Heating Effect of Electric Current, Joule’s law: Solved Example Problems

EXAMPLE 2.28

An electric heater of resistance 10 Ω connected to 220 V power supply is immersed in the water of 1 kg. How long the electrical heater has to be switched on to increase its temperature from 30°C to 60°C. (The specific heat of water is s = 4200 J kg-1)

Solution

According to Joule’s heating law H = I2 Rt

The current passed through the electrical heater = 220V/10Ω = 22 A

The heat produced in one second by the electrical heater H = I2 R

The heat produced in one second H = (22)2 x 10 = 4840 J = 4.84 k J. In fact the power rating of this electrical heater is 4.84 k W.

The amount of energy to increase the temperature of 1kg water from 30°C to 60°C is

Q = ms ∆T (Refer XI physics vol 2, unit 8)

Here m = 1 kg,

s = 4200 J kg-1,

∆T = 30,

so Q = 1 × 4200 x 30 = 126 kJ

The time required to produce this heat energy t = Q/ I2R = 126 ×103 /  4840  â‰ˆ 26 .03 s