A block of metal A is connected in series with another block of metal B such that the two metal blocks have the same area of cross sections. The thermal conductivity of metal A is K and the free end of metal A is at 80^ C . The temperature of the interface is 60^ C and the free end of metal B is at 20^ C . Assuming the two metals have the same thickness, the conductivity of metal B is:

Step-by-step solution

In a steady state, the rate of heat flow through two blocks connected in series is the same. Let A be the cross-sectional area and L be the thickness of each block. The rate of heat flow H is given by H = KA(T₁ - T₂) L . For metal A, the temperature difference is 80^ C - 60^ C = 20^ C . Thus, H_A = KA(80 - 60) L = 20KA L . For metal B, let K_B be its thermal conductivity. The temperature difference is 60^ C - 20^ C = 40^ C . Thus, H_B = K_B A(60 - 20) L = 40K_B A L . Since the blocks are in series, H_A = H_B . Ther