Physics

Watt per meter kelvin is the unit of


ANSWER

Coefficient of thermal conductivity


SOLUTION
$$\dfrac { Watt }{ \left( meter \right) \times \left( kelvin \right)  } =\dfrac { M{ L }^{ 2 }{ T }^{ -3 } }{ LK } =M{ L }{ T }^{ -3 }{ K }^{ -1 }$$
Coefficient of thermal conductivity $$\left( k \right) =\dfrac { \left( \dfrac{q}{A} \right) \times \ t }{ dT } $$

$$\left( q/A \right) =$$ heat transfer per unit area$$\left( W/{ m }^{ 2 } \right) $$

$$dT=$$ temperature difference
$$t=$$ thickness

$$\left[ k \right] =\dfrac { \left[ M{ L }^{ 2 }{ T }^{ -3 } \right] \left[ { L }^{ -2 } \right] \left[ L \right]  }{ \left[ K \right]  } =M{ L }{ T }^{ -3 }{ K }^{ -1 }$$
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