Dimensional formula for thermal conductivity | vedclass

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Question

$\because \frac{ d Q}{ dt }= k A \frac{ d T }{ d x }$
$k =\frac{\left(\frac{ d Q}{ dt }\right)}{ A \left(\frac{ d T }{ d x }\right)}$
$[ k ]=\frac{\

Vedclass · Accepted Answer

$MLT ^{-3} K ^{-1}$

Vedclass · Answer

$\because \frac{ d Q}{ dt }= k A \frac{ d T }{ d x }$
$k =\frac{\left(\frac{ d Q}{ dt }\right)}{ A \left(\frac{ d T }{ d x }\right)}$
$[ k ]=\frac{\left[ ML ^{2} T ^{-3}\right]}{\left[ L ^{2}\right]\left[ KL ^{-1}\right]}=\left[ MLT ^{-3} K ^{-1}\right]$

Dimensional formula for thermal conductivity is (here $K$ denotes the temperature)

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