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 K_AB = 50 J s⁻¹ m⁻¹°C⁻¹, K_BC = 200 J s⁻¹ m⁻¹°C⁻¹ and K_AC = 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

Thermal conductivity of rod AB, K_AB = 50 J/m-s-°C
Temperature of junction at A, T_A = 40°C
Thermal conductivity of rod BC, K_BC = 200 J/m-s-°C
Temperature of junction at B, T_B = 80°C
Thermal conductivity of rod BC, K_CA = 400 J/m-s-°C
Temperature of junction at C, T_C = 80°C
l = 20 cm = 20 × 10⁻² m
A = 1 cm² = 10⁻⁴ m²

(a) $\frac{{\left(\mathrm{\Delta }Q\right)}_{\mathrm{AB}}}{\mathrm{\Delta }t}=\frac{K_{\mathrm{AB}}A\left(T_{\mathrm{B}}-T_{\mathrm{A}}\right)}{l}$
$$\begin{gathered} =\frac{50\times 1\times {10}^{-4}\left(40\right)}{20\times {10}^{-2}} \\ =1\mathrm{W} \end{gathered}$$

(b) $\frac{{\left(\mathrm{\Delta }Q\right)}_{\mathrm{AC}}}{\mathrm{\Delta }t}=\frac{K_{\mathrm{AC}}\mathit{·}A\left(T_{\mathrm{C}}\mathit{-}{T}_{\mathrm{A}}\right)}{l}$
$$\begin{gathered} =\frac{400\times 1\times {10}^{-4}\times \left(40\right)}{20\times {10}^{-2}} \\ =8\mathrm{W} \end{gathered}$$

(c) $\frac{{\left(\mathrm{\Delta }Q\right)}_{\mathrm{BC}}}{\mathrm{\Delta }t}=\frac{\left(K_{\mathrm{BC}}\right)A\left(T_{\mathrm{B}}-T_{\mathrm{C}}\right)}{l}$
= 0