NEET-XII-Physics
28: Heat Transfer
- #25Figure (28-E4) shows an aluminium rod joined to a copper rod. Each of the rods has a length of 20 cm and area of cross section 0.20 cm2. The junction is maintained at a constant temperature 40°C and the two ends are maintained at 80°C. Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached. The conductivites are KAt = 200 W m-1°C-1 and KCu = 400 W m-1°C-1.
Figure
Ans :
Area of cross section, A = 0.20 cm2 = 0.2 × 10-4 m2
Thermal conductivity of aluminium, KAl = 200 W/m °C
Thermal conductivity of copper, KCu = 400 W/m°C
Total heat flowing per second = qAl + qCu
`` =\frac{{K}_{\,\mathrm{\,Al\,}}\mathit{\times }A\mathit{\times }\left(80-40\right)}{l}+\frac{{K}_{\,\mathrm{\,Cu\,}}\times \,\mathrm{\,A\,}\times \left(80-40\right)}{l}``
`` =\frac{200\times 0.2\times {10}^{-4}\times 40}{0.2}+\frac{400\times 0.2\times {10}^{-4}\times 40}{0.2}``
`` =8\times {10}^{-1}+16\times {10}^{-1}``
`` =24\times {10}^{-1}``
`` =2.4\,\mathrm{\,J\,}/\,\mathrm{\,sec\,}``
Heat drawn in 1 minute = 2.4 × 60 = 144 J
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