Question

Heat flow through a wooden board 30mm thick at 0.0086w/cm^3 when one of its side 20°C warmer than the other. Determine the thermal conductivity of the wood.

Answer 1

Solution.

$L=30mm=3cm;$

$\dfrac{Q}{A}=0.0086W/cm^3\sdot3cm=0.0258W/cm^2=258W/m^2;$

$\Delta T=20^oC=20K;$

$\dfrac{\Delta Q}{\Delta t}=$ $\dfrac{\Delta T}{L}\sdot K\sdot A$;

$\dfrac{Q}{A}=\dfrac{\Delta T}{L}\sdot K\implies K=\dfrac{\ Q}{A}\sdot\dfrac{L}{\Delta T}$ ;

$K=258W/m^2\sdot\dfrac{0.03m}{20K}=0.387Js^{-1}m^{-1}K^{-1};$

Answer: $K=0.387JS^{-1}m^{-1}K^{-1}.$