Physics · Thermal Properties Of Matter · NEET
In the steady state no heat is stored in the rod (temperature at each point stops changing). If the sides are insulated, all the heat that enters rod 1 must pass out of it and into rod 2 with none lost. So the rate of heat flow H (heat per second) is identical in rod 1 and rod 2. This is the key rule for series: the heat current H is the same, but the temperature drop across each rod is different.
Set the heat current through rod 1 equal to the heat current through rod 2. Using H = K A (T_hot - T_junction) / L1 for rod 1 and H = K A (T_junction - T_cold) / L2 for rod 2, put them equal and solve for T_junction. For equal area and length with conductivities K1 and K2: T_junction = (K1 T_hot + K2 T_cold) / (K1 + K2). The rod with higher conductivity pulls the junction temperature closer to its own far end.
Add the thermal resistances. R = L/(K A), so total R = L1/(K1 A) + L2/(K2 A). For two rods of equal length L and equal area A, the effective conductivity K_eff of the combined rod (total length 2L) is: K_eff = 2 K1 K2 / (K1 + K2). This is the harmonic-type mean, not the simple average. Notice it is smaller than either single conductivity in most cases.
In SERIES (rods joined end to end) the heat current is the same in both and the resistances add: R = R1 + R2. In PARALLEL (rods welded side by side between the same two hot and cold faces) the temperature difference is the same across both and the heat currents add. For parallel with equal areas, K_eff = (K1 + K2)/2. Mixing these two is the most common exam mistake.
If both rods have the same area, the area cancels out and the junction temperature depends only on the conductivities and lengths. If the areas are different, keep A1 and A2 in the heat-current equations: H = K1 A1 (T_hot - T_j)/L1 = K2 A2 (T_j - T_cold)/L2, then solve for T_j.
Three identical heat-conducting rods are connected in series as shown. The two side rods have thermal conductivity 2K and the middle rod has thermal conductivity K. The left end is maintained at 3T and the right end at T; the rods are insulated from the sides. In steady state the left junction is at T1 and the right junction is at T2. The ratio T1 : T2 is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The heat current (rate of heat flow, in watts) is the same through every rod in a series chain during the steady state, because no heat is stored and none escapes from the insulated sides.
K_eff = 2 K1 K2 / (K1 + K2) for two rods of equal length and equal cross-section joined end to end. It is always less than or equal to the larger of the two conductivities.
The rod with the higher thermal resistance (lower conductivity or greater length) has the larger temperature drop, since the same heat current flows through both and drop = H times R.
Yes. Heat current H is like electric current, temperature difference is like voltage, and thermal resistance R = L/(K A) is like electrical resistance. In series both add: R_total = R1 + R2.
The junction temperature moves closer to the far-end temperature of the better-conducting rod, because a good conductor has a small temperature drop across it.
