Question

If $d$ is the density and $S$ is the specific heat then thermal conductivity $K$ and thermometric conductivity $h$ are related as

• A. $h=\frac{K}{S}$
• B. $h=\frac{dK}{S}$
• C. $h=\frac{K}{dS}$
• D. $h=KSd$

Answer 1

Answer: (C)

$h=\frac{K}{dS}$

For conduction we know that:
$Q$$=\frac{KA\Delta T}{L}$--------(1)
$Q$$=mS\Delta T$ ($m=dV$)-------(2)

And from convection we know that:
$Q$=$hA\Delta T$--------(3)

On equating (1, 2, 3) we get:
$dVS\Delta T$$=\frac{KA\Delta T}{L}$$=hA\Delta T$
$h=\frac{K}{dVS}$ (volume "V" will be constant so it can we taken as unity)

$\Rightarrow h=\frac{K}{dS}$