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A slab of stone of area of $0.36 \mathrm{~m}^2$ and thickness $0.1 \mathrm{~m}$ is exposed on the lower surface to steam at $100^{\circ} \mathrm{C}$. A block of ice at $0^{\circ} \mathrm{C}$ rests on the upper surface of the slab. In one hour $4.8 \mathrm{~kg}$ of ice is melted. The thermal conductivity of slab is
(Given latent heat of fusion of ice $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ )
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(Given latent heat of fusion of ice $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ )
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Verified Answer
The correct answer is:
$1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
$$
\begin{aligned}
\frac{\delta Q}{\delta t} & =\frac{K A}{L}\left(T_1-T_2\right) \\
Q & =\frac{K A}{L}\left(T_1-T_2\right) t \\
Q & =m L_f \\
\frac{K A}{L}\left(T_1-T_2\right) t & =m L_f \\
K & =\frac{m L_f L}{A\left(T_1-T_2\right) t} \\
K & =\frac{4.8 \times 3.36 \times 10^5 \times 0.1}{0.36 \times 100 \times 3600}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{4.8 \times 3.36}{0.36 \times 36} \\
& =1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$
\begin{aligned}
\frac{\delta Q}{\delta t} & =\frac{K A}{L}\left(T_1-T_2\right) \\
Q & =\frac{K A}{L}\left(T_1-T_2\right) t \\
Q & =m L_f \\
\frac{K A}{L}\left(T_1-T_2\right) t & =m L_f \\
K & =\frac{m L_f L}{A\left(T_1-T_2\right) t} \\
K & =\frac{4.8 \times 3.36 \times 10^5 \times 0.1}{0.36 \times 100 \times 3600}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{4.8 \times 3.36}{0.36 \times 36} \\
& =1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$
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