Question

An electric kettle is rated 1000 W – 250 V. It is used to bring water at 20°C to its boiling point. If the kettle is used for 11 minutes and 12 seconds, calculate :

1. Current flowing through the element.
2. Resistance of the element of the kettle.
3. Mass of water in the kettle [SHC of water = 4.2 Jg⁻¹°C⁻¹]

Answer 1

P = 1000 W; V = 250 V; t = 11 min and 12 s = 672 s; $$c = 4.2\frac{\text{J}}{\text{g}^\circ\text{C}}$$

(a) P = I × V

\[\therefore \text{I}  =\frac{\text{P}}{\text{~V}}\]

\[=\frac{1000}{250}\]

= 4 A

\[\text{(b) }\text{R} = \frac{\text{V}}{\text{I}}\]

\[= \frac{250}{4}\]

= 62.5 Ω

(c) Heat energy supplied by the kettle = P × t

= 1000 × 672 

= 672000 J

Heat energy absorbed by the water = m c θ_R

Rise in temperature of water = 100°C − 20°C = 80°C

∴ By the principle of calorimetry

Heat absorbed by the water = Heat supplied by the kettle

m c θ_R = 672000

m × 4.2 × 80 = 672000

m = \[\frac{672000}{4.2 \times 80}\]

= 2000 g