Page 48 - Basic Electrical Engineering
P. 48
One kWh of energy costs 3.50.
The total cost of energy per month = 157.5 × 3.50
= 551.25
Example 1.6 An electric kettle has to raise the temperature of 2 kg of water
from 30°C to 100°C in 7 minutes. The kettle is having an efficiency of 80 per
cent and is supplied from a 230 V supply. What should be the resistance of its
heating element?
Solution:
m = 2 kg = 2000 gms
t − t = 100 − 30 = 70°C
1
2
Specific heat of water = 1
Time of heating = 7 minutes = hours
Output energy of the kettle = m s t
= 2000 × 1 × 70 Calories
= 140 kilo Calories
=
= 0.1627 kWh. [1 kWh = 860 kCal]
Supply voltage, V = 230 Volts.
