Question

2. An electric kettle is rated$1000W-250V$. It is used to bring water at ${20}^{\circ }C$ to its boiling point. If the kettle is used for $11$ minutes and $12$ seconds, calculate :
(a) Current flowing through the element.
(b) Resistance of the element of the kettle.
(c) Mass of water in the kettle $[SHCofwater=4.2J{g}^{-1}°C^{-1}]$

Answer 1

(a) Determine the current flowing through the element :

$\begin{array}{c}P=VI\\ I=\frac{P}{V}\\ =\frac{1000}{250}\\ =4A\\ \end{array}$

The current flowing through the element is $4A$

(b) Determine the resistance of the filament of the element:

$\begin{array}{c}P=\frac{V^2}{R}\\ \text{R}=\frac{{\text{V}}^2}{\text{P}}\\ =\frac{250\times 250}{1000}\\ =\frac{62500}{1000}\\ =62.5\Omega \\ \end{array}$

The resistance of the filament of the element is $62.5\Omega$

(c) Determine the mass of water in the kettle :

Heat energy consumed in$11\min 12\sec s$

$\text{Q}=\text{heat supplied}=mc(\text{T}-t)$

$Q=P\times t=1000\times \left(11\times 60+12\right)$

$=(1000\times 672)J$

$Q=mc(\text{T}-t)$

$(1000\times 672)=m\times 4.2(100-20)$

$m=\frac{672\times 1000}{4.2\times 80}$

$=2000\text{g}$

$=\frac{2000}{1000}$

$=2\text{kg}$

The mass of water in the kettle $=2\text{kg}$