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

The rate of heat developed is given by

$\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}\mathbf{=}\frac{{\mathbf{\left(}\mathbf{210}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{=}\mathbf{2205}\mathit{J}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathit{c}\mathit{a}\mathit{l}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$

Let m be the amount of ice melting per sec and, then

$m\times 80=\frac{2205}{4.2}orm=6.56g{s}^{-1}$