Question

An electric kettle takes 4A current at 220V. How much time will it take to boil 1kg of water from room temperature $20^\circ C$? (the temperature of boiling water is $100^\circ C$ ).
A) 12.6 min
B) 12.8 min
C) 6.3 min
D) 6.4 min

Answer 1

Hint: When water is boiled in the kettle, the energy is supplied in the form of electricity. The amount of electrical energy supplied will be equal to the gain in the amount of heat energy of water.

Complete step by step answer:
Given:
The electric current is $4{\text{A}}$.
The voltage difference is $220{\text{V}}$.
The mass of water is $1{\text{kg}}$.
The room temperature is $20^\circ {\text{C}}$.
The boiling temperature is $100^\circ {\text{C}}$.
The electrical energy produced by the kettle is the energy required to boil the water.
The expression of electrical energy is given as,
$E = VIt$
Here, $V$ is the voltage, $I$ is the current supplied and $t$ is the for current supplied.
While heating the water energy is stored by the water in the form of heat.
The amount of energy stored in the water can be calculated as,
$Q = mc\left( {{T_2} - {T_1}} \right)$
Here, $m$ is mass of water, $c$ is the specific heat of water and is $4180{\text{J/kg}} \cdot {{K^{-1}}}$, ${T_2}$ is the final temperature and ${T_1}$ is the initial temperature.
The amount of electrical energy will be equal to the heat stored in the water according to the energy conservation of law as it converts from electrical energy to heat energy,
$
  VIt = mc\left( {{T_2} - {T_1}} \right) \\
  t = \dfrac{{mc\left( {{T_2} - {T_1}} \right)}}{{VI}} \\
 $
Substitute the values in above expression,
\[
\Rightarrow t = \dfrac{{mc\left( {{T_2} - {T_1}} \right)}}{{VI}} \\
\Rightarrow t = \dfrac{{1 \times 4180\left( {100 - 20} \right)}}{{220 \times 4}} \\
\Rightarrow t = 380\sec \times \dfrac{{1\min }}{{60\sec }} \\
\Rightarrow t = 6.3\min \\
 \]

Thus, the time required to boil the water is $6.3\min $.

Note:
The law of energy conservation is applied as the electrical energy converts into heat energy. Make sure the values are substituted in the equation carefully, so that the final answer comes in proper units.