Two slabs are of the thicknesses d₁ and d₂ . Their thermal conductivities are K₁ and K₂ , respectively. They are in series. The free ends of the combination of these two slabs are kept at temperatures ₁ and ₂ . Assume ₁ > ₂ . The temperature of their common junction is

Step-by-step solution

In a series combination of two slabs, the rate of heat flow H through each slab must be the same in the steady state. The rate of heat flow through the first slab is H = K₁ A ( ₁ - ) d₁ , where A is the cross-sectional area. The rate of heat flow through the second slab is H = K₂ A ( - ₂) d₂ . Equating the two expressions for H : K₁ A ( ₁ - ) d₁ = K₂ A ( - ₂) d₂ Canceling A and rearranging the terms: K₁ d₁ ( ₁ - ) = K₂ d₂ ( - ₂) K₁ ₁ d₁ - K₁ d₁ = K₂ d₂ - K₂ ₂ d₂ ( K₁ d₁ + K₂ d₂ ) = K₁ ₁ d₁ + K₂ ₂ d₂ ( K₁ d₂ + K₂ d₁