# A square iron plate of side 3 cm and thickness 8 mm has its opposite faces at $\;100^{0}C\;$ and $\;60^{0}C\;$ . If K=0.16 C.G.S units , then the amount of heat flow through the rod is 1 minute is

$(a)\;4320\;cal\qquad(b)\;3240\;cal\qquad(c)\;720\;cal\qquad(d)\;1400\;cal$

Explanation :
$\large\frac{dQ}{dt}= \large\frac{KA (T_{1}-T_{2})}{L}$
$=\large\frac{0.16 \times 9 \times (100-60)}{8 \times 10^{-1}}$
$=72\;cal/s=4320\;cal/min\;.$