Question

Three rods of lengths 20 cm each and area of cross section 1 cm2 are joined to form a triangle ABC. The conductivities of the rods are KAB = 50 J s−1 m−1°C−1, KBC = 200 J s−1 m−1°C−1 and KAC = 400 J s−1 m−1°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.

Answer 1

The rate of heat flowing through the rods AB, AC and BC is $1 \mathrm{W}$, $0 \mathrm{W}$, $8 \mathrm{W}$

Explanation:

Given,  

Length of each rod = $20 \mathrm{cm}$

Area of cross section of each rod = $1 \mathrm{cm}^{2}$

Thermal conductivity of rod $A B=50 \mathrm{Js}^{-1} \mathrm{m}^{-1 \circ} \mathrm{C}^{-1}$

Thermal conductivity of rod $\mathrm{BC}=200 \mathrm{Js}^{-1} \mathrm{m}^{-1 \circ} \mathrm{C}^{-1}$

Thermal conductivity of rod $A C=400 \mathrm{Js}^{-1} \mathrm{m}^{-1 \circ} \mathrm{C}^{-1}$

Temperature at junction $A=40^{\circ} \mathrm{C}$

Temperature at junction $A=80^{\circ} \mathrm{C}$

Temperature at junction $A=80^{\circ} \mathrm{C}$

(a) Rate of flow of heat through rod AB

$(\Delta Q / \Delta t)_{A B}=\frac{K_{A B}\left(T_{B}-T_{A}\right) A}{l}$

$=50 \times 40 \times 10^{-4} / 0.2$

$=1 \mathrm{W}$

(b) Rate of flow of heat through rod BC

$(\Delta Q / \Delta t)_{B C}=\frac{K_{B C}\left(T_{C}-T_{B}\right) A}{l}$

$=200 \times 0 \times 10^{-4} / 0.2$

=$0 \mathrm{W}$

(c) Rate of flow of heat through rod AC

$(\Delta Q / \Delta t)_{A C}=\frac{K_{A C}\left(T_{C}-T_{A}\right) A}{l}$

$=400 \times 40 \times 10^{-4} / 0.2$

$=8 W$