NEET-XII-Physics
28: Heat Transfer
- #16Three 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.Ans : Thermal conductivity of rod AB, KAB = 50 J/m-s-°C
Temperature of junction at A, TA = 40°C
Thermal conductivity of rod BC, KBC = 200 J/m-s-°C
Temperature of junction at B, TB = 80°C
Thermal conductivity of rod BC, KCA = 400 J/m-s-°C
Temperature of junction at C, TC = 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}``
`` =\frac{50\times 1\times {10}^{-4}\left(40\right)}{20\times {10}^{-2}}``
`` =1\,\mathrm{\,W\,}``
(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}``
`` =\frac{400\times 1\times {10}^{-4}\times \left(40\right)}{20\times {10}^{-2}}``
`` =8\,\mathrm{\,W\,}``
(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
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