An electric water heater with an internal heating element of resistance 40\,\Omega is conn… — JAMB UTME

An electric water heater with an internal heating element of resistance $40\,\Omega$ is connected to a $200\,\text{V}$ mains power supply. If the heater is operated for $15\,\text{minutes}$ each day, what is the total electrical energy consumed by the heater over a period of $30\,\text{days}$ ?

$7.5\,\text{kWh}$ Answer
B
$30.0\,\text{kWh}$
C
$450.0\,\text{kWh}$
D
$37.5\,\text{kWh}$

Answer

The total electrical energy consumed over 30 days is

7.5\,\text{kWh}

The electrical power rating of the water heater is

P = \frac{V^2}{R} = \frac{200^2}{40} = 1000\,\text{W} = 1.0\,\text{kW}

. Operating for

15\,\text{minutes}

(

0.25\,\text{hours}

) daily for

30\,\text{days}

yields a total time of

7.5\,\text{hours}

. The total energy consumed is

1.0\,\text{kW} \times 7.5\,\text{h} = 7.5\,\text{kWh}

Step-by-Step Solution

Calculate the electric power rating of the heater

P = \frac{V^2}{R} = \frac{200^2}{40} = \frac{40000}{40} = 1000\,\text{W} = 1.0\,\text{kW}

Electric power dissipated in a resistance

R

connected across potential difference

V

is given by

P = \frac{V^2}{R}

Calculate the total operating time in hours

t = 30 \times \frac{15}{60} = 7.5\,\text{hours}

Commercial electrical energy is measured in kilowatt-hours (

\text{kWh}

), so time must be converted from minutes to hours.

Calculate total electrical energy consumed

E = P \times t = 1.0\,\text{kW} \times 7.5\,\text{h} = 7.5\,\text{kWh}

Electrical energy consumed is the product of power in kilowatts and total time in hours.

Key Concept

Calculation of commercial electrical energy consumption in kilowatt-hours using electric power and operating time.

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