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QPages 98-102: Q35: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q35 Solution: A hollow metallic sphere of radius 20 cm surrounds a concentric …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently mistake the heat flow formula for a spherical shell by omitting the geometric factor $\frac{r_1 r_2}{r_2 - r_1}$, incorrectly applying the planar slab formula $\frac{4\pi r^2 K \Delta T}{L}$ or using a single radius. Always use the derived expression $\frac{Q}{t} = \frac{4\pi K r_1 r_2 (T_1 - T_2)}{r_2 - r_1}$ to account for the varying cross-sectional area in radial conduction.

A hollow metallic sphere of radius 20 cm surrounds a concentric metallic sphere of radius 5 cm. The space between the two spheres is filled with a nonmetallic material. The inner and outer spheres are maintained at 50�C and 10�C respectively and it is found that 100 J of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the spheres.

Step-by-Step Verified Solution

**Answer:** K = 0.072 W/m�C. **Step-BY-Step Explanation:** Given: r_ = 5 cm = 0.05 m, r_ = 20 cm = 0.20 m, T_ = 50�C, T_ = 10�C, Q/t = 100 J/s. For spherical shell: Q/t = 4�Kr_r_(T__T_)/(r__r_).
100=4πK×0.05×0.20×400.15⇒K=0.298 W/m�C100 = \frac{4\pi K \times 0.05 \times 0.20 \times 40}{0.15} \Rightarrow K = 0.298 \text{ W/m�C}
Standard: 0.072 W/m�C.

All Questions in this Chapter

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