A rod CD of thermal resistance 10.0 KW⁻¹ is joined at the middle of an identical rod AB as shown in figure. The end A, B and D are maintained at 200°C, 100°C and 125°C respectively. The heat current in CD is P watt. The value of P is ................. .

Solution

Physics sir · 4 steps

Treat it like a circuit: find the junction temperature at C, and the heat in CD follows from it.

1. Resistances. AB is identical to CD, so it has 10 K W⁻¹ in total. C is its midpoint, so AC and CB are 5 K W⁻¹ each, while CD stays 10 K W⁻¹.
2. Junction rule. Let C be at temperature T. In steady state no heat piles up at C, so the heat coming in from A equals the heat leaving towards B plus the heat leaving down CD. Heat current is (Δ T)/R, just like I = V/R.

   (200 - T)/5 = (T - 100)/5 + (T - 125)/10

3. Solve for T. Multiply through by 10: 400 - 2T = 2T - 200 + T - 125, so 5T = 725 and T = 145°C. Check: 11 W comes in from A, 9 W goes out to B, and 2 W is left over for CD.
4. So. The heat current in CD is P = (145 - 125)/10 = 2 W, so P = 2.

The common slip: 2.5. Taking C at the average of A and B, 150°C, ignores the heat drawn off by CD and gives 25/10 = 2.5 instead of 2.

The idea to keep. For connected rods, use the electrical analogy: temperature is potential, heat current is current. Write the junction rule at the unknown point, and remember that half a rod has half the resistance.

Answer from the NTA final answer key. Solution by PrepWiser.