Question

A coil of wire of resistance $50\Omega$ is embedded in a block of ice and a potential difference of $210 \, V$ is applied across it. The amount of ice which melts in $1$ second is [latent heat of fusion of ice = $80calg^{- 1}$ ]

• A. $0.262 \, g$
• B. $2.62 \, g$
• C. $26.2 \, g$
• D. $0.0262 \, g$

Answer 1

Answer: (B)

Heat energy of melting is provided by electric energy. Therefore,
$mL=\frac{V^{2}}{R}t$ , where $m,L,V,R\text{&}t$ represents mass of ice, latent heat of ice, voltage across resistance, resistance and time respectively.
On substituting all the corresponding values in the above relation, we have
$\Rightarrow m\left(80\right)\times 4.2=\frac{\left(210\right)^{2}}{50}\left(1\right)$
$\Rightarrow m=2.625 \, g$