Question

If a wire of resistance $20\Omega$ is covered with ice and a voltage of 210 V is applied across the wire, What will be the rate of melting of ice? [Specific latent heat of fusion of ice = 80 cal/g]

• A. 5.85 g/s
• B. 1.92 g/s
• C. 6.60 g/s
• D. 5.65 g/s

Answer 1

The correct option is C

6.60 g/s

Let us assume that,
Mass of ice melting per second = m gms.

Power loss in the wire = $P=\frac{V^2}{R}$
And, Heat loss due to the wire is = $H=\frac{V^2}{R}\times t$.
Also, amount of heat required to change the phase from ice to water = Q = mL (where, L = latent heat of fusion of water = 334 J/g).
We know, the heat produced in resistance will melt the ice, so,
$\frac{V^2}{R}\times t=mL$
Substituting t= 1 sec, we get:
$\frac{(210{)}^2}{20}=m\times 334J\to m=\frac{(210{)}^2}{20\times 334}$
=6.60 g/s