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Unit Conversions and Dimensional Analysis

Dimensional analysis converts units by multiplying by conversion factors, fractions equal to 1, arranged so unwanted units cancel.

Practice 6: Mathematical Routines

Part 1 · Hook

Why this matters

An electric bill is in kilowatt-hours, a furnace is rated in BTU, a food label lists Calories, and a gas pump measures gallons. The exam mixes these on purpose. Dimensional analysis is one method that handles all of them: multiply by fractions equal to 1 until the units you do not want cancel.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What does "per capita" mean?

  1. Per person
  2. Per year
  3. Per square kilometer
  4. Per city
Show the answer

Per capita divides by the number of people.

  • Correct: Per person:
  • Per year:
  • Per square kilometer:
  • Per city:

2. Which is 7.5×1087.5 \times 10^{8} written out?

  1. 750,000,000
  2. 75,000,000
  3. 7,500,000,000
  4. 750,000
Show the answer

Move the point 8 places right.

  • Correct: 750,000,000:
  • 75,000,000:
  • 7,500,000,000:
  • 750,000:

Part 4 · See it

See it first

A pump runs 6 hours a day for 30 days and draws 2 kilowatts. The chain is 30 days over 1, times 6 hours over 1 day, times 2 kilowatts. Days cancel, and hours times kilowatts gives kilowatt-hours: 360 kWh. Then 360 kWh times $0.15 per kWh, with kWh cancelling, is $54.
A dimensional analysis chain: units cancel like numbers. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Write the starting amount with its unityou know what has to cancel
  2. Multiply by a conversion factor with that unit on the bottomthe old unit cancels and the new unit is left
  3. Repeat until only the answer’s unit is leftthe number is right because each factor equals 1

Part 6 · Key ideas

Key ideas

  • A conversion factor is a fraction equal to 1, such as 1 kWh / 3,400 BTU or 3.79 L / 1 gal.
  • Dimensional analysis: line up factors so each unwanted unit appears once on top and once on the bottom.
  • A watt is power (energy per second); a kilowatt-hour is energy. Energy = power × time.
  • Energy units: joule, kWh, BTU, kilocalorie. 1 kWh = 3.6 million J, about 3,412 BTU.

Part 7 · Misconception

A common mistake

The wrong idea: Kilowatts and kilowatt-hours are the same thing.

What actually happens: Kilowatts measure how fast energy is used (power). Kilowatt-hours measure how much energy was used: a 2-kW pump running 3 hours uses 6 kWh.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. A water pump draws 2 kW and runs 6 hours a day for 30 days. Write the setup, with units, for the energy it uses, then give the answer in kWh.

Type the setup as on the exam, with brackets and units, then the answer.

Show the answer

30days×6h1day×2kW30\,\text{days} \times \frac{6\,\text{h}}{1\,\text{day}} \times 2\,\text{kW}=360= 360 kWh. Days cancel, and kW × h is kWh.

  • Setup: 30days6hday⋅2kW30\,\text{days}\,\frac{6\,\text{h}}{\text{day}} \cdot 2\,\text{kW}; answer 360 kWh

2. Electricity costs $0.15 per kWh. How much does 360 kWh cost?

Type a number, then its unit (dollars). Scale words (2.5 million) and powers of ten (2.5 × 10^6) are read.

Show the answer

360kWh×$0.151kWh360\,\text{kWh} \times \frac{$0.15}{1\,\text{kWh}}=$54= $54.

  • Answer: 54 dollars

3. Which is a unit of power (a rate of energy use) rather than a unit of energy?

  1. Kilowatt (kW)
  2. Kilowatt-hour (kWh)
  3. British thermal unit (BTU)
  4. Joule (J)
Show the answer

A watt is one joule per second, a rate. kWh, BTU and joules are amounts of energy.

  • Correct: Kilowatt (kW): Right: kW is a rate.
  • Kilowatt-hour (kWh): A kWh is energy: a kilowatt used for an hour.
  • British thermal unit (BTU): A BTU is an amount of heat energy.
  • Joule (J): A joule is the SI unit of energy.

Data

Appliances in one home

Power rating and daily use of three appliances in one home (invented data for practice). Electricity costs $0.15 per kWh.
AppliancePower (W)Hours used per day
Space heater1,5004
Refrigerator15024
LED lights (all together)605

4. Write the setup, with units, for the energy the space heater uses in one day, in kWh. Then give the answer.

Type the setup as on the exam, with brackets and units, then the answer.

Show the answer

1,500W×4h×1kW1,000W1{,}500\,\text{W} \times 4\,\text{h} \times \frac{1\,\text{kW}}{1{,}000\,\text{W}}=6= 6 kWh.

  • Setup: 1500W⋅4h⋅1kW1000W\frac{1500\,\text{W} \cdot 4\,\text{h} \cdot 1\,\text{kW}}{1000\,\text{W}}; answer 6 kWh

5. Which appliance uses the most energy in a day?

  1. Space heater
  2. Refrigerator
  3. LED lights
  4. The heater and refrigerator tie
Show the answer

Per day: heater 1,500×4=6,0001{,}500 \times 4 = 6{,}000 Wh; refrigerator 150×24=3,600150 \times 24 = 3{,}600 Wh; lights 60×5=30060 \times 5 = 300 Wh.

  • Correct: Space heater: Right: 6 kWh a day.
  • Refrigerator: The refrigerator runs longest but uses 3.6 kWh a day.
  • LED lights: The lights use 0.3 kWh a day.
  • The heater and refrigerator tie: 6 kWh and 3.6 kWh are not equal.

6. At $0.15 per kWh, how much does it cost to run the refrigerator for 30 days?

Type a number, then its unit (dollars). Scale words (2.5 million) and powers of ten (2.5 × 10^6) are read.

Show the answer

150W×24h=3,600150\,\text{W} \times 24\,\text{h} = 3{,}600 Wh = 3.6 kWh a day; 3.6×30=1083.6 \times 30 = 108 kWh; 108×$0.15=$16.20108 \times $0.15 = $16.20.

  • Answer: 16.2 dollars

7. The family replaces the space heater with one rated 1,000 W, used the same 4 hours a day. How much energy does that save each day?

  1. 2 kWh
  2. 0.5 kWh
  3. 4 kWh
  4. 500 kWh
Show the answer

Old: 6 kWh a day. New: 1,000W×4h=41{,}000\,\text{W} \times 4\,\text{h} = 4 kWh. Saved: 6−4=26 - 4 = 2 kWh a day.

  • Correct: 2 kWh: Right: 6 - 4 = 2 kWh.
  • 0.5 kWh: 0.5 kWh is the power difference (500 W) for one hour.
  • 4 kWh: 4 kWh is what the new heater uses, not the saving.
  • 500 kWh: 500 is the difference in watts, not kWh.

Part 9 · Summary

Summary

Dimensional analysis converts units by multiplying by conversion factors, fractions equal to 1, arranged so unwanted units cancel. Write every factor with units in the setup, then the answer with its unit. Power (watts, kW) is a rate; energy (kWh, J, BTU, kcal) is an amount, and energy = power × time. Cost and emissions follow the same pattern: kWh × price per kWh, kWh × kg CO2 per kWh.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

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