Three identical heat conducting rods are connected in series as shown in the figure. The rods on the side have thermal conductivity 2K while that in the middle has thermal conductivity K . The left end of the combination is maintained at temperature 3T and the right end at T . The rods are thermally insulated from outside. In steady state, temperature at the left junction is T₁ and that at the right junctions is T₂ .

Step-by-step solution

Let the length of each rod be L and the area of cross-section be A . In steady state, the rate of heat flow is the same through all three rods: dQ dt = 2KA(3T - T₁) L = KA(T₁ - T₂) L = 2KA(T₂ - T) L Cancelling KA L from all terms, we get: 2(3T - T₁) = T₁ - T₂ = 2(T₂ - T) From the first and third parts of the equality: 2(3T - T₁) = 2(T₂ - T) 3T - T₁ = T₂ - T T₁ + T₂ = 4T (1) From the second and third parts of the equality: T₁ - T₂ = 2(T₂ - T) T₁ - 3T₂ = -2T (2) Subtracting equation (2) from equation (1): (T₁ + T₂) -