Question

Three rods each of same length and cross - section are joined in series. The thermal conductivity of the materials are K, 2K and 3K respectively. If one end is kept at 200^@C and the other at 100^@C. What would be the temperature of the junctions in the steady state? Assume that no heat is lost due to radiation from the sides of the rods.

Answer 1

Answer:

no no no no no no no no no no no no no no no no no no no no no no

Answer 2

The temperature of the junctions in the steady state are 145.5°C and 118.2°C

• Let us go through the given parameters

       Thermal conductivity of material 1 (K1)= K

       Thermal conductivity of material 2 (K2)= 2K

       Thermal conductivity of material 3 (K3)= 3K

       Temperature at one end  = 200°C

       Temperature at other end = 100°C

• We have the formula

                                H=(T1-T2)/(l/KA)

• By equating them

                                H1=H2=H3

          (200-T1)/(l/KA) = (T1-T2)/(l/2KA) = (T2-100)/(1/3KA)

• By solving the above equations we will get

        T1 = 145.5°C

        T2= 118.2°C