The lower surface of a slab of stone of face-area `3600cm^(2)` and thickness `10cm` is exposed to steam at `100^(@)C` . A block of ice at `0^(@)C` rests on the upper surface of the slab. `4.8g` of ice melts in one hour. Calculate the thermal conductivity of the stone. Latent heat of fusion of ice `=3.36xx10^(5) J kg^(-1)`

Explanation

To solve the problem, we need to find the thermal conductivity (K) of the stone slab given the melting of ice on its surface. Here’s a step-by-step solution:

### Step 1: Convert the given values to SI units
- **Face area (A)**:
\[ A = 3600 \, \text{cm}^2 = 3600 \times 10^{-4} \, \text{m}^2 = 0.36 \, \text{m}^2 \]

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