Question

If a wire of resistance $20\Omega$ is covered with ice and a voltage of 210 V is applied across the wire, then what is the rate of the melting of ice?
[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

Given:
Heating of the wire melts the ice.
Resistance, $R=20\Omega$
Voltage, $V=210V$
Latent heat of fusion for ice, $L=80cal/g$

Power loss in the wire is:
$P=\frac{V^2}{R}$

Amount of heat required to change the phase of mass $m$ from ice to water is:
$H_W=mL$

Rate of absorption of heat by water equals the rate of heat dissipation by the resistor.

$\frac{H_W}{t}=P$
$\frac{mL}{t}=\frac{V^2}{R}$
Rate of melting of ice:
$\frac{m}{t}=\frac{V^2}{RL}$
$\frac{m}{t}=\frac{(210{)}^2}{20\times 80\times 4.2\times {10}^3}$
$\frac{m}{t}=0.00656kg/s$
$\frac{m}{t}=6.56g/s$