Question

A water heater uses a resistor of 16 Ω to boil water. The heater is connected to a 240 V power source. If the heater is used for 2 hours a day, calculate the cost to run the heater for a year (365 days). The electricity cost by TNB is given as 21.8 sen/kWh.

Answer 1

First, calculate the power used each day. Use the following formula:

`P=(V^2)/(R)`

where,

V: voltage = 240V

R: resistor = 16 Ω

Replace the previous values into the formula for P:

`P=((240V)^2)/(16\Omega)=3600W=3.6kW`

Then, consider that the energy used in kWh, each day, is:

`E=P\cdot t=3.6kW\cdot2h=7.2\text{kWh}`

And in a year, the energy used is:

`E^(\prime)=7.2\text{kWh}\cdot365=2628kWh`

And the cost will be:

`\text{cost}=(2628kWh)(\frac{21.8sen}{\text{kWh}})=57290.4sen`

Hence, the cost to run the heater in one year is 57290.4sen