Two metallic plates of equal thickness and equal cross-sectional area, with thermal conductivities K1 and K2, are placed together face to face to form a composite slab. What is the equivalent thermal conductivity K of the combined slab?

Introduction / Context:
Heat conduction through composite slabs is a standard topic in thermal physics and engineering. When two or more layers of different materials are placed together, the effective or equivalent thermal conductivity depends on how they are arranged. This question involves two metallic plates of equal thickness and equal area, placed face to face, and asks for the equivalent thermal conductivity of the resulting composite slab.

Given Data / Assumptions:

• Two plates with thermal conductivities K1 and K2 are used.
• The plates have equal thickness and equal cross-sectional area.
• They are placed face to face, so heat flows perpendicular to the layers, passing through one then the other.
• The temperature difference is applied across the total thickness of the composite slab.
• Steady state one-dimensional heat conduction is assumed, with no heat loss to surroundings.

Concept / Approach:
When layers of different materials are arranged in series with respect to heat flow, the total thermal resistance is the sum of the individual resistances. Thermal resistance R for a layer is given by R = L / (K * A), where L is thickness, K is thermal conductivity and A is cross-sectional area. For equal thickness L and equal area A, the equivalent thermal conductivity K_equiv for two layers in series can be obtained by writing the total resistance as the sum of the two and then expressing it in the same form as a single slab of thickness 2L and area A.

Step-by-Step Solution:
Step 1: Let each plate have thickness L and area A. The conductivities are K1 and K2. Step 2: Thermal resistance of plate 1 is R1 = L / (K1 * A). Thermal resistance of plate 2 is R2 = L / (K2 * A). Step 3: Since they are in series, total resistance R_total = R1 + R2 = L / (K1 * A) + L / (K2 * A). Step 4: Factor out L / A to get R_total = (L / A) * (1 / K1 + 1 / K2). Step 5: For an equivalent single slab of thickness 2L, area A and conductivity K, the resistance would be R_equiv = 2L / (K * A). Step 6: Set R_equiv equal to R_total: 2L / (K * A) = (L / A) * (1 / K1 + 1 / K2). Step 7: Cancel L / A on both sides to obtain 2 / K = 1 / K1 + 1 / K2. Step 8: Rearranging gives 2 / K = (K1 + K2) / (K1 * K2), so K = 2*K1*K2 / (K1 + K2).

Verification / Alternative check:
The result has the form of a harmonic mean scaled by a factor of 2, which is typical for two equal layers in series. As a quick check, if K1 = K2, say both equal to K0, then K = 2*K0*K0 / (2*K0) = K0, which makes physical sense because the composite of two identical materials should have the same conductivity as each individually. If one conductivity is much smaller than the other, the equivalent conductivity is closer to the smaller value, which also matches physical intuition that the poorer conductor dominates the resistance. These logical checks confirm the correctness of the formula.

Why Other Options Are Wrong:
K1*K2 / (K1 + K2): This would arise if the total thickness were L instead of 2L, so it does not match the given geometry of equal thickness plates forming a slab of total thickness 2L.
(K1 + K2) / (K1*K2): This expression has the wrong dimensional form for conductivity and actually represents the reciprocal of the harmonic mean, not the conductivity itself.
None: Incorrect because a valid and standard expression for the equivalent conductivity definitely exists.

Common Pitfalls:
A common error is to treat the two plates as if they were in parallel for heat flow, which leads to an incorrect arithmetic mean rather than the series combination formula. Another mistake is forgetting that the total thickness is 2L, which changes the equivalent resistance. Students should always draw a simple diagram, mark the thickness of each layer and write thermal resistance expressions correctly before combining them. This habit reduces algebraic mistakes and helps in more complex composite wall problems as well.

Final Answer:
The equivalent thermal conductivity of the composite slab is 2*K1*K2 / (K1 + K2).