Question

If a wire of resistance 20 ohm is covered with ice and a voltage of 210V is applied across the wire,then the rate of melting of ice is.....please help the answer is 6.56gs^-1

Answer 1

hey newton,
see the above image.

Answer 2

Answer:

Rate of melting of ice = 6.56 g s⁻¹

Explanation:

As per the question,

We know that,

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,

$\frac{Q}{t}=\frac{210^{2}}{20}$

$\frac{Q}{t}=2205Js^{-1}$

Since,

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

Therefore,

Rate of melting of ice is given by

$Rate\ of\ melting\ of\ ice=\frac{2205}{336}$

Or

Rate of melting of ice = 6.56 g s⁻¹