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QPages 98-102: Q12: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q12 Solution: A cubical box of volume is made up of 0.1 cm thick wood. The…

⚠️ Common Exam Traps & Conceptual Pitfalls

In steady‑state heat conduction the flux depends on the cross‑sectional area, not the volume. For a cube the area is $6a^2$, not $6a$. Always convert all dimensions to SI units before inserting into $Q= rac{K,A,Delta T}{d}$ to avoid a factor of $a$ error.

A cubical box of volume 216 cm3216 \text{ cm}^3 is made up of 0.1 cm thick wood. The inside is heated electrically by a 100 W heater. It is found that the temperature difference between the inside and the outside surface is 5�C in steady state. Assuming that the entire electrical energy spent appears as heat, find the thermal conductivity of the material of the box.

Step-by-Step Verified Solution

**Answer:** K = 0.926 W/m�C. **Step-by-Step Explanation:** Given: V = 216 cm_, edge a = 6 cm, surface area = 6a_ = 216 _ 10__ m_, thickness = 0.1 cm = 10__ m, _T = 5�C, P = 100 W.
100=K×216×10−4×510−3⇒K=0.926 W/m�C100 = \frac{K \times 216 \times 10^{-4} \times 5}{10^{-3}} \Rightarrow K = 0.926 \text{ W/m�C}
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