🌡️ Temperature & Temperature Scales

Part 1 of 7 — What Is Temperature, Really?

You feel "hot" and "cold" every day — but what does temperature actually measure at the molecular level? Understanding this connection is the key to thermodynamics.

Temperature Is About Motion

At the microscopic level, all matter consists of particles (atoms, molecules) in constant random motion. Temperature is a measure of the average translational kinetic energy of these particles.

• Higher temperature → particles move faster on average
• Lower temperature → particles move slower on average

⚠️ Temperature measures average kinetic energy, not total energy. A cup of boiling water has higher temperature than a swimming pool at 25°C, but the pool has far more total thermal energy because it contains vastly more molecules.

The Three Temperature Scales

Celsius (°C)

• Based on the properties of water at 1 atm
• 0°C = freezing point of water
• 100°C = boiling point of water
• Used worldwide in science and daily life

Fahrenheit (°F)

• 32°F = freezing point of water
• 212°F = boiling point of water
• 180 Fahrenheit degrees span the same range as 100 Celsius degrees
• Used in the United States for everyday purposes

Kelvin (K)

• The SI unit of temperature — used in all physics equations
• 0 K = absolute zero (the lowest possible temperature)
• 273.15 K = freezing point of water
• 373.15 K = boiling point of water
• No degree symbol: we write "300 K," not "300°K"
• No negative values — you cannot go below absolute zero

Absolute Zero

At 0 K = −273.15°C = −459.67°F, all classical molecular motion ceases. This is the theoretical lower limit of temperature. In practice, scientists have cooled matter to within billionths of a kelvin above absolute zero, but never reached it exactly.

Conversion Formulas

Celsius ↔ Kelvin

$T_K = T_C + 273.15$ $T_C = T_K - 273.15$

A change of 1°C equals a change of 1 K. These scales differ only by an offset.

Celsius ↔ Fahrenheit

$T_F = \frac{9}{5}T_C + 32$ $T_C = \frac{5}{9}(T_F - 32)$

A change of 1°C equals a change of 9/5 = 1.8°F.

Key Reference Points

Fun Fact

−40° is the same on both Celsius and Fahrenheit scales!

$T_F = \frac{9}{5}(-40) + 32 = -72 + 32 = -40°F \checkmark$

Temperature Concepts Quiz 🧠

Temperature Conversion Drill 🌡️

Convert the following temperatures. Round to the nearest whole number.

1. 68°F = ___ °C
2. 25°C = ___ K
3. 500 K = ___ °C

Exit Quiz

⚖️ Thermal Equilibrium & the Zeroth Law

Part 2 of 7 — The Foundation of Temperature Measurement

Before we can measure temperature, we need a fundamental principle that tells us what temperature means when two objects interact. This principle is so basic that it was added to thermodynamics after the first and second laws — and numbered "zero."

The Zeroth Law of Thermodynamics

Thermal Equilibrium

When two objects are placed in thermal contact, energy (heat) flows from the hotter object to the cooler one. Eventually, the net heat flow stops and both objects reach the same temperature. They are then in thermal equilibrium.

🔑 Two objects in thermal equilibrium have the same temperature. No net heat flows between them.

The Zeroth Law

$\text{If } A \leftrightarrow C \text{ and } B \leftrightarrow C, \text{ then } A \leftrightarrow B$

Where $\leftrightarrow$ means "is in thermal equilibrium with."

In words: If object A is in thermal equilibrium with object C, and object B is also in thermal equilibrium with object C, then A and B are in thermal equilibrium with each other (they have the same temperature).

Why It Matters

This law is what makes thermometers possible! A thermometer (object C) reaches thermal equilibrium with your body (object A). Later it reaches equilibrium with a pot of water (object B). If the readings match, A and B have the same temperature — even though A and B never touched each other.

Direction of Heat Flow

Heat always flows spontaneously from higher temperature to lower temperature:

$T_{\text{hot}} \xrightarrow{\text{heat}} T_{\text{cold}}$

This continues until $T_{\text{hot}} = T_{\text{cold}}$ (thermal equilibrium). Heat never flows from cold to hot on its own — that would require external work (refrigerators, heat pumps).

How Thermometers Work

All thermometers exploit some thermometric property — a physical quantity that changes predictably with temperature:

The Constant-Volume Gas Thermometer

This is the most fundamental thermometer. It measures the pressure of a fixed amount of gas at constant volume:

$T = (273.16 \text{ K}) \frac{P}{P_{\text{triple}}}$

where $P_{\text{triple}}$ is the pressure at the triple point of water (273.16 K by definition). As the amount of gas is reduced toward zero, all gases give the same reading — this defines the ideal gas temperature scale, which is identical to the Kelvin scale.

Zeroth Law & Equilibrium Quiz 🧠

Thermal Equilibrium Reasoning 🎯

Complete each statement about thermal equilibrium and heat flow.

Exit Quiz

📏 Linear Thermal Expansion

Part 3 of 7 — Why Bridges Have Gaps

When you heat a solid, its atoms vibrate with greater amplitude around their equilibrium positions. This causes the material to expand in all directions. For long, thin objects, the most noticeable change is in length — this is linear thermal expansion.

The Linear Expansion Formula

$\Delta L = \alpha L_0 \Delta T$

where:

• $\Delta L$ = change in length (m)
• $\alpha$ = coefficient of linear expansion (1/°C or 1/K)
• $L_0$ = original length at the initial temperature (m)
• $\Delta T = T_f - T_i$ = change in temperature (°C or K)

The new length is:

$L = L_0 + \Delta L = L_0(1 + \alpha \Delta T)$

Coefficients of Linear Expansion

🔑 Invar has an extremely low $\alpha$, making it ideal for precision instruments, clock pendulums, and scientific equipment where dimensional stability matters.

Important Notes

• $\Delta T$ is the same whether in °C or K (same size degree)
• The formula is valid for moderate temperature changes
• $\alpha$ is approximately constant over typical temperature ranges
• Expansion occurs in all directions, but for thin rods or rails, the length change dominates

Engineering Applications

Expansion Joints in Bridges

Bridges have gaps (expansion joints) at regular intervals to allow the roadway to expand in summer and contract in winter without buckling or cracking.

Example: A 500 m steel bridge warms from 0°C to 40°C:

$\Delta L = (12 \times 10^{-6})(500)(40) = 0.24 \text{ m} = 24 \text{ cm}$

That's nearly a foot of expansion! Without joints, the bridge would buckle.

Railroad Tracks

Old-style railroad tracks were laid with gaps between sections to allow expansion. Modern continuously welded rail is pre-stressed so that it can handle thermal expansion without buckling (up to about 60°C above the stress-free temperature).

Bimetallic Strips

Two metals with different $\alpha$ values are bonded together. When heated, the metal with the higher $\alpha$ expands more, causing the strip to bend toward the lower-$\alpha$ side.

Applications: thermostats, circuit breakers, oven thermometers

Shrink Fitting

A metal ring is heated so its inner diameter expands. It's slipped over a shaft, then cooled. As it contracts, it grips the shaft with enormous force — no bolts or welds needed!

Linear Expansion Quiz 🧠

Linear Expansion Drill 📏

Use $\alpha_{\text{steel}} = 12 \times 10^{-6}$ /°C and $\alpha_{\text{Al}} = 23 \times 10^{-6}$ /°C.

1. A steel bridge is 200 m long at 10°C. Find $\Delta L$ in cm when it heats to 40°C.

2. An aluminum rod is 3.0 m at 20°C. Find $\Delta L$ in mm when it heats to 120°C.

3. A steel rail is 25.0 m long at 15°C. What gap (in mm) must be left between rails to prevent buckling if temperature can reach 50°C?

Round all answers to 3 significant figures.

Exit Quiz

📦 Volume & Area Thermal Expansion

Part 4 of 7 — Expansion in Two and Three Dimensions

A solid expands in all directions when heated. For flat objects (plates, sheets), we care about area expansion. For bulk objects (tanks, spheres, liquids), we care about volume expansion.

Area Expansion

For a flat surface of original area $A_0$:

$\Delta A = 2\alpha \, A_0 \, \Delta T$

$A = A_0(1 + 2\alpha \, \Delta T)$

The factor of 2 arises because a surface has two linear dimensions, and each expands by factor $\alpha$.

Volume Expansion

For a solid or liquid of original volume $V_0$:

$\Delta V = \beta \, V_0 \, \Delta T$

$V = V_0(1 + \beta \, \Delta T)$

where $\beta$ is the coefficient of volume expansion.

The Key Relationship

For isotropic solids:

$\boxed{\beta \approx 3\alpha}$

This is because a volume has three linear dimensions. Each expands by $\alpha$, giving:

$(1 + \alpha \Delta T)^3 \approx 1 + 3\alpha \Delta T$ for small $\alpha \Delta T$.

Volume Expansion Coefficients

🔑 Liquids generally have much larger $\beta$ values than solids. This is why a full gasoline tank can overflow on a hot day!

The Anomalous Expansion of Water 🌊

Water is one of the most unusual substances in nature. Most materials contract when cooled and expand when heated. Water does this above 4°C, but between 0°C and 4°C, water expands as it cools.

Key Facts

• Water has its maximum density at 4°C ($\rho = 1{,}000.0$ kg/m³)
• From 4°C → 0°C, water expands (density decreases)
• Ice at 0°C is about 9% less dense than liquid water at 0°C

Why This Matters for Life on Earth

1. Lakes freeze from the top down: In winter, surface water cools below 4°C and becomes less dense, so it stays on top. The densest water (4°C) sinks to the bottom.

2. Ice floats: Since ice is less dense than liquid water, it forms an insulating layer on the surface, protecting aquatic life below.

3. Deep lakes maintain ~4°C at the bottom: Even when the surface freezes, the bottom stays at 4°C, allowing fish and other organisms to survive winter.

Practical Consequence

If water behaved like a "normal" liquid (contracting all the way to freezing), lakes would freeze from the bottom up, likely killing all aquatic life in cold climates. Water's anomalous expansion is crucial for life on Earth!

Volume Expansion Quiz 🧠

Volume Expansion Drill 📦

1. A steel container holds 50.0 L of gasoline at 10°C. How much gasoline (in mL) overflows when heated to 35°C? Use $\beta_{\text{gas}} = 950 \times 10^{-6}$ /°C and $\beta_{\text{steel}} = 36 \times 10^{-6}$ /°C. (Answer = ΔV_gas − ΔV_tank)

2. A glass flask ($\beta = 27 \times 10^{-6}$ /°C) has a volume of 200.0 cm³ at 20°C. Find its volume increase $\Delta V$ (in cm³) when heated to 120°C. Round to 3 significant figures.

3. An aluminum cube ($\alpha = 23 \times 10^{-6}$ /°C) has a side of 10.0 cm at 20°C. Find the increase in its volume $\Delta V$ (in cm³) when heated to 220°C. Use $\beta = 3\alpha$. Round to 3 significant figures.

Exit Quiz

🔧 Problem-Solving Workshop

Part 5 of 7 — Multi-Step Thermal Expansion Problems

Now that you know the linear and volume expansion formulas, let's tackle the classic problem types that appear on the AP Physics 2 exam: gap problems, ring-on-rod problems, and compound expansion problems.

Problem-Solving Strategy

Step-by-Step Approach

1. Identify what is expanding (length, area, or volume?)
2. List known quantities: $L_0$ (or $V_0$), $\alpha$ (or $\beta$), $T_i$, $T_f$
3. Calculate $\Delta T = T_f - T_i$
4. Apply the appropriate formula: $\Delta L = \alpha L_0 \Delta T$ or $\Delta V = \beta V_0 \Delta T$
5. Check units and reasonableness of answer

Gap Problems

A gap between two rails or slabs must accommodate the expansion of one or both pieces. If two rails each of length $L_0$ share a gap:

$\text{Minimum gap} = 2 \times \alpha L_0 \Delta T$

(Each rail expands toward the gap from its end.)

Ring-on-Rod Problems

Goal: Slip a ring over a rod (or shaft) by heating the ring or cooling the rod.

Ring (hole) expansion: The inner diameter of a ring expands just like a solid piece of the same material:

$d_{\text{ring}} = d_0(1 + \alpha_{\text{ring}} \Delta T)$

Rod expansion:

$d_{\text{rod}} = d_0'(1 + \alpha_{\text{rod}} \Delta T)$

Condition for fit: $d_{\text{ring}} \geq d_{\text{rod}}$ (ring hole must be at least as large as the rod diameter).

Compound Problems

When a liquid is inside an expanding container, the apparent volume change of the liquid is:

$\Delta V_{\text{apparent}} = (\beta_{\text{liquid}} - \beta_{\text{container}}) V_0 \Delta T$

The liquid's true expansion minus the container's expansion gives the observable overflow or level change.

Worked Example: Ring on a Rod

A brass ring has an inner diameter of 4.000 cm at 20°C. A steel rod has a diameter of 4.010 cm at 20°C. To what temperature must you heat the ring so it slips over the rod?

Given: $\alpha_{\text{brass}} = 19 \times 10^{-6}$ /°C, $d_{\text{ring}} = 4.000$ cm, $d_{\text{rod}} = 4.010$ cm

Need: Find $T$ where $d_{\text{ring}}(T) = d_{\text{rod}}$ (rod stays at 20°C)

$d_{\text{ring}}(T) = d_0(1 + \alpha \Delta T) = d_{\text{rod}}$

$4.000(1 + 19 \times 10^{-6} \cdot \Delta T) = 4.010$

$1 + 19 \times 10^{-6} \cdot \Delta T = \frac{4.010}{4.000} = 1.0025$

$19 \times 10^{-6} \cdot \Delta T = 0.0025$

$\Delta T = \frac{0.0025}{19 \times 10^{-6}} = 131.6°\text{C}$

$T = 20 + 131.6 = 151.6°\text{C} \approx 152°\text{C}$

You must heat the brass ring to about 152°C for it to slip over the steel rod.

Problem-Solving Concepts Quiz 🧠

Multi-Step Problem Drill 🔧

1. A steel rod has diameter 2.000 cm at 25°C. A copper ring has inner diameter 1.990 cm at 25°C. To what temperature (°C) must the copper ring be heated to just fit over the rod? ($\alpha_{\text{Cu}} = 17 \times 10^{-6}$ /°C, assume rod stays at 25°C.) Round to the nearest °C.

2. A concrete sidewalk slab is 5.00 m long at 5°C. What gap (in mm) is needed between slabs if the maximum temperature is 45°C? ($\alpha_{\text{concrete}} = 12 \times 10^{-6}$ /°C)

3. A glass flask ($\beta = 27 \times 10^{-6}$ /°C) holds exactly 500.0 mL of mercury ($\beta = 182 \times 10^{-6}$ /°C) at 20°C. How many mL of mercury overflow when heated to 100°C?

Round all answers to 3 significant figures.

Exit Quiz

🔬 Kinetic Theory Connection

Part 6 of 7 — Temperature at the Molecular Level

We started Part 1 by saying temperature measures average kinetic energy. Now let's make that precise with the kinetic theory of gases, which connects macroscopic temperature to microscopic molecular motion.

Average Kinetic Energy

For an ideal gas, the average translational kinetic energy per molecule is:

$\boxed{KE_{\text{avg}} = \frac{3}{2} k_B T}$

where:

• $k_B = 1.38 \times 10^{-23}$ J/K (Boltzmann's constant)
• $T$ = absolute temperature in Kelvin

Key Implications

1. $KE_{\text{avg}}$ depends ONLY on temperature — not on the type of gas, not on the mass of the molecule, not on pressure or volume.

2. At the same temperature, a helium atom and a nitrogen molecule have the same average kinetic energy.

3. At absolute zero ($T = 0$ K), the classical kinetic energy is zero — all molecular translational motion ceases.

For One Mole of Gas

The total translational kinetic energy of one mole ($N_A$ molecules) is:

$KE_{\text{total}} = N_A \cdot \frac{3}{2}k_BT = \frac{3}{2}RT$

where $R = N_A k_B = 8.314$ J/(mol·K) is the universal gas constant.

Root-Mean-Square (RMS) Speed

Since $KE_{\text{avg}} = \frac{1}{2}m v_{\text{rms}}^2 = \frac{3}{2}k_BT$, we can solve for the rms speed:

$\boxed{v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}}$

where:

• $m$ = mass of one molecule (kg)
• $M$ = molar mass (kg/mol)
• $R = 8.314$ J/(mol·K)

Example: Nitrogen at Room Temperature

$M_{N_2} = 28 \times 10^{-3}$ kg/mol, $T = 300$ K:

$v_{\text{rms}} = \sqrt{\frac{3(8.314)(300)}{28 \times 10^{-3}}} = \sqrt{\frac{7{,}483}{0.028}} = \sqrt{267{,}250} \approx 517 \text{ m/s}$

That's over 1,100 mph! Gas molecules move incredibly fast.

Comparing Different Gases at Same Temperature

Since $KE_{\text{avg}}$ is the same for all gases at a given $T$:

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_2 v_2^2$

$\frac{v_1}{v_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{M_2}{M_1}}$

Lighter molecules move faster. Helium ($M = 4$) moves $\sqrt{28/4} = \sqrt{7} \approx 2.65$ times faster than nitrogen ($M = 28$) at the same temperature.

Maxwell-Boltzmann Distribution

Not all molecules in a gas move at the same speed. The Maxwell-Boltzmann distribution describes the spread:

• Most molecules cluster near a most probable speed $v_p = \sqrt{\frac{2k_BT}{m}}$
• The average speed $\bar{v} = \sqrt{\frac{8k_BT}{\pi m}}$ is slightly higher
• The rms speed $v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}}$ is highest of the three

$v_p < \bar{v} < v_{\text{rms}}$

At higher temperatures, the distribution flattens and shifts right — more molecules move faster, and the spread of speeds increases.

Kinetic Theory Quiz 🧠

Kinetic Theory Drill 🔬

Use $k_B = 1.38 \times 10^{-23}$ J/K and $R = 8.314$ J/(mol·K).

1. Find the average translational KE (in units of $10^{-21}$ J) of a gas molecule at $T = 400$ K. Round to 3 significant figures.

2. Find the rms speed (in m/s) of oxygen molecules ($M = 32 \times 10^{-3}$ kg/mol) at $T = 300$ K. Round to the nearest whole number.

3. At what temperature (in K) would hydrogen molecules ($M = 2 \times 10^{-3}$ kg/mol) have an rms speed of 2{,}500 m/s? Round to the nearest whole number.

Exit Quiz

🎯 Synthesis & AP Review

Part 7 of 7 — Putting It All Together

This final part connects all the concepts from the topic — temperature scales, thermal equilibrium, linear and volume expansion, and kinetic theory — into a unified review with AP-style problems.

Concept Map

$\text{Temperature} \xrightarrow{\text{measures}} KE_{\text{avg}} = \frac{3}{2}k_BT$

$\text{Temperature change} \xrightarrow{\text{causes}} \begin{cases} \Delta L = \alpha L_0 \Delta T & \text{(linear)} \\ \Delta A = 2\alpha A_0 \Delta T & \text{(area)} \\ \Delta V = \beta V_0 \Delta T & \text{(volume, } \beta \approx 3\alpha\text{)} \end{cases}$

$\text{Zeroth Law} \xrightarrow{\text{enables}} \text{Temperature measurement (thermometers)}$

$v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}$

Common AP Mistakes to Avoid ⚠️

1. Using °C in Kinetic Theory Equations

$KE_{\text{avg}} = \frac{3}{2}k_BT$ and $v_{\text{rms}} = \sqrt{3RT/M}$ require $T$ in Kelvin. Using Celsius gives a completely wrong answer.

2. Confusing Temperature and Heat

• Temperature = average KE per molecule (intensive property)
• Heat = energy transferred due to temperature difference (not a property of a system)

3. Thinking Holes Shrink When Heated

A hole in a heated plate expands as if it were filled with the same material. The hole gets larger, not smaller.

4. Forgetting That $\beta \approx 3\alpha$

If given $\alpha$ and asked about volume expansion, you need $\beta = 3\alpha$. Many students use $\alpha$ directly for volume problems.

5. Ignoring Container Expansion

When a liquid heats up in a container, the apparent expansion is $(\beta_{\text{liquid}} - \beta_{\text{container}})V_0 \Delta T$, not $\beta_{\text{liquid}} V_0 \Delta T$.

6. Assuming All Molecules Move at the Same Speed

The Maxwell-Boltzmann distribution shows a range of speeds. The rms speed is a statistical measure, not the speed of every molecule.

7. Confusing $v_p$, $\bar{v}$, and $v_{\text{rms}}$

Always: $v_p < \bar{v} < v_{\text{rms}}$. The rms speed is the one that appears in $KE = \frac{1}{2}mv_{\text{rms}}^2$.

Comprehensive Concept Quiz 🧠

Mixed Review Drill 🎯

1. Convert 350 K to °F. Round to the nearest whole number.

2. An iron rod ($\alpha = 12 \times 10^{-6}$ /°C) is 1.500 m at 20°C. A brass rod ($\alpha = 19 \times 10^{-6}$ /°C) is 1.500 m at 20°C. Both are heated to 120°C. Find the difference in their lengths $|L_{\text{brass}} - L_{\text{iron}}|$ in mm.

3. Find the rms speed (in m/s) of helium atoms ($M = 4.0 \times 10^{-3}$ kg/mol) at 200 K. Use $R = 8.314$ J/(mol·K). Round to the nearest whole number.

AP-Style Free Response Preview 📝

Here is the type of multi-part question you might see on the AP Physics 2 exam:

A steel railroad rail is 25.0 m long at 20.0°C. The coefficient of linear expansion for steel is $\alpha = 12.0 \times 10^{-6}$ /°C.

(a) Calculate the change in length of the rail when the temperature increases to 50.0°C.

$\Delta L = \alpha L_0 \Delta T = (12.0 \times 10^{-6})(25.0)(30.0) = 9.00 \times 10^{-3}$ m $= 9.00$ mm

(b) If no expansion gap is provided, describe what happens to the rail and explain using physics principles.

Without a gap, the rail cannot expand freely. The constraint produces a compressive thermal stress. If the stress exceeds the yield strength of steel, the rail buckles — it bends laterally out of alignment. This is an application of $\sigma = Y \alpha \Delta T$, where $Y$ is Young's modulus.

(c) A train wheel is a steel ring with inner diameter 0.8500 m at 20.0°C. The axle has diameter 0.8510 m. To what minimum temperature must the wheel be heated to slip onto the axle?

$d(1 + \alpha \Delta T) = 0.8510$ $0.8500(1 + 12.0 \times 10^{-6} \cdot \Delta T) = 0.8510$ $\Delta T = \frac{0.8510/0.8500 - 1}{12.0 \times 10^{-6}} = \frac{0.001176}{12.0 \times 10^{-6}} = 98.0$°C $T = 20.0 + 98.0 = 118.0$°C

(d) Once the wheel cools back to 20°C on the axle, explain the microscopic reason it grips the axle tightly.

As the wheel cools, its atoms lose kinetic energy and vibrate with smaller amplitudes. The average interatomic spacing decreases, causing the ring to contract. Since it cannot contract past the axle diameter, internal stresses develop that create a strong compressive fit (interference fit).

Final Comprehensive Quiz 🏆