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Question

The left end of a copper rod (length = 20 cm, area of cross section = 0.20 cm2) is maintained at 20°C and the right end is maintained at 80°C. Neglecting any loss of heat through radiation, find (a) the temperature at a point 11 cm from the left end and (b) the heat current through the rod. Thermal conductivity of copper = 385 W m−1°C−1.

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Solution



Area of cross section, A = 0.2 cm2 = 0.2 × 10−4 m2
Thermal conductivity, k = 385 W m–1 °C–1

Rate of flow of heat=Temperature differenceThermal resistanceΔQΔt = kA(T1-T2)lΔQΔt = 80-200.2×385×0.2×10-4 = 2310×10-3 = 2.31 J/sec

Let te temperature of point C be T, which is at a distance of 11 cm from the left end.
Rate of flow of heat is given by

ΔQΔt = kAΔTlΔTl = ΔQΔt×1kAT-2011×10-2 = 2.31383×0.2×10-4T = 33+20T = 53°C

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