The unit of thermal conductivity is

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Question

$\frac{\mathrm{dQ}}{\mathrm{dt}}=-(\mathrm{K}) \mathrm{A} \frac{\mathrm{dT}}{\mathrm{dx}}$

$\frac{\mathrm{J}}{\mathrm{s}}=(\mathrm{K}) \mathrm{m}^{

Vedclass · Accepted Answer

$\mathrm{Wm}^{-1} \mathrm{K}^{-1}$

Vedclass · Answer

$\frac{\mathrm{dQ}}{\mathrm{dt}}=-(\mathrm{K}) \mathrm{A} \frac{\mathrm{dT}}{\mathrm{dx}}$

$\frac{\mathrm{J}}{\mathrm{s}}=(\mathrm{K}) \mathrm{m}^{2} \frac{\mathrm{kelvin}}{\mathrm{m}}$

$(\mathrm{K})= watt \;\mathrm{m}^{-1} \mathrm{K}^{-1}$

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