Two slabs of equal thickness and cross-sectional area have thermal conductivities K1 and K2. If they are joined in series, the equivalent thermal conductivity of the combination is:
Explanation:
For slabs in series, the rate of heat flow (H) is the same. Total resistance R_eq = R1 + R2. Since thermal resistance R = L / (KA), and total length is 2L: (2L) / (K_eq * A) = (L / K1A) + (L / K2A). Canceling L/A yields 2/K_eq = 1/K1 + 1/K2. Solving gives K_eq = 2K1K2 / (K1 + K2).
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