If a wire of resistance $20\,\Omega $ is covered with ice and a voltage of $210\, V$ is applied across the wire, then the rate of melting of ice is .................. $g/s$
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$H = \frac{{{V^2}}}{R} \times t = \frac{{{{(210)}^2}}}{{20}} \times 1 = mL$
$\frac{{{{(210)}^2}}}{{20}} = m \times 80 \times 4.2$ $ \Rightarrow $ $m = 6.56\,\,g/s\,$
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