Question

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

• A. 0.85 gm/s
• B. 1.92 gm/s
• C. 6.56 gm/s
• D. None of these

Answer 1

The correct option is C 6.56 gm/s

According to Joule's law of heating, the rate of heat developed across wire of resistance $R$ is given by,

$\frac{Q}{t}=\frac{V^2}{R}$

Now,

$\frac{Q}{t}=\frac{{210}^2}{20}=2205J{s}^{-1}$

Since,

Latent heat of ice $=80cal/g=336J/g$

Therefore,

Rate of melting $=\frac{2205}{336}=6.56g{s}^{-1}$