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Question: Answered & Verified by Expert
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
PhysicsThermal Properties of MatterMHT CETMHT CET 2021 (23 Sep Shift 1)
Options:
  • A $0.02 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
  • B $0.04 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
  • C $0.05 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
  • D $0.004 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
Solution:
1706 Upvotes Verified Answer
The correct answer is: $0.02 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
$\begin{aligned} & \frac{\mathrm{Q}}{\mathrm{At}}=\frac{\mathrm{k} \Delta \theta}{\mathrm{d}} \\ & \therefore 10=\mathrm{k} \times \frac{9}{1.8 \times 10^2} \\ & \therefore \mathrm{k}=\frac{18 \times 10^{-2}}{9}=2 \times 10^{-2} \mathrm{kcal} / \mathrm{ms}^{\circ} \mathrm{C}\end{aligned}$

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