Explanation

Heat $\mathrm{\Delta}Q$ transferred through a rod of length $L$ and area $A$ in time $\mathrm{\Delta}t$ is $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}Q = KA\left( \frac{T_{1} - T_{2}}{L} \right)\mathrm{\Delta}t$ $\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K = \frac{\mathrm{\Delta}Q \times L}{A\left( T_{1} - T_{2} \right)\mathrm{\Delta}t}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$ Substituting dimensions for corresponding quantities in Eq. (i), we have $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack K\rbrack = \frac{\left\lbrack ML^{2}T^{- 2} \right\rbrack\lbrack L\rbrack}{\left\lbrack L^{2} \right\rbrack\lbrack\theta\rbrack\lbrack T\rbrack}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left\lbrack MLT^{- 3}\theta^{- 1} \right\rbrack$