The temperature difference between two sides of an iron plate, $1.8 \mathrm{~cm}$ thick is $9^{\circ}…

The temperature difference between two sides of an iron plate, $1.8 \mathrm{~cm}$ thick is $9^{\circ} \mathrm{C}$. Heat is transmitted through the plate $10 \mathrm{k} \mathrm{cal} / \mathrm{s} . \mathrm{m}^2$ at steady state. The thermal conductivity of iron is
  1. $0.02 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
  2. $0.04 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
  3. $0.05 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
  4. $0.004 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$

Solution

$\begin{aligned} & \frac{\mathrm{Q}}{\mathrm{At}}=\frac{\mathrm{k} \Delta \theta}{\mathrm{d}} \\ & \therefore 10=\mathrm{k} \times \frac{9}{1.8 \times 10^{-2}} \\ & \therefore \mathrm{k}=\frac{18 \times 10^{-2}}{9}=2 \times 10^{-2} \mathrm{kcal} / \mathrm{ms}^{\circ} \mathrm{C}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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