Question

The electrical resistivity of copper at 27˚C is 1.72*10^-8Ωm. compute its thermal conductivity if the

Lorentz number is 2.26*10^-8 WΩK^-2​

Answer 1

Answer:

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Answer 2

Given: The electrical resistivity of copper at 27˚C is 1.72*10^-8 Ωm. Lorentz number is 2.26*10^-8 WΩK^-2

To find: Thermal conductivity of copper

Explanation: Let the electrical resistivity of the copper be p and its conductivity be s.

Conductivity is the reciprocal of resistivity.

s= 1/p

$s = \frac{1}{1.72 \times {10}^{ - 8} }$

$s = 5.81 \times {10}^{7} {ohm}^{ - 1} {m}^{ - 1}$

Let the temperature be t and Lorentz number be l.

t = 27° C

= 27+273 K

= 300 K

$l = 2.26 \times {10}^{ - 8}$

The relation between Lorentz number l, electrical conductivity s, temperature t and thermal conductivity k is given by:

k = s × l × t

$= > k = 5.81 \times {10}^{7} \times 2.26 \times {10}^{ - 8} \times 300$

=> k = 393.918 W / m K

Therefore, the thermal conductivity of copper is 393.18 W / m K.