The temperature difference between two sides of metal plate, $3 \mathrm{~cm}$ thick is $15^{\circ}…

The temperature difference between two sides of metal plate, $3 \mathrm{~cm}$ thick is $15^{\circ} \mathrm{C}$. Heat is transmitted through plate at the rate of $900 \mathrm{kcal}$ per minute per $\mathrm{m}^2$ at steady state. The thermal conductivity of metal is
  1. $1.8 \times 10^{-2} \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
  2. $4.5 \times 10^{-2} \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
  3. $3 \times 10^{-2} \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$
  4. $6 \times 10^{-2} \frac{\mathrm{kcal}}{\mathrm{ms}^{\circ} \mathrm{C}}$

Solution

$\begin{aligned} & \frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{kA} \Delta \theta}{\mathrm{d}} \\ & \therefore \mathrm{k}=\frac{\mathrm{Q}}{\mathrm{tA}} \cdot \frac{\mathrm{d}}{\Delta \theta} \\ & \frac{\mathrm{Q}}{\mathrm{tA}}=900 \mathrm{kcal} \text { per minute per } \mathrm{m}^2=\frac{900}{60}=15 \mathrm{kcal} / \mathrm{s} \cdot \mathrm{m}^2 \\ & \mathrm{~d}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}, \Delta \theta=15^{\circ} \mathrm{C} \\ & \therefore \mathrm{k}=\frac{15 \times 3 \times 10^{-2}}{15}=3 \times 10^{-2} \mathrm{kcal} / \mathrm{s} \cdot \mathrm{m} .{ }^{\circ} \mathrm{C} \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 1)

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