Question

An electric kettle takes 4 amperes current at $250$ volts. How much time will it take to boil 1Kg of water from$20^{\circ}$? The temperature of boiling water is $100^{\circ}$.

Answer 1

To solve this problem first we will calculate the power of kettle from given information about kettle then calculate the energy required to boil water from$20^{\circ}$.Finally by comparing both energies we will get our final answer.

Complete step by step solution:

Given
Electric kettle takes 4 ampere current at 250 V.
So , power of kettle is
$P = VI$
$P = 250 \times 4$
$P = 1000 W = 1000 J/S$....(1)
Now kettle can generate heat by $1000 J/S$
Heat energy required to boil 1Kg of water from$20^{\circ}$
$H = MC \vartriangle T$
Where
H is heat energy.
M mass of water.
$\vartriangle T$ is a change in temperature.
C is the specific heat capacity of water which is $4200 J/Kg K$.
Now heat energy required to boil 1Kg of water from$20^{\circ}$
$H = 1\times 4200 \times 80$
$H = 336000 J$........(2)
And we know
$P = \dfrac{{H}}{t}$
$t = \dfrac{{H}}{P}$
Now substituting the value of H and P from equation (1) and (2).
$t = \dfrac{{336000}}{1000}$
$t = 336 sec$
$t = 5.6 min$
**Therefore time required to boil 1Kg of water from $20^{\circ}$ is 5.6 min.
**

Note: When the electricity travels through any wire, it causes the flow of Electrons or Charge.
And if we calculate “rate of flow of charge(charge/time)” in that wire then this calculated rate of current flowing in that wire, Voltage is referred to as the energy per coulomb of charge here, volts have the dimension of Joules per Coulomb.