Question

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 the melting of ice is ___. [Specific latent heat of fusion of ice = 80 cal/g]

• A. 0.85 g/s
• B. 1.92 g/s
• C. 6.56 g/s
• D. 5.65 g/s

Answer 1

The correct option is C 6.56 g/s

Power loss in the wire can be easily found as $P=\frac{V^2}{R}$
Heat loss due to the wire is :$H=\frac{V^2}{R}\times t$. This heat is used to melt the ice.
Amount of heat required to change the phase from ice to water = mL .
And this heat is equal to $\frac{V^2}{R}t$.
Hence, $mL=\frac{V^2}{R}t$.
We know, the heat produced in resistance will melt the ice. For ice to melt, Q =mL and L =80 cal/g
=$80\times 4.2J/g$...... [As , 1 calorie = 4.2 Joules]
So, Mass of ice melting per second $\times 80\times 4.2J$$=\frac{(210{)}^2}{20}$
or , Mass of ice melting per second (m) =$\frac{(210{)}^2}{20\times 80\times 4.2}$
=6.56 g/s