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Measurement, STD2 M7 2025 HSC 21

A house has a reverse-cycle air conditioner which uses 2.5 kW of power for cooling and 3.2 kW of power for heating. The cost of electricity is 29 cents per kWh .

  1. Find the cost, in dollars and cents, of cooling the house for 6 hours.   (1 mark)

  2. The cost of operating the air conditioner to heat the house during winter last year was $640. There are 92 days in winter.
  3. Find the number of hours, to 1 decimal place, that the air conditioner was used on average per day.   (2 marks)

Show Answers Only

a.    \(\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35\)

b.    \(7.5\ \text{hours}\)

Show Worked Solution

a.    \(\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35\)
 

b.    \(\text{Let \(h\) = hours used per day}\)

\(\text{Cost per day}\ = h \times 3.2 \times 0.29\)

\(\text{Cost (92 days )}\ = 92 \times h \times 3.2 \times 0.29\)

\(\text{Find \(h\) when cost = \$640:}\)

\(640\) \(=92 \times h \times 3.2 \times 0.29\)  
\(h\) \(=\dfrac{640}{92 \times 3.2 \times 0.29}=7.5\ \text{hours (1 d.p.)}\)  

Measurement, STD2 M7 2025 HSC 10 MC

An electricity company charges customers 37 cents per kWh for electricity used, \( U\), and pays customers 5 cents per kWh for electricity produced, \(P\).

The electricity company also charges customers a fee of 71 cents per day.

Which formula should be used to calculate a customer's daily cost of electricity, \(C\), in cents?

  1. \( C=71+37 U-5 P \)
  2. \( C=71+37 U+5 P \)
  3. \( C=71-37 U-5 P \)
  4. \( C=71-37 U+5 P \)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Customer pays:}\ 71 + 37 \times U\)

\(\text{Customer receives:}\ 5 \times P\)

\(\text{Daily cost}\ = 71 + 37 \times U-5 \times P\)

\(\Rightarrow A\)

v1 Measurement, STD2 M7 2021 HSC 27

The price and the power consumption of two different models of air purifiers are shown.

\[ \begin{array}{|l|l|} \hline \text{Air Purifier X} & \text{Air Purifier Y} \\ \hline \text{Price: \$480} & \text{Price: \$462.40} \\ \hline \text{Power: 95 W} & \text{Power: 88 W} \\ \hline \end{array} \]

The average cost for electricity is 30c/kWh. A household runs an air purifier for an average of 10 hours a day.

  1. The annual cost of electricity for Air Purifier X for this household is \$104.03.
  2. For this household, what is the difference in the annual cost of electricity between Air Purifier X and Air Purifier Y? (2 marks)
  3. For this household, how many years will it take for the total cost of buying and using Air Purifier X to be equal to the cost of buying and using Air Purifier Y? (2 marks)
Show Answers Only
  1. `$7.66`
  2. `2.3\ \text{years}`
Show Worked Solution
a. `text{Annual power usage (Y)}` `= 88 \times 10 \times 365 = 321200\ \text{Wh}`
    `= 321.2\ \text{kWh}`
  `text{Annual cost (Y)}` `= 321.2 \times 0.30 = \$96.36`
  `text{Annual cost (X)}` `= 95 \times 10 \times 365 / 1000 \times 0.30 = \$104.03`
  `text{Difference}` `= 104.03-96.36 = \$7.66`
b. `text{Price difference}` `= 480-462.40 = \$17.60`
  `text{Years to equal total cost}` `= 17.60 / 7.66 ≈ 2.3\ \text{years}`

v1 Measurement, STD2 M7 2018 HSC 28c

A 900-watt air purifier runs for 2 hours per day at 60% power. Electricity costs $0.30 per kWh.

What is the total cost of running the air purifier for 150 days? (3 marks)

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`$48.60`

Show Worked Solution
`text(Daily usage)` `= 900 xx 2 xx 60text(%)`
  `= 1080\ \text{watts}`

 

`text(Total usage over 150 days)` `= 150 xx 1080`
  `=162 \ 000\ \text{watts}`
  `= 162\ \text{kWh}`

 

`∴ \  text(Cost)` `= 162 xx 0.30`
  `= $48.60`

Measurement, STD2 M7 2024 HSC 17

The cost of electricity is 30.13 cents per kWh .

Calculate the cost of using a 650 W air conditioner for 6 hours.   (2 marks)

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\(\text{Cost} =\$ 1.18\)

Show Worked Solution

\(\text{Usage}=6 \times 650=3900\, \text{W}=3.9\, \text{kW}\)

\(\text{Cost} =3.9 \times 30.13=117.507 \,\text{c}=\$ 1.18 \, \text {(nearest cent)}\)

Measurement, STD2 M7 2021 HSC 27

The price and the power consumption of two different brands of television are shown.

The average cost for electricity is 25c/kWh. A particular family watches an average of 3 hours of television per day.

  1. The annual cost of electricity for Television A for this family is $48.18.
  2. For this family, what is the difference in the annual cost of electricity between Television A and Television B?  (2 marks)

  3. For this family, how many years will it take for the total cost of buying and using Television A to be equal to the cost of buying and using Television B?  (2 marks)

Show Answers Only
  1. `$4.38`
  2. `5\ text(years)`
Show Worked Solution
a.   `text{Annual power usage (B)}` `= 160 xx 3 xx 365`
    `=175\ 200`
    `=175.2\ text(kWh)`

♦ Mean mark part (a) 49%.
  `text{Annual cost (B)}` `= 175.2 xx 0.25`
    `=$43.80`

 

  `text{Difference in cost}` `= 48.18 – 43.80`
    `=$4.38`

♦♦ Mean mark part (b) 23%.
b.   `text{Difference in price}` `= 921.90 – 900`
    `=$21.90`

 

  `text(Years to even out cost)` `=21.90/4.38`
    `=5\ text{years}`

Measurement, STD2 M7 SM-Bank 12

Francis is buying a fridge. The energy usage of his two choices are below:

Fridge 1 (5 star rating): 227 kWh per year

Fridge 2 (2 star rating): 492 kWh per year

Each fridge has an expected life of 8 years.

If electricity costs 32.4 cents per kWh, how much will Francis save in electricity over the expected life of the more energy efficient fridge, to the nearest dollar?  (2 marks)

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`$687`

Show Worked Solution

`text(Energy cost of fridge 1)`

`= 227 xx 0.324 xx 8`

`= $588.38`

`text(Energy cost of fridge 2)`

`= 492 xx 0.324 xx 8`

`= $1275.26`

`:.\ text(Energy saving)` `= 1275.26 – 588.38`
  `= 686.88`
  `= $687`

Measurement, STD2 M7 2018 HSC 28c

Every day, a 1200-watt microwave oven is used for 45 minutes at 40% power. Electricity is charged at $0.25 per kWh.

What is the cost of running this microwave oven for 180 days?  (3 marks)

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`$16.20`

Show Worked Solution
`text(Daily usage)` `= 1200 xx 45/60 xx 40text(%)`
  `= 360\ text(watts)`

 

`text(180 day usage)` `= 180 xx 360`
  `= 64\ 800\ text(watts)`
  `= 64.8\ text(kW)`

 

`:.\ text(C)text(ost over 180 days)` `= 64.8 xx 0.25`
  `= $16.20`

Measurement, STD2 M7 SM-Bank 1

Bikram runs a hot yoga studio.

If it costs 34 cents for 1-kilowatt (1000 watts) for 1 hour, how much does it cost him to run three 3200-watt heaters from 9:00 am to 12:30 pm on a single day? (Give your answer to the nearest cent)  (2 marks)

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`$11.42`

Show Worked Solution

`text(Total energy usage)`

`= 3 xx 3200 xx 3.5\ text(hours)`

`= 33\ 600\ text(watts)`

 

`:.\ text(C)text(ost)` `= (33\ 600)/1000 xx 0.34`
  `= 11.424`
  `= $11.42\ \ \ text{(nearest cent)}`

Measurement, STD2 M1 2016 HSC 28b

The cost of buying a new heater is $990. It has an energy consumption of 505 kWh per year.

Energy is charged at the rate of $0.35 kWh.

How much will it cost in total to purchase and then run this heater for five years?  (2 marks)

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`$1873.75`

Show Worked Solution

`text(C)text(ost to run heater for 5 years)`

`= 5 xx 505 xx 0.35`

`= $883.75`
 

`:.\ text(Total purchase and running cost)`

`= 883.75 + 990`

`= $1873.75`