MHT CET · Physics · Thermal Properties of Matter
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
- 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 \(0.004 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}\)
Answer & Solution
Correct Answer
(A) \(0.02 \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{\mathrm{Q}}{\mathrm{At}}=\frac{\mathrm{k} \Delta \theta}{\mathrm{d}} \\ & \therefore 10=\mathrm{k} \times \frac{9}{1.8 \times 10^2} \\ & \therefore \mathrm{k}=\frac{18 \times 10^{-2}}{9}=2 \times 10^{-2} \mathrm{kcal} / \mathrm{ms}^{\circ} \mathrm{C}\end{aligned}\)
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