Question

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

• A. $8.85g{s}^{-1}$
• B. $1.92g{s}^{-1}$
• C. $6.56g{s}^{-1}$
• D. $none$

Answer 1

Answer: (C)

$6.56g{s}^{-1}$

We know that heat ,$H=I^{2}RT=\frac{V^2}{R}t$

The rate of heat developed in wire $=H/t=\frac{V^2}{R}=\frac{(210{)}^2}{20}=2205J{s}^{-1}=\frac{2205}{4.2}cal{s}^{-1}$ as $(1cal=4.2J)$

Let m be the amount of ice melting per sec and, then $mL=H/t$ where

$L=80cal/g=$ latent heat of ice.

thus, $m=\frac{\left(H/t\right)}{L}=\frac{2205}{4.2\times 80}=6.56g{s}^{-1}$