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One end of a $2.35\,\,m$ long and $2.0\,\,cm$ radius aluminium rod$(K = 235 \,\,W.m^{-1}K^{-1})$ is held at $20^0\,\,C$. The other end of the rod is in contact with a block of ice at its melting point. The rate in $kg.s^{-1}$ at which ice melts is
[Take latent heat of fusion for ice as $\frac{{10}}{3} ×10^5 J.kg^{-1} $]
$48\pi × 10^{-6}$
$24\pi × 10^{-6}$
$2.4\pi × 10^{-6}$
$4.8\pi × 10^{-6}$
Solution
$l=2.35 m, r=2 c m$
$K=235, T=20^{\circ} C, \Delta T=20^{\circ} C$
$\frac{d H}{d t}=\left(\frac{d m}{d t}\right) L_{f}$
$\left(\frac{d m}{d t}\right)=\left(\frac{d H}{d t}\right)_{L_{f}}^{\perp}=\frac{K A}{l}(\Delta T) \frac{1}{L_{f}}$
$\left(\frac{d m}{d t}\right)=\frac{235 \times \pi\left(2 \times 10^{-2}\right)^{2} \times 20}{2.35 \times \frac{10}{3} \times 105}$
$=\frac{100 \pi \times 4 \times 10^{-4} \times 20}{10 \times 105} \times 3$
$=2.4 \pi \times 10^{-6} \mathrm{ks} / \mathrm{sec}$
