The temperatures are given as:
- \( T_1 = 100 \, \degree C \) (temperature of steam)
- \( T_2 = 0 \, \degree C \) (temperature of ice)

### Step 4: Calculate the temperature difference
\[ \Delta T = T_1 - T_2 = 100 \, \degree C - 0 \, \degree C = 100 \, \degree C \]

### Step 5: Convert thickness to meters
Given that the thickness is 10 cm, we convert it to meters:

\[ d = 10 \, \text{cm} = 0.1 \, \text{m} \]

### Step 6: Substitute values into the heat transfer equation
Now we can rearrange the heat transfer equation to solve for \( K \):

\[ K = \frac{Q \cdot d}{A \cdot \Delta T \cdot t} \]

Where:
- Area \( A = 0.34 \, m^2 \)

Substituting the values:

\[ K = \frac{122400 \, \text{J} \cdot 0.1 \, \text{m}}{0.34 \, m^2 \cdot 100 \, \degree C \cdot 3600 \, s} \]

### Step 7: Calculate K
Calculating the denominator:

\[ 0.34 \cdot 100 \cdot 3600 = 1224000 \]

Now substituting back to find \( K \):

\[ K = \frac{12240}{1224000} = 0.01 \, \text{Js}^{-1} \text{m}^{-1} \degree C^{-1} \]

### Final Answer
Thus, the thermal conductivity of the stone is:

\[ K = 0.01 \, \text{Js}^{-1} \text{m}^{-1} \degree C^{-1} \]