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MHT CET 2021 (24 Sep, Shift 1) · Heat Conduction

Heat Conduction question from MHT CET 2021 (24 Sep, Shift 1)

Q32MHT CET 2021 · 24 SEP · SHIFT 1Easy

Heat Conduction

The temperature difference between two sides of an iron plate, 1.8 cm1.8\text{ cm} thick is 9∘C9^\circ\text{C}. Heat is transmitted through the plate 10 kcal/s m210\text{ kcal/s m}^2 at steady state. The thermal conductivity of iron is

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Answer: (a)

  1. Heat flux HA=KΔTd⇒K=(H/A) dΔT\frac{H}{A} = \frac{K\Delta T}{d} \Rightarrow K = \frac{(H/A)\,d}{\Delta T}
  2. K=10×0.0189=0.02 kcal m−1s−1 ∘C−1K = \frac{10 \times 0.018}{9} = 0.02\ \text{kcal m}^{-1}\text{s}^{-1}\,{}^\circ\text{C}^{-1}
  3. So the answer is (a).
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