If you are given the amount of energy transferred in kilojoules, you need to convert it to joules to calculate the power in watts.
joules (J) | kilojoules (kJ) |
120 | |
34 | |
1 230 | |
24,6 |
joules (J) | kilojoules (kJ) |
120 | 0,12 |
34 000 | 34 |
1 230 | 1,23 |
24 600 | 24,6 |
Chapter overview
2 weeks
This chapter deals with the costs involved when using electricity. Some learners will have prepaid electricity meters in their homes, while others will be billed monthly according to their usage. Which ever way they are billed, they need to have an understanding of how the charges are calculated. In order to simplify the calculations we need to assign a tariff. Eskom uses a sliding tariff scale and so it can sometimes be tricky to calculate exact costs.
IMPORTANT NOTE:
This chapter has rather been called 'Cost of electrical energy' and not 'Cost of electrical power' as in CAPS. We do not pay for power, we pay for electrical energy.
The following statements in CAPS are incorrect:
consumers pay for the quantity of power they use
quantity of electrical power used is measured in kWh (kilowatt hour)
The points should rather read:
consumers pay for the quantity of energy they use
quantity of electrical energy used is measured in kWh (kilowatt hour)
The kilowatt hour is not a measure of power (which is measured in watts). 1 kWh is equal to 3,6 million joules. Joules is the SI unit of energy, but it is a small unit and is therefore not a suitable unit to use on electricity bills. We therefore use the kilowatt hour to measure the energy consumed.
7.1 What is electrical power? (1 hour)
Tasks | Skills | Recommendation |
Activity: Power rating of different appliances | Observing, researching, comparing, listing, calculating | CAPS suggested |
7.2 The cost of energy consumption (5 hours)
Tasks | Skills | Recommendation |
Activity: Calculating energy consumption | Comparing, calculating | CAPS suggested |
Activity: Calculating the cost of electrical energy | Comparing, calculating, analysing, justifying, explaining | CAPS suggested |
Activity: Home survey | Researching, collecting data, calculating, analysing | Suggested |
Activity: Comparing the energy efficiency of different light bulbs | Comparing, describing, calculating, explaining | Suggested |
Activity: Career research | Researching, working in groups, writing, presenting | CAPS suggested |
Electrical power is the rate of electrical energy supply. It is the amount of energy supplied per unit of time. In simpler terms, it is how fast the electrical energy is supplied.
Power is measured in watts (W).
A rate is ratio where one quantity is compared to time, for example km/h which is comparing distance (in kilometers) to time (in hours)
We can calculate the power using the formula:
Power = energy/time
Energy is measured in joules and so this means that power is the amount of joules supplied in a certain period of time. When doing calculations of power, you need to have the energy measured in joules and time measured in seconds.
1 watt is the same as 1 joule of energy transferred in a second. (1 watt = 1 joule per second)
There are 1000 watts in 1 kilowatt (kW).
Different appliances use different amounts of power, depending on their function. All electrical appliances have a stamp or a sticker which indicates the power rating. If you look at your hairdryer or kettle you should find it easily.
This video has a few great tips on how you can save electricity, money and the planet!
Which of the above two appliances uses more power to operate?
The electric pan uses much more power than the fan.
If possible, bring some different appliances to class, such as a kettle, toaster or iron, to show learners the labels with the power ratings on them. You can also walk around the school and identify the different power ratings on appliances around the school. Bring newspapers to school to use the advertisements section to allow learners to also study the appliances and to identify the power ratings. Bring in different light bulbs which have different power ratings to show these to learners.
INSTRUCTIONS:
Fill the power ratings of the various appliances into the following table.
Appliance | Power (W) |
Toaster | |
Electric beater | |
Television | |
Urn | |
Below are the power ratings for the appliances given here. Learners must also identify others.
Appliance | Power (W) |
Toaster | 700 |
Electric beater | 175 |
Television | 54 |
Urn | 1500 |
QUESTIONS:
Complete the following table to convert between joules and kilojoules:
If you are given the amount of energy transferred in kilojoules, you need to convert it to joules to calculate the power in watts.
joules (J) | kilojoules (kJ) |
120 | |
34 | |
1 230 | |
24,6 |
joules (J) | kilojoules (kJ) |
120 | 0,12 |
34 000 | 34 |
1 230 | 1,23 |
24 600 | 24,6 |
Complete the following table to convert between watts and kilowatts:
watts (W) | kilowatts (kW) |
1 760 | |
4,56 | |
25 | |
0,56 |
watts (W) | kilowatts (kW) |
1 760 | 1,76 |
4 560 | 4,56 |
25 | 0,025 |
560 | 0,56 |
Sequence the appliances listed in your table above from those that use the most power to those that use the least power.
Learner-dependent answer, depending on the appliances listed. Learners will find that appliances which provide heat use much more energy than appliances, like a fan or radio, which supply movement and sound respectively.
Did you record the power ratings of any other appliances which involve heating? What do you notice about the power for these appliances?
Appliances involving heating, such as kettles, toasters, irons, heaters, use a lot of power.
The following questions involve calculations based on the equation:
power = energy/timeIf 180 kJ of energy is transferred to your bedroom lamp in half an hour, what is the power rating of your lamp? Show your calculations.
100 000 kJ of energy passes through a power station every minute. What is the power rating of the power station? Show your calculations.
power = energy/time = 180 000 / 1 800 = 100 W
power = energy/time = 100 000 000 / 60 = 1 666 666,67 W
We pay for the electricity that we use in our homes. How do we calculate how much we pay based on our energy consumption?
Eskom charges us for the electrical energy we use in our homes. Eskom charges us based on our energy consumption. The more electrical energy we use to run our household electrical appliances, the more Eskom charges us.
How do we work out how much energy we use? Think for example of using a 1000 W microwave to warm your food for 1 minute. How much energy is transferred? We can rearrange the following equation:
power = energy/time
So that we have:
energy = power × time
In this formula, energy is measured in joules, time is measured in seconds and power in watts.
Therefore to calculate the energy consumption of using a 1000 W microwave, we can calculate it as follows:
energy = power × time
= 1000 W × 60 s
= 60 000 J
Eskom now wants to work out our energy consumption for the whole month for all the appliances in your home. If 60 000 J of energy were used to warm food for 1 minute, then you can see that we would calculate an extremely large number for our energy consumption for the whole month in joules. This is not practical for electricity bills. We therefore have an alternative unit for energy consumption.
The quantity used for energy consumption is the kilowatt-hour (kWh). 1 kWh is the energy used if a 1000 W appliance is used for 1 hour.
1 kWh is equal to 3 600 000 joules.
The kilowatt-hour is a measure of energy consumption as it is calculated by multiplying power in kilowatts by time in hours.
We can calculate the energy consumption of different appliances by multiplying the power rating by the amount of time it was used in hours.
INSTRUCTIONS:
A kilowatt-hour requires that the unit of the time is in hours. You must therefore convert any times that you are given into hours for your calculations.
Seconds (s) | Minutes (min) | Hours (h) |
620 | ||
120 | ||
127 | ||
940 | ||
4,5 | ||
12,25 |
Seconds (s) | Minutes (min) | Hours (h) |
620 | 10,33 | 0,172 |
120 | 2 | 0,033 |
7 620 | 127 | 2,12 |
56 400 | 940 | 15,67 |
16 200 | 270 | 4,5 |
44 100 | 735 | 12,25 |
QUESTIONS:
An oven with a power rating of 3600 W is used to bake a cake for 1 hour. What is the energy consumption?
3600 W = 3,6 kW
energy consumption = power x time = 3,6 x 1 = 3,6 kWh
A kettle with a power rating of 2200 W is used to boil water for 6 minutes. What is the energy consumption?
2200 W = 2,2 kW
6 minutes = 0.1 hours
energy consumption = 2,2 x 0.1 = 0,22 kWh
You use a 3600 W oven to bake a cake for 1,5 hours. What is the energy consumption?
3600 W = 3,6 kW
energy consumption = 3,6 X 1,5 = 5,4 kWh
A 120 W light bulb is left on for 2 hours. A 60 W light bulb is left on for 3.5 hours. Which light bulb has a higher energy consumption? Show your calculations.
120 W light bulb:
120 W = 0,12 kW
energy consumption = 0,12 x 2 = 0,24 kWh.
60 W light bulb:
60 W = 0,06 kW
energy consumption = 0,06 x 3,5 = 0,21 kWh
Therefore, the 120 W light bulb uses more power.
We are charged for the number of kilowatt-hours that we use. The cost of energy consumption is charged in cents per kilowatt-hour (c/kWh). The following table gives us the rates at which homeowners are charged for purchasing their power directly from Eskom. As you can see, there are different 'blocks'. The more energy you use per month, the more you pay per kilowatt-hour. This is called a tiered tariff system.
Eskom Homepower Tariffs 2013
Different energy consumptions per month | Energy charge (c/kWh) | Environmental levy charge (c/kWh)l | Total (c/kWh) |
Block 1 [≤ 50 kWh] | 67.07 | 2.28 | 69.35 |
Block 2 [51 - 350 kWh] | 83.32 | 2.28 | 85.60 |
Block 3 [351 - 600 kWh] | 124.74 | 2.28 | 127.02 |
Block 4 [> 600 kWh] | 137.03 | 2.28 | 139.31 |
This table is adapted from http://www.eskom.co.za/c/article/145/tariffs/
The tiered tariff system is used to encourage people to save electricity and use it wisely as the less you use, the less you pay per unit of electricity.
In order to calculate your electricity costs, choose the block that applies to your home. For example, if your home uses 252 kWh of electricity in a month then you fall into Block 2.
Let's calculate the cost if your home used 252 kWh in April 2013. The first 50 kWh are charged at the lower rate so: 50 × 69,35 = 3 467,5 cents The rest of the units are charged at the block 2 rate: (252 - 50) = 202 Therefore, 202 × 85,60 = 17 291,2 cents So, in total you would have to pay 3 467,5 + 17 291,2 = 20 758,7 cents Remember, the tariffs are quoted in cents, not rands, so you need to do a conversion. 20 758,7/100 = R207,59 This means that your total bill would be R207,59 for the month of April 2013 |
What if you do not want to work out your entire bill, just how much one particular appliance is costing you? The average per unit price of electricity in 2013 is 71,65 c/kWh. This is the per unit price we will use for our calculations.
The unit price of electricity varies with consumption. The price indicated here was as it was in 2013 for a particular usage. You can also use other rates in your calculations with your class and specify them upfront.
The actual cost per unit depends on whether or not you buy your electricity directly from Eskom or from your local municipality. In fact, the tariff differs from municipality to municipality as well.
If we want to know how much we will pay for using a particular appliance we would use the following calculation:
cost = power rating of appliance × number of hours it was used for × unit price of electricity
Remember that we can find the power rating for an appliance on the labels. Did you notice that both of the labels indicated power in watts (W) and not kilowatts (kW)? That means that if you use them to calculate the cost of electricity you must first convert them to kilowatts.
Let's try an example calculation for the microwave.
We want to work out the cost of using a small oven (1500 W) for a total of 1 hour in a day. The following steps outline what you should do. Step 1: Write down the formula cost = power rating × time × price Step 2: List all the given values in a problem power rating = 1500 W = 1,5 kW time = 1 hour price = 71,65 c/kWh Step 3: Substitute the given values into the formula to find the unknown cost = 1,5 kW × 1 hour × 71,65 c/kWh = 107,475 cents = R1,07 Step 4: Write down the solution on its own line with the units. The cost is R1,07 to run a small oven for 1 hour. |
Lets's try another example.
Have you ever notice your fridge start humming after a period of silence? Fridges are extremely energy-expensive electrical appliances. To keep the temperature at a constant cool temperature, fridges contain a thermostat that measure how cool the air is inside your fridge. When the temperature inside the fridge warms beyond a certain point the thermostat will switch on the energy-expensive compressor and condenser. Fridges are specially insulated so try keep cool air inside, and the energy demands of a fridge varies greatly depending on how often the doors are opened, and what is kept inside.
Imagine now, that you left the fridge door open by accident as you rushed to school and didn't notice until the next day! We now want to work out how much it costs to run a fridge with a power rating of 2200 W for a day.
cost = power rating × time × price power rating = 2200 W= 2,2 kW time = 24 hours price = 71,65 c/kWh cost = 2,2 kW × 24 hours × 71,65 c/kWh = 3 783,12 cents = R37,83 per day The cost is R37,83 to run a fridge for 1 day. |
INSTRUCTIONS:
Appliance | Power rating |
Microwave oven | 1 360 W |
Conventional oven | 6 000 W |
Television | 105 W |
Geyser | 4 800 W |
Incandescent light bulb | 100 W |
Fluorescent light tube | 40 W |
Vacuum cleaner | 1600 W |
Washing machine | 2200 W |
QUESTIONS:
Imagine that your family has used 320 kWh of electricity this month. Calculate the cost of the 320 kWh.
cost = 320 kWh x 71,65
= 22928 cents
= R229,28
A potato takes about 1 hour to cook in a conventional oven. In a microwave it takes approximately 12 minutes. Calculate the cost of cooking the potato in each appliance and write down which one is the cheaper option.
12 minutes = 0,2 hours
1 360 W = 1,36 kW
6 000 W = 6 kW
microwave oven: cost = 1,36 x 0,2 x 71,65 = 19,49 cents = R0,19
conventional oven: cost = 6 x 1 x 71,65 = 429,9 cents = R4,30
The microwave is cheaper to run than a conventional oven.
Which light bulb is cheaper to run for an hour, the incandescent light bulb or the fluorescent light bulb? Justify your answer with a calculation.
100 W = 0,1 kW
40 W = 0,04 kW
incandescent light bulb = 0,1 x 1 x 71,65 = 7,165 cents = R0,07
fluorescent light bulb = 0,04 x 1 x 71,65 = 2,866 cents = R0,03
The fluorescent light bulb is cheaper.
If you have a prepaid electricity voucher for R15, how long could you watch TV?
R15 = 1 500 cents
105 W = 0,105 kW
Number of kWh = 1500/71,65 = 20,94 kWh
number of hours = 20,94/0,105 = 199,43 hours
You need to vacuum your room and it takes you 30 minutes to do this. What does this cost?
1600 W = 1,6 kW
30 minutes = 0.5 hours
cost = 1,6 x 0.5 x 71,65
= 57,32 cents
= R0,57
It takes the geyser to two and a half hours to heat water from 20 °C to 65 °C. How much does it cost to heat the water?
4800 W = 4,8 kW
cost = 4,8 x 2,5 x 71,65
= 537,375 cents
= R5,37
What alternative appliance could a family use to heat water which would not demand such a high use of electricity?
A family can install a solar water heater to use to heat water instead of a geyser. This uses solar power and therefore reduces the family's energy consumption, saving electricity and money.
We can see that different appliances have different power ratings and so require more electricity to run. This means that some appliances are more expensive to run than others. An incandescent light bulb, for example, is more expensive to use than a fluorescent light bulb. If you remember, an incandescent light bulb loses most of its energy as heat, instead of light.
Do you know how much electricity your family consumes? Let's find out.
INSTRUCTIONS:
Use the following space for your table.
The length of the table will depend on the number of appliances in each learner's home. Learners answers will vary. They should perform a separate calculation for each appliance in their home and then add the amounts together to get the total cost of electricity.
Here is an example of what a calculation might look like for a light bulb of 120 W which runs for 2 hours per day. You should encourage learners to follow these steps:
Step 1: Write down the formula
cost = power rating x time x price
Step 2: List all the given values in a problem
power rating = 120 W = 0,12 kW
time = 2 hours
price = 71,65 c/kWh
Step 3: Substitute the given values into the formula to find the unknown
cost = 0,12 x 2 x 71,65
= 17,196 cents
Step 4: Write down the solution on its own line with the units.
Cost of electricity is 17 cents (R0,17)
QUESTIONS:
All of these answers will depend of the electrical appliances used in the learner's home. It is important to check that they are using the formula correctly and that they are converting watts to kilowatts and cents into rands.
What was your total estimated electricity cost for one day?
Learner-dependent answer.
If there are 30 days in an average month, what would your estimated monthly bill be?
Learner-dependent answer.
Can you think of any ways that your family could reduce the amount of electricity you use?
This answer is learner-dependant. Some learners might only have a few electrical appliances and would not be in a position to reduce usage. Some learners may indicate that they could be more efficient in their use of electricity. They could switch off unnecessary lighting. Use blankets and other forms of insulation rather than using electrical heaters in winter etc. You may find that some learners are already using geyser blankets and solar panels. Those learners may indicate that they are already doing what they can to reduce electricity use.
There are many different ways in which we can try to conserve electricity in order to save money. Our choice of light bulbs can affect our electricity bills. LED light bulbs are the most energy efficient. They have lower power ratings than normal light bulbs but most of their energy is transferred as light and very little as heat (unlike incandescent light bulbs). Light transmitted from the bulb is measured in lumens.
Some more tips on conserving electricity http://www.eskom.co.za/AboutElectricity/ElectricityTips/Pages/Conserving_Electricity.aspx
More information comparing these three types of lights is available here: http://www.designrecycleinc.com/led%20comp%20chart.html
INSTRUCTIONS:
Read the information in the table and use it to answer the questions.
LED | Compact Fluorescent Light bulbs (CFLs) | Incandescent light bulbs | |
Example | |||
Average life span (in hours) | 50 000 | 8 000 | 1 200 |
Watts | 8 | 15 | 60 |
Lumens | 800 | 800 | 800 |
Which of the three light bulbs will last the longest?
LED
Which of the three light bulbs has the highest power rating?
incandescent
How do the three light bulbs compare in terms of how much light they can provide?
These three light bulbs all emit the same number of lumens, so they can all provide the same amount of light.
Calculate how much it would cost to run each light bulb in a house for 5 hours a day for an entire year (365 days). Use 71,65 cents/kWh.
LED: cost = 0,008 x 5 x 365 x 71,65 = 1046,09 cents = R10,46
Fluorescent: cost = 0,015 x 5 x 365 x 71,65 = 1962,42 cents = R19,61
Incandescent: cost = 0,06 5 x 365 x 71,65 = 784,68 cents = R78,46
Which light bulb would you choose to use? Explain your choice.
The answer is learner-dependant. Learners should recognise that the LEDs are cheaper to run from day-to-day, although their initial cost is relatively expensive.
Which bulb is the best for the environment? Explain your choice.
The LED light uses less electricity for the same light output. This means that less electricity is consumed. If everyone is using LED lights then the overall demand for electricity would be lower and so if less electricity is produced there would be less pollution.
LEDs and incandescent light bulbs also do not contain mercury, which is poisonous, whilst fluorescent light bulbs do.
Using light bulbs which use less electricity can have a knock-on effect. If everyone is using less electricity, then there is less demand for electricity to be produced in our coal-powered power stations in South Africa. If less electricity is produced then fewer fossil fuels are burnt and this would lead to a reduction in the production of excess greenhouse gases.
Learn more about the Joule, an electric car made right here in South Africa.
There are many important and rewarding careers in the electrical energy sector. Let's take some time to research some of them.
This is an opportunity for the learners to be made aware of the many different career paths available in the electrical energy sector. There is a short list of suggested careers to research, but encourage the learners to find more. Have the learners work in small groups to do their research and then have then report their findings back to the class.
INSTRUCTIONS:
QUESTIONS:
Write down the career path that you found the most interesting.
Learner-dependent answer.
What did you like about that particular career path?
Learner-dependent answer.
The Zooniverse website provides a great overview of the various citizen science projects that learners can get involved in. There is a huge variety of projects, from helping to identify possible planets around stars, analysing real life cancer data, looking at tropical cyclone data, or listening to the calls from whales or bats. And there are also many others. Citizen science is scientific research which is conducted in whole or in part by nonprofessional scientists, specifically the general public. Encouraging learners to get involved in some of these projects will open their eyes to the possibilities out there, and also add meaning and value to what they learn within the Natural Sciences classroom. https://www.zooniverse.org/
'Citizen science' is when the general public takes part in and conducts scientific research.
Want to take part in some real science research? Check out these citizen science projects to get involved easily. https://www.zooniverse.org/
Concept map
Use the following page to design your own concept map to summarise this chapter on the cost of electrical energy.
Teacher's version
What is the power rating on the following two appliances? [2 marks]
A frying pan.
A fan.
1400 W
120 W
Refer to the table of power ratings for common appliances. List the appliances in sequence from those that use the least power to those that use the most. [2 marks]
Appliance | Power rating |
Stove | 3600 W |
Microwave | 1200 W |
Washing machine | 2200 W |
Kettle | 2200 W |
Fridge | 230 W |
Toaster | 750 W |
Energy saver globe | 40 W |
Incandescent light bulb | 120 W |
Vacuum cleaner | 1600 W |
Energy saver globe, incandescent light bulb, fridge, toaster, microwave, vacuum cleaner, washing machine, kettle, stove.
What is electrical power? Explain in your own words. [2 marks]
Learners should explain that it is the amount of electrical energy transferred per second.
Explain what is meant by 1 watt of power. [2 marks]
1 watt of power is equal to 1 joule of energy supplied in 1 second.
What does it mean that a stove has a power rating of 3600 W and a microwave has a power rating of 1200 W? Compare these two appliances in terms of the energy supplied. [3 marks]
This means that the stove uses more power than the microwave as the stove uses 3600 joules of energy per second, whereas a microwave uses 1200 joules of energy per second.
Joules (J) | Kilojoules (kJ) |
145 | |
134 | |
1 650 | |
32,12 | |
Watts (W) | Kilowatts (kW) |
1 850 | |
3,79 | |
32 | |
0,485 |
Joules (J) | Kilojoules (kJ) |
145 | 0,145 |
134 000 | 134 |
1 650 | 1,65 |
32 120 | 32,12 |
Watts (W) | Kilowatts (kW) |
1 850 | 1,85 |
3 790 | 3,79 |
32 | 0,032 |
485 | 0,485 |
An electric iron is rated at 1500 W. If the iron is used for 3 hours every day, find the number of units of electrical energy it consumes in the month of February. [3 marks]
power rating = 1500 / 1000 = 1,5 kW
time = 3 x 28 = 84 hours
energy consumption = 1,5 x 84 = 126 kWh
An electric kettle is rated at 1000 W. If it is used for 1 hour every day, find the number of units of electrical energy it consumes for the month of August. [3 marks]
power rating = 1000/1000 = 1 kW
time = 1 x 31 = 31 hours
energy consumption = 1 x 31 = 31 kWh
Your house's electricity meter in March was 3456 and in April it was 4566.
How much electrical energy did your household use in that one month period? [2 marks]
If electrical energy is charged at 71,65 c/kWh, what will your bill be for that one month period? [3 marks]
Energy consumed = 4566 - 3456 = 1110 kWh
energy = 1110 kWh
price = 71,65 c/kWh
Cost = 1110 x 71,65 = 79531,5 cents = R795,32
A 120 W electric blanket is left on for 8 hours
How many kilowatt hours of electrical energy is used by the blanket? [3 marks]
If the unit cost of electricity is 71,65 cents, what is the cost, in rands, of using the electric blanket for 8 hours? [3 marks]
power rating = 120 W = 0,12 kW
time = 8 hours
energy = power x time = 0,12 x 8 = 0,96 kWh
Cost = 0.96 x 71,65 = 68,784 cents = R0,69
A 2600 W kettle in a school staffroom is used 8 times a day for five minutes each time.
What is the total time that the kettle is switched on during each 5 day school week? [2 marks]
How much energy is consumed to run the kettle for this period (in kilowatt hours)? [2 marks]
If the cost of a unit is 71,65 cents, what is the cost of running the kettle for this period? [3 marks]
time = 8 x 5 x 5 = 200 minutes = 3,33 hours per 5 day school week.
power rating = 2 600 W = 2,6 kW
time = 3,33 hours
energy = 2,6 x 3,33 = 8,66 kWh
cost = 8,66 x 71,65 = 620,35 cents = R6,20
If you had a prepaid electricity voucher with a value of R35, calculate the following.
How long could you run a 230 W fridge for if electricity costs 71,65 cents per kWh. [5 marks]
How long could you run six 60 W incandescent light bulbs? [5 marks]
cost = R35 = 3500 cents
power rating = 230 W = 0,23 kW
price of electricity = 71,65 c/kWh
Number of available units (kWh) on voucher = 3500/71,65 = 48,85 kWh
time = energy/power = 48,85 kWh / 0,23 kW = 212 hours
Number of available units (kWh) on voucher = 3500/71,65 = 48,85 kWh
power rating for one bulb = 60 W = 0,06 kW
power rating for six bulbs = 0,06 x 6 = 0,36 kW
time = energy/power = 48,85 / 0,36 = 135,69 hours
Which light bulb, 15 W CFL or an 8 W LED, would you choose to use? Explain your answer. [3 marks]
Both light bulbs produce the same amount of light but because the power rating of the LED is lower, the cost of the electricity to run the light is less. The LED is cheaper to use. Learners may also mention that CFLs contain mercury while LEDs do not.
A tumble dryer has a power rating of 4 500 W. How long did it take to dry a load of wet washing if electricity costs 71,65 cents per kWh and the cost of running the dryer was R4,84? [6 marks]
cost = R4,84 = 484 cents
power rating = 4 500 W = 4,5 kW
cost = energy consumed x price per unit
energy consumed = cost/price per unit = 484/71,65 = 6,755 kWh
time = energy/power = 6,755/4,5 = 1,5 hours
Total [62 marks]