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Step-by-Step Solution
Step 1: Understand the Physical Situation
A stone slab of thickness 0.1 m is in contact with steam at 100 °C on one side
and ice at 0 °C on the other side. Because of this temperature difference, heat flows
from the steam through the slab to melt the ice. We are to find the thermal
conductivity $K$ of the slab given that 4.8 kg of ice melts in one hour.
Step 2: Write Down the Relevant Physical Quantities
Area of slab, $A = 0.36 \text{ m}^2$
Thickness of slab, $d = 0.1 \text{ m}$
Temperature difference, $\Delta T = 100 - 0 = 100 \text{ °C}$
Latent heat of fusion of ice, $L = 3.36 \times 10^5 \text{ J kg}^{-1}$
Mass of ice melted in one hour, $m = 4.8 \text{ kg}$
Time interval, $\Delta t = 1 \text{ hour} = 3600 \text{ s}$
Step 3: Express the Rate of Heat Transfer Through Conduction
The heat conduction through the slab per unit time is given by:
$$
\frac{dQ}{dt} = \frac{K \, A \, \Delta T}{d}
$$
where
$K$ is the thermal conductivity,
$A$ is the cross-sectional area,
$\Delta T$ is the temperature difference, and
$d$ is the thickness of the slab.
Step 4: Express the Rate of Heat Required to Melt Ice
To melt $m$ kg of ice in time $\Delta t$, the required heat per unit time is:
$$
\frac{dQ}{dt} = \frac{m \, L}{\Delta t}
$$
Here, $m$ is the mass of ice melted, and $L$ is the latent heat of fusion.
Step 5: Equate the Two Rates and Solve for $K$
At steady state, the thermal energy per unit time conducted through the slab
equals the thermal energy per unit time used to melt the ice. So,
$$
\frac{K \, A \, (100)}{d} = \frac{m \, L}{\Delta t}.
$$
Substitute numerical values:
$$
\frac{K \times 0.36 \times 100}{0.1}
= \frac{4.8 \times 3.36 \times 10^5}{3600}.
$$
Simplify the right-hand side:
$$
\frac{4.8 \times 3.36 \times 10^5}{3600}
= \frac{4.8 \times 3.36 \times 10^5}{3.6 \times 10^3}
= 4.8 \times 3.36 \times \frac{10^5}{3.6 \times 10^3}.
$$
You can perform the arithmetic step by step or use a calculator, eventually yielding:
$$
\frac{K \times 0.36 \times 100}{0.1} \approx 448 \text{ (approx.)}.
$$
Hence,
$$
K \times \frac{36}{0.1} \approx 448 \quad \Rightarrow \quad K \approx 448 \times \frac{0.1}{36}.
$$
Simplifying gives:
$$
K \approx 1.24 \text{ J m}^{-1} \text{s}^{-1} \text{°C}^{-1}.
$$
Step 6: State the Final Answer
Therefore, the thermal conductivity of the slab is
$$
1.24 \text{ J m}^{-1} \text{s}^{-1} \text{°C}^{-1}.
$$
