Question

A $10\text{H}$ inductor carries a current of $20\text{A}$. How much ice at $0{}^{\circ }\text{C}$ could be melted by the energy stored in the magnetic field of the inductor. Latent heat of fusion of ice is $3.4\times {10}^5\text{J/Kg}$.

• A. $0.88\text{kg}$
• B. $0.59\times {10}^{-2}\text{kg}$
• C. $1.28\times {10}^{-2}\text{kg}$
• D. $2.38\text{Kg}$

Answer 1

The correct option is B $0.59\times {10}^{-2}\text{kg}$

Energy stored in the inductor is,

$U=\frac{1}{2}L{i}^2$

This energy is completely used in meting the ice. If $L_f$ is the latent heat of fusion, then,

$\frac{1}{2}L{i}^2=m{L}_f$

$\Rightarrow m=\frac{L{i}^2}{2L_f}$

Where, $m$ is the mass of ice melted.

Here, $L=10\text{H};i=20\text{A};L_f=3.4\times {10}^5\text{J/kg}$

$\therefore m=\frac{10\times {20}^2}{2\times (3.4\times {10}^5)}=0.59\times {10}^{-2}\text{kg}$

Hence, $\left(\text{B}\right)$ is the correct answer.

Why this question ? Key concept:- In such type of problem, apply conservation of energy, Energy released by inductor = energy absorbed by ice.