A slab of stone of area of $0.36 \mathrm{~m}^2$ and thickness $0.1 \mathrm{~m}$ is exposed on the lower…

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 $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ )
  1. $1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
  2. $1.29 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
  3. $2.05 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$
  4. $1.02 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}$

Solution

$\begin{aligned} \frac{\delta Q}{\delta t} & =\frac{K A}{L}\left(T_1-T_2\right) \\ Q & =\frac{K A}{L}\left(T_1-T_2\right) t \\ Q & =m L_f \\ \frac{K A}{L}\left(T_1-T_2\right) t & =m L_f \\ K & =\frac{m L_f L}{A\left(T_1-T_2\right) t} \\ K & =\frac{4.8 \times 3.36 \times 10^5 \times 0.1}{0.36 \times 100 \times 3600} \end{aligned}$ $\begin{aligned} & =\frac{4.8 \times 3.36}{0.36 \times 36} \\ & =1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C} \end{aligned}$

Asked in: NEET 2012 (Mains)

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