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QPages 98-102: Q22: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q22 Solution: A composite slab is prepared by pasting two plates of thickness …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently mistake the equivalent conductivity of slabs in series for the arithmetic mean or the parallel formula $K_{eq} = \frac{K_1 K_2}{K_1 + K_2}$. Remember that for series arrangements, thermal resistances add up, leading to a harmonic mean-like structure where the total thickness is in the numerator and the sum of thickness-to-conductivity ratios is in the denominator.

A composite slab is prepared by pasting two plates of thickness L1L_1 and L2L_2 and thermal conductivities K1K_1 and K2K_2. The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

Step-by-Step Verified Solution

**Answer:** K_eq = (L_ + L_)/(L_/K_ + L_/K_). **Step-by-Step Explanation:** For composite slab in series, thermal resistances add:
Req=R1+R2=L1K1A+L2K2AR_{eq} = R_1 + R_2 = \frac{L_1}{K_1A} + \frac{L_2}{K_2A}
Equivalent conductivity: K_eq = (L_ + L_)/(L_/K_ + L_/K_).

All Questions in this Chapter

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