6.2 Metric System & Unit Conversions

6.2 Metric System & Unit Conversions

Quick Summary: Metric measurement fluency is tested extensively on the non-calculator and calculator-permitted sub-sections of the Ontario MPT. This section explains the base-10 structure of the International System of Units (SI), details conversion mechanics for 1D linear lengths, 2D surface areas, and 3D volumetric space, and clarifies the physical equivalences linking cubic volume ($ ext{cm}^3$), liquid capacity ($ ext{mL}$, $ ext{L}$), and water mass ($ ext{g}$, $ ext{kg}$). Dimensional analysis techniques ensure error-free problem solving.

---

The International System of Units (SI) Framework

The metric system is a base-10 positional system. All units for a given physical quantity are built by attaching standardized prefixes to a core base unit:

• Length: Metre ($ ext{m}$)
• Mass: Gram ($ ext{g}$) or Kilogram ($ ext{kg}$ as SI base)
• Capacity / Volume: Litre ($ ext{L}$) or Cubic Metre ($ ext{m}^3$)

Metric Prefixes Hierarchy

Mnemonic: "King Henry Died By Drinking Chocolate Milk" helps candidates instantly recall the prefix order from largest ($ ext{kilo-}$) to smallest ($ ext{milli-}$).

---

Linear Unit Conversions (1D)

Linear measurement measures distance along a single dimension. Converting between linear metric units involves shifting the decimal point by multiplying or dividing by powers of $10$.

$\text{km} \xrightarrow{\times 1000} \text{m} \xrightarrow{\times 100} \text{cm} \xrightarrow{\times 10} \text{mm}$ $\text{mm} \xrightarrow{\div 10} \text{cm} \xrightarrow{\div 100} \text{m} \xrightarrow{\div 1000} \text{km}$

Key Linear Equivalencies:

• $1 \text{ kilometre (km)} = 1,000 \text{ metres (m)}$
• $1 \text{ metre (m)} = 100 \text{ centimetres (cm)} = 1,000 \text{ millimetres (mm)}$
• $1 \text{ centimetre (cm)} = 10 \text{ millimetres (mm)}$

---

Area Unit Conversions (2D)

Area measures surface coverage across two perpendicular dimensions ($length \times width$). When converting area units, the linear scale factor must be squared!

Deriving Area Scale Factors:

• Since $1 \text{ m} = 100 \text{ cm}$, a $1 \text{ m} \times 1 \text{ m}$ square has an area of: $1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm} = 10,000 \text{ cm}^2 = 10^4 \text{ cm}^2$
• Since $1 \text{ cm} = 10 \text{ mm}$, a $1 \text{ cm} \times 1 \text{ cm}$ square has an area of: $1 \text{ cm}^2 = 10 \text{ mm} \times 10 \text{ mm} = 100 \text{ mm}^2 = 10^2 \text{ mm}^2$
• Since $1 \text{ km} = 1,000 \text{ m}$: $1 \text{ km}^2 = 1,000 \text{ m} \times 1,000 \text{ m} = 1,000,000 \text{ m}^2 = 10^6 \text{ m}^2$

Special Land Area Unit: The Hectare (ha)

In Ontario land measurement and EQAO curriculum standards, large land areas are expressed in hectares (symbol: $\text{ha}$): $1 \text{ hectare (ha)} = 100 \text{ m} \times 100 \text{ m} = 10,000 \text{ m}^2$ $1 \text{ square kilometre (km}^2\text{)} = 100 \text{ hectares (ha)}$

---

Volume Unit Conversions (3D)

Volume measures space enclosed across three dimensions ($length \times width \times height$). When converting 3D volume units, the linear scale factor must be cubed!

Deriving Volume Scale Factors:

• Since $1 \text{ m} = 100 \text{ cm}$: $1 \text{ m}^3 = 100 \text{ cm} \times 100 \text{ cm} \times 100 \text{ cm} = 1,000,000 \text{ cm}^3 = 10^6 \text{ cm}^3$
• Since $1 \text{ cm} = 10 \text{ mm}$: $1 \text{ cm}^3 = 10 \text{ mm} \times 10 \text{ mm} \times 10 \text{ mm} = 1,000 \text{ mm}^3 = 10^3 \text{ mm}^3$

Common Trap: A common candidate error on the MPT is asserting that $1 \text{ m}^2 = 100 \text{ cm}^2$ or $1 \text{ m}^3 = 100 \text{ cm}^3$. Remember: area requires squaring the factor ($100^2 = 10,000$) and volume requires cubing the factor ($100^3 = 1,000,000$).

---

Connecting Volume, Capacity, and Mass

One of the most elegant features of the SI metric system is the direct physical bridge connecting geometric volume (cubic space), liquid capacity (fluid measure), and mass (for pure water at standard temperature and pressure).

graph LR
    VOL["Cubic Volume<br/>1 cm³"] <==> CAP["Liquid Capacity<br/>1 mL"] <==> MASS["Water Mass<br/>1 g"]
    VOL2["Cubic Volume<br/>1,000 cm³ (1 dm³)"] <==> CAP2["Liquid Capacity<br/>1 L (1,000 mL)"] <==> MASS2["Water Mass<br/>1 kg (1,000 g)"]
    VOL3["Cubic Volume<br/>1 m³ (1,000,000 cm³)"] <==> CAP3["Liquid Capacity<br/>1,000 L (1 kL)"] <==> MASS3["Water Mass<br/>1 Metric Tonne (1,000 kg)"]

The Core Metric Equivalencies:

1. Small-Scale Bridge: $1 \text{ cm}^3 = 1 \text{ mL} \quad \Longleftrightarrow \quad 1 \text{ mL of pure water} = 1 \text{ gram (g)}$
2. Medium-Scale Bridge: $1,000 \text{ cm}^3 = 1 \text{ dm}^3 = 1 \text{ Litre (L)} \quad \Longleftrightarrow \quad 1 \text{ L of pure water} = 1 \text{ kilogram (kg)}$
3. Large-Scale Bridge: $1 \text{ m}^3 = 1,000 \text{ Litres (L)} \quad \Longleftrightarrow \quad 1 \text{ m}^3 \text{ of pure water} = 1 \text{ metric tonne (1,000 kg)}$

---

Dimensional Analysis (Unit Factor Method)

To perform multi-step conversions reliably, use dimensional analysis. Set up conversion factors as fractions equal to 1, arranging units so that unwanted units cancel out diagonally.

$\text{Target Quantity} = \text{Given Quantity} \times \left( \frac{\text{Desired Unit}}{\text{Given Unit}} \right)$

---

Multi-Step Worked Conversion Problem

Problem Statement: A rectangular municipal water storage reservoir has the following internal dimensions:

• Length = $15 \text{ metres}$
• Width = $8 \text{ metres}$
• Depth = $2.5 \text{ metres}$

Calculate:

1. The volume of the reservoir in cubic metres ($\text{m}^3$) and cubic centimetres ($\text{cm}^3$).
2. The total liquid capacity of the reservoir when fully filled, expressed in Litres ($\text{L}$).
3. The total mass of the water in kilograms ($\text{kg}$) and metric tonnes ($\text{t}$).
4. If water is pumped out at a rate of $500 \text{ millilitres per second}$ ($\text{mL/s}$), how many hours will it take to completely empty the reservoir?

Step-by-Step Solution:

1. Step 1: Calculate Volume in $\text{m}^3$ and $\text{cm}^3$ $\text{Volume} = l \times w \times h = 15 \text{ m} \times 8 \text{ m} \times 2.5 \text{ m} = 300 \text{ m}^3$ To convert $\text{m}^3$ to $\text{cm}^3$, multiply by $1,000,000$ ($100^3$): $\text{Volume} = 300 \times 1,000,000 \text{ cm}^3 = 300,000,000 \text{ cm}^3$

2. Step 2: Calculate Liquid Capacity in Litres Using the equivalence $1 \text{ m}^3 = 1,000 \text{ L}$: $\text{Capacity} = 300 \text{ m}^3 \times 1,000 \text{ L/m}^3 = 300,000 \text{ Litres (L)}$

3. Step 3: Calculate Water Mass Since $1 \text{ L}$ of water has a mass of $1 \text{ kg}$: $\text{Mass in kg} = 300,000 \text{ kg}$ Since $1 \text{ metric tonne} = 1,000 \text{ kg}$: $\text{Mass in tonnes} = \frac{300,000 \text{ kg}}{1,000 \text{ kg/t}} = 300 \text{ metric tonnes (t)}$

4. Step 4: Calculate Pumping Time in Hours First convert capacity to millilitres: $300,000 \text{ L} = 300,000,000 \text{ mL}$. Pumping rate = $500 \text{ mL/s}$. $\text{Time in seconds} = \frac{300,000,000 \text{ mL}}{500 \text{ mL/s}} = 600,000 \text{ seconds}$ Convert seconds to hours using dimensional analysis: $\text{Time in hours} = 600,000 \text{ s} \times \frac{1 \text{ min}}{60 \text{ s}} \times \frac{1 \text{ hr}}{60 \text{ min}} = \frac{600,000}{3,600} \approx 166.67 \text{ hours}$

Final Answer Summary: Volume = $300 \text{ m}^3$ ($300,000,000 \text{ cm}^3$), Capacity = $300,000 \text{ L}$, Water Mass = $300,000 \text{ kg}$ ($300 \text{ t}$), Pumping time = $166.67 \text{ hours}$.