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In a certain experiment, cylindrical samples of diameter 4 cm and length 7 cm are used. The two thermocouples in each sample are placed 0.12 m apart. After initial transients, the electric heater is observed to draw 0.6 A at 110 V, and both differential thermometers read a temperature difference of 8°C. Determine the thermal conductivity of the sample. The thermal conductivity of the sample is:

Respuesta :

Answer:

K = .3941 × 10³ W/m.K

Explanation:

Qcond = K A ΔT÷ L

∴K = Qcond ×L ÷ A ΔT

J ÷ S = P

P = I × V =Qcond

∴Qcond = I × V

               = 0.6 A × 110 V

               =66 W

L = 0.12 m

ΔT = 8 °C

Qcond =33 V

Area = (πD²) ÷ 4

       = [π (4 × 10⁻² )²] ÷  4

        = 1.256 × 10⁻³ m²

∴A = 1.256 × 10 ⁻³³ m²

So K = ( Qcond × L ) ÷ A ΔT

         = (33) (0.12 ) ÷ (1.256 ×10⁻³ ) × 8

         = 0.3941 × 10³ W/m .K

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