Question

A heater coil rated $1000\,W$ is connected to a $110\,V$ mains. How much time will it take to melt $625\,gm$ of ice at ${0^ \circ }C$? (${\text{L = 80calg}}{{\text{m}}^{{\text{ - 1}}}}$)
A. $100\,s$
B. $150\,s$
C. $200\,s$
D. $210\,s$

Answer 1

Learn the Joule’s law of heating for resistive circuits and find latent heat of ice to find the time required. The Joule’s heating for a resistive circuit is proportional to the square of current flowing, the resistance of the circuit and the time.

Formula used:
The Joule’s law of heating in an electric circuit is given by,
$H = {I^2}Rt$
where, $I$ is the current through the circuit, $R$ is the resistance of the circuit and $t$ is the time passed.
The latent heat of mass $m$ is given by,
$H = mL$
where, $L$ is the specific latent heat of the object at a fixed temperature (melting or boiling or freezing point)

Complete step by step answer:
We have given here 635gm of ice at ${0^ \circ }C$ and it will be melted using a heater of 1000W. Now, the energy required to melt ice at ${0^ \circ }C$ is equal to the latent heat of the ice. Now, the latent heat of mass m is given by, $H = mL$. So, the heat required to melt 625gm of ice at ${0^ \circ }C$ will be,
$H = 625 \times {\text{80 = 50000cal}}$
$\Rightarrow H = 21 \times {10^4}\,J$

Now, this heat will be required to melt the ice now the energy loss of a circuit is given by the joule’s law of heating, $H = {I^2}Rt = Pt$ where, $I$ is the current through the circuit, $R$ is the resistance of the circuit and $t$ is the time passed, where P is the power of the element.Now, putting the value of $H$ in the equation we will have,
$21 \times {10^4} = {10^3}t$
$\therefore t = 210\,s$
Hence the time required to melt $625\,gm$ of ice is $210\,s$.

Hence, option D is the correct answer.

Note: The heat energy required to melt ice from the freezing or melting point is equal to the latent heat of the material. But if it is melted from a higher temperature we have to find the heat energy required to drop the temperature to freezing point and then calculate the latent heat.