A wall has two layers A and B , each made of different materials. Both layers are of same thickness. But, the thermal conductivity of material A is twice that of B . If in the steady state, the temperature difference across the wall is 24^ C , then the temperature difference across the layer B is

Step-by-step solution

Let K_A and K_B be the thermal conductivity of material A and B , respectively. Let T₁ be the temperature of A and T₂ be the temperature of B . At steady state, rate of flow of heat is given by, aligned Q t & = Q_A t = Q_B t K^ A (T₂-T₁ ) l_A+l_B & = K_A A (T-T₁ ) l_A & = K_B A (T₂-T ) l_B aligned where K is thermal conductivity of both material and A and l are area and length of the material respectively. Using Eq. (i), we get aligned & K_A A l_A (T-T₁ )= K_B A (T₂-T ) l_B & 2 K_B (T-T₁ )=K_B (T₂-T ) & ( l_A=l_B a