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QPages 98-102: Q16: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q16 Solution: Three rods of lengths 20 cm each and area of cross section are…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often attempt to calculate heat flow through rod BC by assuming it must conduct heat due to the network configuration, but since $T_B = T_C = 80^circ ext{C}$, the temperature gradient $Delta T$ is zero, resulting in zero heat flow regardless of the rod's high conductivity.

Three rods of lengths 20 cm each and area of cross section 1 cm21 \text{ cm}^2 are joined to form a triangle ABC. The conductivities of the rods are KAB=50 J s−1 m−1 �C−1K_{AB} = 50 \text{ J s}^{-1} \text{ m}^{-1} \text{ �C}^{-1}, KBC=200 J s−1 m−1 �C−1K_{BC} = 200 \text{ J s}^{-1} \text{ m}^{-1} \text{ �C}^{-1} and KAC=400 J s−1 m−1 �C−1K_{AC} = 400 \text{ J s}^{-1} \text{ m}^{-1} \text{ �C}^{-1}. The junctions A, B and C are maintained at 40�C, 80�C and 80�C respectively. Find the rate of heat flowing through the rods AB, AC and BC.

Step-by-Step Verified Solution

**Answer:** Heat flow through AB = 1 W, AC = 8 W, BC = 0. **Step-by-Step Explanation:** Given: l = 20 cm = 0.2 m, A = 1 cm_ = 10__ m_, T_A = 40�C, T_B = 80�C, T_C = 80�C. K_AB = 50, K_BC = 200, K_AC = 400.
ΔQABΔt=50×10−4×(80−40)0.2=1 W\frac{\Delta Q_{AB}}{\Delta t} = \frac{50 \times 10^{-4} \times (80-40)}{0.2} = 1 \text{ W}
ΔQACΔt=400×10−4×(80−40)0.2=8 W\frac{\Delta Q_{AC}}{\Delta t} = \frac{400 \times 10^{-4} \times (80-40)}{0.2} = 8 \text{ W}
ΔQBCΔt=200×10−4×(80−80)0.2=0\frac{\Delta Q_{BC}}{\Delta t} = \frac{200 \times 10^{-4} \times (80-80)}{0.2} = 0
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