In a composite slab there are two materials having coefficients of thermal conductivity K and 2K, thickness x and 4x repectively. The temperature of the two outer surfaces of a composite slab are T₂ and T₁(T₂ > T₁) . T₂ is on side K and T₁ is on side 2K. The rate of heat transfer through the slab in a steady state is [ A(T₂ - T₁)K x ] f , where f is equal to

Step-by-step solution

The thermal resistance of a slab is given by R = L KA , where L is the thickness, K is the thermal conductivity, and A is the area of cross-section. For the first material, the thermal resistance is R₁ = x KA . For the second material, the thermal resistance is R₂ = 4x 2KA = 2x KA . Since the two materials form a composite slab in series, the equivalent thermal resistance is: R_ eq = R₁ + R₂ = x KA + 2x KA = 3x KA The rate of heat transfer through the slab in a steady state is given by: H = T₂ - T₁ R_ eq = T₂ - T₁