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 `=3.36 xx 10^(5) J kg^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the mass of ice from grams to kilograms
Given that the mass of ice melted is 4.8 g, we convert this to kilograms:
\[ \text{Mass of ice} = \frac{4.8 \, \text{g}}{1000} = 0.0048 \, \text{kg} \]

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