Two slabs A and B of different materials but of the same thickness are joined as shown in fig. (2). The thermal conductivities of A and B are $$K1$$ and $$K2$$ respectively. The thermal conductivity of the composite slab is

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The thermal resistance of a slab of length I, area A and thermal conductivity K is given by $$R=l/(KA)For$$ slab A, $$R=(l/2)/K1AFor$$ slab B, $$R2=(l/2)/K2AThe$$ two rods are joined in series. Hence, their thermal resistance $$R=R1+R2or$$    $$lKA=(l/2)K1A+(l/2)K2A$$  or $$1K=121K1+1K2or$$    $$K=2K1K2K1+K2$$

Two slabs A and B of different materials but of the same thickness are joined as shown in fig. (2). The thermal conductivities of A and B are K₁ and K₂ respectively. The thermal conductivity of the composite slab is

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Two slabs A and B of different materials but of the same thickness are joined as shown in fig. (2). The thermal conductivities of A and B are K1 and K2 respectively. The thermal conductivity of the composite slab is

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Two slabs A and B of different materials but of the same thickness are joined as shown in fig. (2). The thermal conductivities of A and B are K₁ and K₂ respectively. The thermal conductivity of the composite slab is