Energy use in the home
Energy is used in homes and businesses for running different appliances. We use energy for heating, cooling, cooking, and running machinery. Energy has units of joules (J), Calories (Cal) and kilowatt hours (kWh). We generally use kilowatt hours when describing the use of energy by electrical appliances in the home and we use joules and Calories to describe energy related to food and exercise.
The energy of $1$1 joule is defined as the amount of energy transferred to an object when we apply $1$1 newton of force over a $1$1 metre distance-this is roughly equivalent to lifting a medium sized ($100$100 g) tomato $1$1 metre. A Calorie is defined to be the energy required to heat $1$1 kg of water by $1^\circ$1° Celsius. $1$1 Calorie is equivalent to $4184$4184 joules. But what about kilowatt hours?
Every appliance has a power rating. This is the rate at which energy is consumed or generated and is measured in watts (W), which is equal to $1$1 joule per second. So a $40$40 watt light-globe would require $40$40 joules of energy each second to run. $1$1 kilowatt hour is the energy supplied by $1000$1000 watts of power sustained for $1$1 hour - this is equivalent to $3600000$3600000 joules. Additional units of kilojoules (kJ) and megajoules (MJ) are useful for large amounts of energy, where $1$1 kWh is equal to $3.6$3.6 MJ or $3600$3600 kJ.
Energy consumption within the home is generally measured in kilowatt hours (kWh) as this is more convenient than joules, especially when appliances consume high amounts and we are calculating energy use over hours or days. For example, a full kettle on average may use $0.1$0.1 kWh to boil water and a washing machine could use $0.75$0.75 kWh per use. From the conversion diagram above we can see a single kWh is equivalent to $3.6\times10^6$3.6×106 joules, so the washing mashing requires $2700000$2700000 joules per load. You can see how it could sometimes be more useful to think of the amount of energy used by a washing machine in as $0.75$0.75 kWh rather than $2700000$2700000 J.
Practice question
question 1
Use the fact that $1$1 kWh = $3.6\times10^6$3.6×106 J to calculate how many joules $104.23$104.23 kWh is equal to.
Express your answer as a basic numeral.
Energy consumption of appliances
The energy consumption of an appliance is the amount of energy it uses. We measure energy consumption of electrical appliances in kilowatt-hours (kWh). We can also express it in megawatt-hours when describing larger amounts.
We can calculate the energy consumption of an appliance over time by using:
$\text{Energy}=\text{Power}\times\text{Time}$Energy=Power×Time
- If we have power in kilowatts and time in hours, this will calculate the energy in kWh.
- Remember you can convert from watts to kilowatts by dividing by $1000$1000.
Electricity suppliers usually charge households and businesses in cents per kWh.
Worked example
Belle has a $80$80 W LED television that she watches $4$4 hours every day.
(a) How many kWh of energy will the television use in a day?
Think: To obtain the energy used in kWh we require the power to be in kilowatts and the time to be in hours and then multiply.
Do: First convert $80$80 W to kW: $80\div1000=0.08$80÷1000=0.08 kW
Hence, energy used in $4$4 hours is:
| Energy | $=$= | $0.08\times4$0.08×4 kWh |
| $=$= | $0.32$0.32 kWh |
Therefore, the television will use $0.32$0.32 kWh in a day.
(b) How much will the electricity cost if the average cost of electricity is $\$0.18$$0.18 per kWh?
Think: To calculate cost, multiply the usage by the rate.
Do:
| Cost | $=$= | $\$0.18\times0.32$$0.18×0.32 |
| $=$= | $\$0.0576$$0.0576 |
Therefore, the cost is approximately $\$0.058$$0.058 (or $5.8$5.8 cents).
Practice questions
Question 2
Valentina uses a $550$550-watt iron for $8$8 hours every week.
How much energy (in kWh) does it use per week?
What is the cost of operating the iron for one year if the price of electricity is $\$0.2785$$0.2785/kWh, correct to the nearest cent?
Assume there are $52$52 weeks in a year.
Question 3
A fridge consumes energy at a rate of $210$210 watts. The fridge remains on all year. The average rate for electricity is $\$0.14$$0.14 per kWh.
Given that there are $365$365 days in a year, what is the running cost of the fridge for an entire year? Give your answer in dollars and round to the nearest cent.
Energy rating of appliances
We rely on electrical appliances for so many of our daily activities. The consequences of our need for electricity is high electricity bills but also high carbon dioxide emissions.
One way we can reduce our energy consumption is to use more energy efficient appliances. In Australia an energy rating label is provided for various appliances, such as refrigerators, televisions, dishwashers, and air conditioners. The label tells you how much energy an appliance will use in a year and the star rating tells you how energy efficient the appliance is.
The lower the number in the red box the less it will cost you to run. The higher the number of stars, the more energy efficient the appliance is.
Practice questions
Question 4
A heater consumes $185$185 kWh of energy per year. Calculate the cost of operating the heater for a year if the average cost of electricity is $\$0.19$$0.19 per kWh.
Question 5
A family is comparing the price and energy consumption of dishwashers in a store. They expect that over a year they will use the dishwasher for $320$320 loads. They find a $2$2-star rated dishwasher for $\$320$$320 that uses $1.9$1.9 kWh per load, and compare this to an energy saving $4$4-star rated dishwasher which costs $\$460$$460 and uses $1.1$1.1 kWh per load.
Use the fact that $1$1 kWh costs $17$17 cents to answer the following questions.
Over a one year period, how much will they expect to save in electricity costs using the $4$4-star dishwasher rather than the $2$2-star dishwasher? Round your answer to the nearest cent.
Will the electricity saving in the first year offset the difference in purchase price of the dishwashers?
Yes
ANo
B