Question

An electric heater is rated 750 W – 200 V. Calculate (a) current flowing through it, (b) resistance of its element. (c) If the heater is used to heat 400 g of water at 20°C and switched on for 5 minutes, calculate the final temperature of the water. [sp. heat capacity of water is \[4.2\frac{\text{J}}{\text{g}^\circ\text{C}}\]].

Answer 1

m = 400 g; p.d = 200 V; t = 5 min = 300 s, Sp. heat capacity = \[4.2\frac{\text{J}}{\text{g}^\circ\text{C}}\]

Let the rise in temperature of water = θ_R

(a) P = I × V

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

\[=\frac{750}{200}\]

= 3.75 A

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

\[= \frac{200}{3.75}\]

= 53.33 Ω

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

= 750 × 300

= 225000 J

Heat energy absorbed by the water = m c θ_R

= 400 × 4.2 × θ_R

∴ By the principle of calorimetry

Heat absorbed by water = Heat supplied by the heater

400 × 4.2 × θ_R = 225000

\[\therefore \theta_{R}=\frac{225000}{400 \times 4.2}\]

= 13.39°C

∴ Final temperature = initial temperature + rise in temperature (θ_R)

= 20°C + 13.39°C

= 33.39°С