Chapter 3: Matter and Energy – Vocabulary Flashcards

Classification of Matter (Section 1)

Matter: anything that has mass and occupies space.
Classified by composition:
- Pure Substance
- Mixture
Pure Substances
- Have a fixed or definite composition; the constituent makeup is the same every time.
- Elements: simplest type of pure substance; made up of atoms; each element is made up of one type of atom.
- Compounds: contain atoms from two or more elements; combination is the same every time; chemical formula represents the composition (e.g.,
- Water = \mathrm{H_2O} (2 hydrogen atoms + 1 oxygen atom)
- Salt = \mathrm{NaCl}
- Can be separated into the individual elements only by chemical reaction; cannot be separated by physical methods.
Mixtures
- Combination of two or more pure substances.
- Most everyday matter is a mixture (examples: air, steel/brass, tea/coffee/soda, ocean water).
- Mixed physically, not chemically; no new bonds formed; separation does not require breaking chemical bonds.
- Separated by physical methods (e.g., size separation, magnetic separation, filtration).
- Lab note: Filtration is a common method.
Homogeneous vs Heterogeneous Mixtures
- Homogeneous mixtures (also called solutions): uniform composition; separation not visible; examples include air, seawater, brass (copper + zinc).
- Heterogeneous mixtures: nonuniform composition; appears as two or more regions (e.g., oil and water, chocolate chip cookie).
Quick practice (identify the following as pure substance or mixture):
- a) Copper in wire → pure substance (element)
- b) A blueberry muffin → mixture
- c) Neon gas → pure substance (element)
- d) A soft drink → mixture
- e) Hydrogen peroxide (H₂O₂) → pure substance (compound)

States and Properties of Matter (Section 2)

States of matter (characteristics)
- Solid
- Shape: definite shape
- Volume: definite volume
- Particle arrangement: fixed, very close
- Interactions: very strong
- Particle movement: very slow
- Examples: Ice, salt, iron
- Liquid
- Shape: takes shape of container
- Volume: definite volume
- Particle arrangement: random, close
- Interactions: strong
- Particle movement: moderate
- Examples: Water, oil, vinegar
- Gas
- Shape: takes shape of container
- Volume: takes shape of container
- Particle arrangement: random, far apart
- Interactions: essentially none
- Particle movement: very fast
- Examples: Water vapor, helium, air
Physical properties vs Physical changes
- Physical properties: characteristics observable or measurable without changing the sample's identity (e.g., shape, color, odor, luster, size, density, melting point).
- Physical changes: changes in a physical property that do not alter the identity of the sample; no new substance is produced (e.g., shape change, size change, phase changes like melting/freezing/boiling, conductivity changes, physical state changes).
Chemical properties vs Chemical changes
- Chemical properties: describe the ability of a sample to change into a new substance (e.g., combustibility, corrosion, reactivity).
- Chemical changes: a process that produces one or more new substances, with new physical and chemical properties (e.g., wood burns, iron rusts, gas bubbles when two liquids are mixed).
Summary: Distinguishing features
- Physical properties/changes do not alter the substance’s identity (though phase changes are changes of state).
- Chemical properties/changes involve transformation into a different substance with new properties.

Temperature (Section 3)

Temperature scales
- Celsius (°C): used for most chemistry contexts; water freezes at 0°C and boils at 100°C at sea level.
- Kelvin (K): SI unit; 0 K is absolute zero; water freezes at 273 K and boils at 373 K.
- Fahrenheit (°F) used in some contexts (e.g., everyday temperature in the U.S.); water freezes at 32°F and boils at 212°F.
- No degree sign in Kelvin (e.g., 273 K).
Conversions
- Fahrenheit to Celsius: $TF = 1.8\,TC + 32$
- Celsius to Fahrenheit: $TC = \frac{TF - 32}{1.8}$
- Celsius to Kelvin: $TK = TC + 273.15$
- Kelvin to Celsius: $TC = TK - 273.15$
Notable values and references
- Water freezing: 0°C = 273 K = 32°F
- Water boiling: 100°C = 373 K = 212°F
- Absolute zero: 0 K = -273.15°C = -459.67°F
Temperature conversion table (examples shown in slides)
- 37.0°C = 98.6°F = 310 K (approx.)
- 0°C = 273 K = 32°F
- 100°C = 373 K = 212°F

Energy (Section 4)

Energy: the ability to do work; two main types
- Kinetic Energy: energy of motion; any object in motion has kinetic energy (examples: water flowing over a dam, swimming, exercising).
- Potential Energy: stored energy; determined by position or chemical bonds (examples: water at top of dam before falling; compressed spring; chemical bonds).
Energy units
- SI unit: Joules (J); larger commonly used: kilojoules (kJ)
- Calorie system (historical in nutrition): calories (cal)
- Definition: energy needed to raise the temperature of 1 g of water by 1°C
- 1 cal = 4.184 J
- 1 kcal = 4184 J = 4.184 kJ
- Food energy uses Calories (Cal) with a capital C
- 1 Cal = 1 kcal = 1000 cal = 4.184 kJ
Converting energy units (example)
- If a pat of butter provides 150 J, how many calories is that?
- Using: 1 cal = 4.184 J → calories = $\frac{150}{4.184} \approx 3.59\,\text{cal}$ ≈ 35.9 cal (2 s.f. shown as 36 cal)
Energy and nutrition connection
- Typical daily diet example: 2100 Calories ≈ 8800 kJ
- Conversion: $1\,\text{Cal} = 1\,\text{kcal} = 1000\,\text{cal} = 4.184\,\text{kJ}\,.$ $1\,\text{Cal} = 4.184\,\text{kJ} = 4184\,\text{J}$
Calorimeters
- Instrument used to measure heat transfer
- Energy values measured in kcal/g or kJ/g
- How it works: food sample combusted in a chamber with oxygen; heat released to surrounding water; temperature rise of water used to calculate energy value
- Components pictured: insulated container, steel combustion chamber, water, thermometer, ignition wires, stirrer
Caloric values of foods
- Carbohydrates: 4 kcal/g; 17 kJ/g
- Fat: 9 kcal/g; 38 kJ/g
- Protein: 4 kcal/g; 17 kJ/g
Example: energy content from food
- Whole milk: 13 g carbohydrates, 9.0 g fat, 9.0 g protein
- Energy: $13\,g\times 4\,\text{kcal/g} + 9.0\,g\times 9\,\text{kcal/g} + 9.0\,g\times 4\,\text{kcal/g} = 52 + 81 + 36 = 169\,\text{kcal}$

Energy and Nutrition (Section 5)

Nutritional energy concept
- Food provides energy for work such as muscle contraction, cellular growth/repair, and metabolism steps.
- Primary fuel: Carbohydrates; fats and proteins can serve as fuel if carbohydrate is low.
- Energy units for nutrition: Calories (Cal) with a capital C; 1 Cal = 1 kcal = 1000 cal = 4.184 kJ = 4184 J.
Calorimeters and energy values
- Calo values used to estimate energy content of foods via calorimetry.
Practice exercise references (examples given in slides)
- Energy content calculations for foods, e.g., Milk example above.

Specific Heat (Section 6)

Specific Heat (SH or c)
- A physical property that tells how well a substance absorbs heat.
- Definition: the amount of heat required to raise the temperature of exactly 1 g of a substance by 1°C.
- Higher SH means slower temperature change; lower SH means faster temperature change.
Key values (examples from slides)
- Water: SH = 4.180 J/kg·K (equivalently 4.184 J/g·K or 1 cal/g·°C)
- Sand: SH ≈ 670 J/kg·K
- Asphalt: SH ≈ 900 J/kg·K
SH table (selected substances)
- Elements: Aluminum 0.897 J/g·°C; Copper 0.385 J/g·°C; Gold 0.129 J/g·°C; Iron 0.452 J/g·°C; Silver 0.235 J/g·°C; Titanium 0.523 J/g·°C
- Compounds: Ammonia 2.04 J/g·°C; Ethanol 2.46 J/g·°C; Sodium chloride 0.864 J/g·°C; Water (liquid) 4.184 J/g·°C; Water (solid) 2.03 J/g·°C
Heat transfer equation
- The heat gained or lost by a substance can be calculated using: $q = m \times c \times \Delta T$
- Where: $q$ = heat (J), $m$ = mass (g), $c$ = specific heat (J/g·°C or J/g·K), and $\Delta T$ = change in temperature (°C or K).
- Sign convention: heat gained (positive $q$ ); heat lost (negative $q$ ).
Alternative form
- Specific heat can be calculated when heat and mass and temperature change are known: $S_H = \frac{q}{m \times \Delta T}$
Practice and examples
- Example 1: Heat transferred to a 24.8 g sample (SH = 2.6 J/g·°C) from 20.2°C to 24.5°C
- $\Delta T = 24.5 - 20.2 = 4.3\,^{\circ}C$
- $q = m c \Delta T = 24.8 \text{ g} \times 2.6 \text{ J/g·°C} \times 4.3^{\circ}\text{C} \approx 2.80\times 10^2\text{ J}$
- Example 2: A 10.0 g sample absorbs 24.75 J from 21.4°C to 26.9°C
- $\Delta T = 26.9 - 21.4 = 5.5^{\circ}C$
- SH = \frac{q}{m \Delta T} = \frac{24.75}{10.0 \times 5.5} = 0.45\text{ J g^{-1}°C^{-1}}
Practice problems (q = m × SH × ΔT)
- Example prompts: Determine the calories or kJ needed to raise a given mass by a given ΔT; or determine SH given q, m, ΔT.

Changes of State (Section 7)

What is a change of state?
- A change of matter from one state to another (solid, liquid, gas).
- Processes that require heat input (endothermic):
- Melting (solid → liquid)
- Evaporation (liquid → gas)
- Sublimation (solid → gas)
- Processes that release heat (exothermic):
- Freezing (liquid → solid)
- Condensation (gas → liquid)
- Deposition (gas → solid)
Diagram note (text description)
- The process graph shows solid, liquid, gas phases with arrows indicating transitions (Melting, Freezing, Vaporization/Evaporation, Condensation, Sublimation, Deposition).
Melting and Freezing details
- Melting: solid to liquid; occurs at the melting point, the temperature where particles have enough energy to overcome attractive forces.
- Freezing: liquid to solid; occurs at the freezing point, the temperature where particles do not have enough energy to overcome attractive forces.
- The melting and freezing points are the same for a given substance (e.g., water freezes at 0°C and melts at 0°C when heat is removed or added, respectively).
Sublimation and Deposition
- Sublimation: solid to gas (e.g., dry ice, CO₂(s), at -78°C under certain conditions).
- Deposition: gas to solid (e.g., deposition of water vapor as frost crystals).
- Note: Sublimation is used to freeze-dry foods; water can sublimate under certain conditions; freezer burn is related to sublimation/dehydration effects.
Evaporation, Condensation, and Boiling
- Evaporation: surface phenomenon where liquid molecules at the surface gain enough energy to enter the gas phase; cooling of the remaining liquid occurs.
- Boiling: occurs at the boiling point; gas bubbles form inside the liquid and rise to the surface.
- Condensation: gas → liquid; releases energy.
Heat of phase changes
- Heat of vaporization (and condensation): energy required to convert exactly 1 g of a liquid to a gas at the boiling point; for water, exactly 1 g requires $540\,\text{cal} = 2260\,\text{J}$ .
- Use these to calculate energy for evaporation/condensation: $\text{Heat} = m \times \text{heat of vaporization}$ (for vaporization) or $\text{Heat} = m \times \text{heat of condensation}$ (for condensation).
Heats of fusion, vaporization, condensation – practice
- Fusion (melting) of ice at 0°C: $q = m \times \Delta H{\text{fusion}}$ with \Delta H{\text{fusion}} = 334\,\text{J g^{-1}}
- Example: Melt 32.0 g ice at 0°C
- q = 32.0\,\text{g} \times 334\,\text{J g^{-1}} = 10,688\,\text{J} \approx 1.07\times 10^4\,\text{J} → 10.7 kJ (3 s.f.)
- Condensation energy example: 50.0 g steam at 100°C condenses
- Heat released: q = m \times 2260\,\text{J g^{-1}} = 50.0 \text{ g} \times 2260\,\text{J g^{-1}} = 1.13\times 10^5\,\text{J} = 113\,\text{kJ}
Heating and cooling curves
- Visualize changes of state as a substance is heated or cooled
- Temperature on the y-axis; heat input on the x-axis (heating curves) or heat removal (cooling curves)
- Diagonal segments: temperature changes; horizontal segments: changes of state (plateaus at phase transition temperatures)
- Example features:
- Boiling point of water at 100°C
- Melting/Freezing point at 0°C
Curve interpretation practice
- From the heating/cooling curve:
- What temperature does water condense? → 100°C (condensation occurs at the boiling point when vapor is present at equilibrium)
- What happens to liquid water at 0°C? → It freezes (freezing point at 0°C)
- At 40°C, water is a liquid
- When water freezes, heat is removed (exothermic process)

Key equations to remember

Temperature conversions:
- $TF = 1.8\,TC + 32$
- $TC = \frac{TF - 32}{1.8}$
- $TK = TC + 273.15$
- $TC = TK - 273.15$
Energy, heat, and specific heat:
- $q = m\,c\,\Delta T$
- Where $q$ is heat (J), $m$ is mass (g), $c$ is specific heat (J g^{-1} °C^{-1}), and $\Delta T$ is change in temperature (°C).
- If using specific heat $SH$ : $q = m\,SH\,\Delta T$
- Specific heat capacity ( examples ):
- Water: c{H2O(l)} = 4.184\ \text{J g^{-1} °C^{-1}} (often written as 4.184\ \text{J g^{-1} K^{-1}})
Energy units and conversions:
- $1\ \text{cal} = 4.184\ \text{J}$
- $1\ \text{kcal} = 4184\ \text{J} = 4.184\ \text{kJ}$
- Food energy: $1\ \text{Cal} = 1\ \text{kcal} = 1000\ \text{cal} = 4.184\ \text{kJ}$
Heats of phase changes (per gram):
- Fusion (ice → water) at 0°C: \Delta H_{\text{fusion}} = 334\ \text{J g^{-1}}
- Vaporization (liquid → gas) at boiling point (water): \Delta H_{\text{vap}} = 2260\ \text{J g^{-1}}
- Condensation (gas → liquid) at boiling point: same magnitude as vaporization per gram (2260 J g^{-1})

Connections and implications

Pure substances vs mixtures underpin material identity in chemistry; separation methods distinguish physical vs chemical changes.
Phase changes involve latent heat (no temperature change during the phase transition) which is captured by fusion/vaporization/condensation enthalpies.
Temperature scales and conversions are essential for laboratory measurements and data interpretation across contexts (environmental data, food energy labeling, calorimetry).
Specific heat governs how substances respond to heating or cooling, affecting calorimetry results and energy budgeting in biological and environmental systems.
Nutritional energy concepts link chemistry to real-world dietary planning; calorimetry provides a quantitative bridge between food mass and energy content.