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

Hint: 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.