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A slab of stone of area $3600 \mathrm{~cm}^2$ and thickness $10 \mathrm{~cm}$ is exposed on the lower surface to steam at $100^{\circ} \mathrm{C}$. A block of ice at $0^{\circ} \mathrm{C}$ rests on upper surface of the slab. In one how $4.8 \mathrm{~kg}$ of ice is melted. The thermal conductivity of the stone in $\int \mathrm{s}^{-1} \mathrm{~m}^{-1} \mathrm{k}^{-1}$ is (Latent heat of ice $=3.36 \times 10^5 \mathrm{~J} / \mathrm{kg}$ )
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Verified Answer
The correct answer is:
1.24
Given, area of slab of stone $(A)=3600 \mathrm{~cm}^2$
$$
=3600 \times 10^{-4} \mathrm{~m}^2
$$
Thickness of stone slab $=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}$
Temperature difference $(\Delta \theta)=100^{\circ} \mathrm{C}$
We know that,
$$
\begin{array}{rlrl}
& \frac{Q}{t} & =\frac{K A \Delta \theta}{1} \\
\Rightarrow & \frac{4.8 \times 3.36 \times 10^5}{60 \times 60} & =\frac{K \times 3600 \times 10^{-4} \times 100}{10 \times 10^{-2}} \\
\Rightarrow & K & =\frac{4.8 \times 3.36 \times 10^5 \times 10 \times 10^{-2}}{3600 \times 10^{-4} \times 100 \times 60 \times 60} \\
\Rightarrow & & K & =\frac{161280}{129600} \\
\Rightarrow & & K & =1.24 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}
\end{array}
$$
$$
=3600 \times 10^{-4} \mathrm{~m}^2
$$
Thickness of stone slab $=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}$
Temperature difference $(\Delta \theta)=100^{\circ} \mathrm{C}$
We know that,
$$
\begin{array}{rlrl}
& \frac{Q}{t} & =\frac{K A \Delta \theta}{1} \\
\Rightarrow & \frac{4.8 \times 3.36 \times 10^5}{60 \times 60} & =\frac{K \times 3600 \times 10^{-4} \times 100}{10 \times 10^{-2}} \\
\Rightarrow & K & =\frac{4.8 \times 3.36 \times 10^5 \times 10 \times 10^{-2}}{3600 \times 10^{-4} \times 100 \times 60 \times 60} \\
\Rightarrow & & K & =\frac{161280}{129600} \\
\Rightarrow & & K & =1.24 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}
\end{array}
$$
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