🔥 Heat vs. Temperature

Part 1 of 7 — They're NOT the Same Thing!

People use "heat" and "temperature" interchangeably in everyday language — but in physics, they mean very different things. Confusing them is one of the most common mistakes on the AP exam.

Temperature — A Property

Temperature is a measure of the average translational kinetic energy of the molecules in a substance.

• It describes the state of a system at a given moment
• It is a scalar quantity measured in kelvins (K), degrees Celsius (°C), or degrees Fahrenheit (°F)
• The SI unit for physics is the kelvin (K)

Heat — A Process

Heat (symbol $Q$) is energy transferred between systems due to a temperature difference.

• Heat is NOT something a system "has" — it is energy in transit
• It only exists during the process of transfer
• Once energy is transferred, it becomes internal energy of the receiving system
• SI unit: joules (J)

🧠 Key insight: A hot cup of coffee doesn't "contain heat." It has high internal energy. When it sits on a table, energy flows as heat from the coffee to the cooler surroundings.

Units of Heat Energy

Joule (J) — The SI Unit

The joule is the standard SI unit of energy: $1 \text{ J} = 1 \text{ kg} \cdot \text{m}^2/\text{s}^2$

Calorie (cal)

Originally defined as the energy needed to raise 1 gram of water by 1°C: $1 \text{ cal} = 4.186 \text{ J}$

Kilocalorie (kcal) / Food Calorie (Cal)

The "Calorie" on food labels (capital C) is actually a kilocalorie: $1 \text{ kcal} = 1 \text{ Cal} = 1{,}000 \text{ cal} = 4{,}186 \text{ J}$

British Thermal Unit (BTU)

Used in heating/cooling systems in the US: $1 \text{ BTU} = 1{,}055 \text{ J}$

Sign Convention for Heat

In physics, we define:

• $Q > 0$: energy flows into the system (system gains energy)
• $Q < 0$: energy flows out of the system (system loses energy)

This sign convention is consistent with the first law of thermodynamics: $\Delta U = Q - W$.

The Direction of Heat Flow

Heat always flows spontaneously from a higher-temperature object to a lower-temperature object. This continues until both objects reach the same temperature — a state called thermal equilibrium.

The Zeroth Law of Thermodynamics

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.

This is the principle behind thermometers! The thermometer (C) reaches equilibrium with the object being measured (A), and the reading tells us A's temperature.

Important Distinctions

Common Misconception

"Adding heat always raises temperature." FALSE! During a phase change (like melting ice at 0°C), you add heat but the temperature stays constant. The energy goes into breaking molecular bonds instead.

Heat vs. Temperature Concept Quiz 🧠

Unit Conversion Drill 🔢

Convert between heat energy units. Round to the nearest whole number.

1. 500 calories = ___ joules (use 1 cal = 4.186 J)
2. 8{,}372 J = ___ kilocalories
3. 3 food Calories (kcal) = ___ joules

Exit Quiz — Heat Fundamentals

🧊 Specific Heat Capacity

Part 2 of 7 — Why Does Water Take So Long to Boil?

Different substances require different amounts of energy to change temperature by the same amount. This property is called specific heat capacity — and it explains everything from why beaches are comfortable to why car engines use water-based coolant.

The Fundamental Equation

$Q = mc\Delta T$

Where:

• $Q$ = heat energy transferred (J)
• $m$ = mass of the substance (kg)
• $c$ = specific heat capacity (J/(kg·°C) or J/(kg·K))
• $\Delta T = T_f - T_i$ = change in temperature (°C or K)

Understanding the Variables

Specific heat capacity $c$ tells you: "How many joules does it take to raise 1 kg of this substance by 1°C?"

Sign of $Q$ and $\Delta T$

• If $T_f > T_i$: $\Delta T > 0$, so $Q > 0$ → substance absorbs heat
• If $T_f < T_i$: $\Delta T < 0$, so $Q < 0$ → substance releases heat

The sign convention is built right into the equation!

Why Water Is Special

Water has an exceptionally high specific heat: $c = 4{,}186$ J/(kg·°C). This is about:

• 9.3× higher than iron
• 10.7× higher than copper
• 32.7× higher than lead

Consequences of Water's High Specific Heat

1. Climate Moderation Coastal cities have milder climates than inland cities. Water absorbs and releases large amounts of energy with relatively small temperature changes, so oceans act as thermal buffers.

2. Biological Importance The human body is ~60% water. This high specific heat helps us maintain a stable body temperature — our bodies can absorb or release significant energy without dangerous temperature swings.

3. Cooling Systems Car radiators, nuclear power plants, and computer liquid cooling systems all use water because it can carry away large amounts of heat energy per degree of temperature rise.

4. Cooking A pot of water takes a long time to boil — but once heated, it stays hot for a long time too. This makes water excellent for cooking (even heat distribution) and terrible for fast heating.

Why is water's specific heat so high?

Water molecules form hydrogen bonds with each other. Much of the energy you add goes into breaking and rearranging these bonds rather than increasing molecular speed (temperature). This extra "energy sink" gives water its unusually high specific heat.

Specific Heat Concept Quiz 🧠

Specific Heat Calculation Drill 🔢

Use $Q = mc\Delta T$. Round answers to the nearest whole number.

1. How much heat is needed to raise 2.0 kg of water from 20°C to 80°C? ($c_w = 4{,}186$ J/(kg·°C)). Answer in kJ.

2. A 0.50 kg block of aluminum ($c = 900$ J/(kg·°C)) absorbs 9{,}000 J of heat. What is the temperature change in °C?

3. A 0.30 kg piece of copper ($c = 390$ J/(kg·°C)) cools from 200°C to 50°C. How much heat does it release? Answer as a positive number in J.

Exit Quiz — Specific Heat Mastery

⚖️ Calorimetry

Part 3 of 7 — Tracking Heat Exchange

When a hot object is placed in contact with a cold object inside an insulated container, energy flows as heat until both reach the same final temperature. Calorimetry is the science of measuring these heat exchanges — and conservation of energy makes the math elegant.

The Calorimetry Principle

In an isolated system (no heat escapes to the environment):

$Q_{\text{lost}} + Q_{\text{gained}} = 0$

Or equivalently:

$|Q_{\text{lost by hot}}| = |Q_{\text{gained by cold}}|$

Setting Up Calorimetry Problems

1. Identify all objects exchanging heat
2. Write $Q = mc\Delta T$ for each object
3. Apply conservation: $\sum Q_i = 0$ (all Q's add to zero)
4. Solve for the unknown

Sign Convention Reminder

• Object cooling: $\Delta T < 0$ → $Q < 0$ (releases heat)
• Object warming: $\Delta T > 0$ → $Q > 0$ (absorbs heat)

Worked Example

A 0.50 kg iron ball ($c = 450$ J/(kg·°C)) at 300°C is dropped into 2.0 kg of water ($c = 4{,}186$ J/(kg·°C)) at 20°C. Find the final temperature $T_f$.

$Q_{\text{iron}} + Q_{\text{water}} = 0$ $m_{\text{Fe}} c_{\text{Fe}} (T_f - 300) + m_w c_w (T_f - 20) = 0$ $(0.50)(450)(T_f - 300) + (2.0)(4{,}186)(T_f - 20) = 0$ $225(T_f - 300) + 8{,}372(T_f - 20) = 0$ $225 T_f - 67{,}500 + 8{,}372 T_f - 167{,}440 = 0$ $8{,}597 T_f = 234{,}940$ $T_f = 27.3°\text{C}$

Notice: the final temperature is much closer to the water's initial temperature because water has both more mass AND a much higher specific heat.

Calorimetry Concept Quiz 🧠

Finding Unknown Specific Heat

Calorimetry is commonly used to measure the specific heat of an unknown substance.

Method

1. Heat a known mass of the unknown substance to a known temperature
2. Drop it into a known mass of water at a known temperature
3. Measure the final equilibrium temperature
4. Solve $Q_{\text{lost}} + Q_{\text{gained}} = 0$ for $c_{\text{unknown}}$

Formula

$m_{\text{unk}} c_{\text{unk}} (T_f - T_{\text{unk,i}}) + m_w c_w (T_f - T_{w,i}) = 0$

Solving for $c_{\text{unk}}$:

$c_{\text{unk}} = \frac{-m_w c_w (T_f - T_{w,i})}{m_{\text{unk}} (T_f - T_{\text{unk,i}})}$

Example

A 0.200 kg metal sample at 150°C is placed in 0.400 kg of water at 22°C. The final temperature is 25°C.

$c_{\text{metal}} = \frac{-(0.400)(4{,}186)(25 - 22)}{(0.200)(25 - 150)}$

$c_{\text{metal}} = \frac{-(0.400)(4{,}186)(3)}{(0.200)(-125)}$

$c_{\text{metal}} = \frac{-5{,}023.2}{-25} = 201 \text{ J/(kg·°C)}$

This is close to the specific heat of tin (210 J/(kg·°C)), so the metal is likely tin.

Calorimetry Problem Drill 🔢

A 0.40 kg block of copper ($c = 390$ J/(kg·°C)) at 250°C is placed in 1.5 kg of water ($c = 4{,}186$ J/(kg·°C)) at 18°C in an insulated calorimeter.

1. What is the thermal mass ($mc$) of the copper in J/°C?

2. What is the thermal mass ($mc$) of the water in J/°C?

3. What is the final equilibrium temperature? Round to one decimal place (°C).

Exit Quiz — Calorimetry Mastery

🌊 Heat Transfer Mechanisms

Part 4 of 7 — Conduction, Convection & Radiation

There are three ways heat can transfer from one place to another. Each operates by a different physical mechanism, and understanding them is essential for AP Physics 2.

Conduction — Energy Through Contact

Conduction is heat transfer through direct molecular collisions within a material or between materials in contact.

The Rate Equation

$\frac{Q}{t} = \frac{kA\Delta T}{L}$

Where:

• $Q/t$ = rate of heat flow (W = J/s)
• $k$ = thermal conductivity (W/(m·K))
• $A$ = cross-sectional area perpendicular to heat flow (m²)
• $\Delta T$ = temperature difference across the material (K or °C)
• $L$ = thickness of the material (m)

Thermal Conductivity Values

Key Insights

• Metals are excellent conductors (high $k$) because free electrons carry energy quickly
• Insulators (wood, foam, air) have low $k$ and resist heat flow
• Trapped air is one of the best insulators — this is why puffy jackets and double-pane windows work!

Convection — Bulk Fluid Motion

Convection is heat transfer by the physical movement of a fluid (liquid or gas). It cannot occur in solids.

Natural (Free) Convection

Driven by buoyancy — warm fluid rises, cool fluid sinks, creating circulation.

• Warm air rising from a heater
• Ocean currents driven by temperature differences
• Mantle convection inside Earth
• Boiling water in a pot

The mechanism: When fluid is heated, it expands and becomes less dense. The less-dense warm fluid rises, while cooler, denser fluid sinks to replace it. This creates a convection current.

Forced Convection

Driven by an external force (fan, pump, wind).

• A fan blowing air over your skin
• A car radiator with a fan
• Forced-air heating systems
• Blood pumped through your body

Why Convection Is Faster Than Conduction

In convection, energy is physically carried by moving fluid parcels — much faster than waiting for molecule-by-molecule energy transfer. This is why a fan cools you faster than still air, and why stirring soup helps it heat evenly.

Radiation — Electromagnetic Waves

Radiation is heat transfer via electromagnetic waves. It requires no medium — this is how the Sun heats the Earth across the vacuum of space.

Stefan-Boltzmann Law

The power radiated by an object:

$P = \sigma \varepsilon A T^4$

Where:

• $P$ = radiated power (W)
• $\sigma = 5.67 \times 10^{-8}$ W/(m²·K⁴) — Stefan-Boltzmann constant
• $\varepsilon$ = emissivity (0 to 1, dimensionless)
• $A$ = surface area (m²)
• $T$ = absolute temperature (K — must use kelvins!)

Emissivity

• $\varepsilon = 1$: perfect blackbody (absorbs and emits all radiation)
• $\varepsilon = 0$: perfect reflector (emits no radiation)
• Dark, rough surfaces: $\varepsilon$ close to 1
• Shiny, polished surfaces: $\varepsilon$ close to 0

Key Features

• Radiation intensity depends on $T^4$ — doubling the temperature increases radiated power by $2^4 = 16$ times!
• All objects above 0 K emit thermal radiation
• Net radiation: $P_{\text{net}} = \sigma \varepsilon A (T_{\text{object}}^4 - T_{\text{surroundings}}^4)$
• Hotter objects radiate at shorter wavelengths (Wien's law)

Heat Transfer Mechanism Quiz 🧠

Identify the Heat Transfer Mechanism 🎯

Exit Quiz — Heat Transfer Mechanisms

🧱 Conduction Problem Solving

Part 5 of 7 — Thermal Resistance and Composite Walls

Now that you know the conduction equation, let's master the calculations. We'll work through single-layer problems, then build up to composite (multi-layer) walls — a favorite AP topic.

Single-Layer Conduction Review

$\frac{Q}{t} = \frac{kA\Delta T}{L}$

This equation gives the steady-state rate of heat flow — the rate after the system has settled into a constant temperature gradient.

Worked Example

A glass window ($k = 0.84$ W/(m·K)) is 1.2 m wide, 1.5 m tall, and 6.0 mm thick. The inside surface is at 20°C and the outside surface is at 5°C. What is the rate of heat loss?

$A = (1.2)(1.5) = 1.8 \text{ m}^2$ $L = 6.0 \text{ mm} = 0.006 \text{ m}$ $\Delta T = 20 - 5 = 15 \text{ K}$

$\frac{Q}{t} = \frac{(0.84)(1.8)(15)}{0.006} = \frac{22.68}{0.006} = 3{,}780 \text{ W}$

That's nearly 4 kilowatts — through a single window! This is why insulation matters.

Thermal Resistance (R-Value)

Just like electrical resistance opposes current flow, thermal resistance opposes heat flow.

Defining R-Value

$R = \frac{L}{k}$

Where:

• $R$ = thermal resistance per unit area (m²·K/W)
• $L$ = thickness (m)
• $k$ = thermal conductivity (W/(m·K))

The conduction equation becomes:

$\frac{Q}{t} = \frac{A \Delta T}{R}$

Composite Walls — Series

For layers in series (heat passes through one layer after another):

$R_{\text{total}} = R_1 + R_2 + R_3 + \cdots$

This is exactly like resistors in series in electric circuits!

$\frac{Q}{t} = \frac{A \Delta T_{\text{total}}}{R_{\text{total}}}$

Example: Double-Pane Window

A double-pane window has two glass panes ($k = 0.84$, $L = 4$ mm each) with an air gap ($k = 0.024$, $L = 10$ mm) between them.

$R_{\text{glass}} = \frac{0.004}{0.84} = 0.00476 \text{ m}^2\text{·K/W each}$

$R_{\text{air}} = \frac{0.010}{0.024} = 0.417 \text{ m}^2\text{·K/W}$

$R_{\text{total}} = 0.00476 + 0.417 + 0.00476 = 0.426 \text{ m}^2\text{·K/W}$

The air gap provides about 98% of the total thermal resistance, despite being only slightly thicker than the glass! This demonstrates why trapped air is such an effective insulator.

Conduction & R-Value Quiz 🧠

Conduction Calculation Drill 🔢

A Styrofoam cooler wall ($k = 0.033$ W/(m·K)) is 3.0 cm thick and has a total surface area of 0.80 m². The inside is at 2°C and the outside is at 32°C.

1. What is the thermal resistance $R = L/k$ of the wall? (m²·K/W, round to 3 significant figures)

2. What is the rate of heat flow into the cooler? (in watts, round to 3 significant figures)

3. How much heat enters the cooler in 6 hours? (in kJ, round to nearest whole number)

Composite Wall Drill 🔢

A wall consists of 10 cm of brick ($k = 0.60$ W/(m·K)) and 5 cm of insulation ($k = 0.040$ W/(m·K)) in series. The wall area is 12 m². The inside temperature is 22°C and the outside is $-8$°C.

1. $R_{\text{brick}}$ in m²·K/W (round to 3 significant figures)

2. $R_{\text{insulation}}$ in m²·K/W (round to 3 significant figures)

3. Total rate of heat loss through the wall in watts (round to nearest whole number)

Exit Quiz — Conduction Problem Solving

🌍 Applications & Real-World Heat Transfer

Part 6 of 7 — Physics in Everyday Life

The principles of heat transfer are at work everywhere — in your kitchen, your house, and across the entire planet. Understanding these applications solidifies your physics intuition and shows up frequently on the AP exam as conceptual questions.

The Thermos (Vacuum Flask) — Engineering Perfection

A thermos is brilliantly designed to block ALL three heat transfer mechanisms:

Conduction → Blocked by Vacuum

• A vacuum between double walls eliminates molecular collisions
• No molecules = no conduction
• Small contact points at the neck are made of low-conductivity material

Convection → Blocked by Vacuum + Sealed Lid

• No air in the vacuum gap = no convection currents
• The sealed lid prevents warm air from escaping at the top
• Without the lid, convection would be the dominant loss mechanism

Radiation → Reduced by Silvered Walls

• The inner walls are coated with reflective silver
• Shiny surfaces have low emissivity ($\varepsilon \approx 0.02$)
• Most infrared radiation is reflected back rather than absorbed/emitted
• This reduces radiative loss by ~98%

Result

A good thermos can keep coffee hot for 12+ hours because all three pathways are severely restricted. The small remaining heat loss comes from conduction through the lid and neck, and the tiny amount of radiation not reflected by the silvered walls.

Physics of Cooking

Boiling Water

• Water boils at 100°C (at 1 atm) — no matter how high you turn the flame
• A rolling boil doesn't cook food faster than a gentle boil (same temperature!)
• Adding a lid reduces convective heat loss, so water boils faster
• At high altitude, lower pressure → lower boiling point → food takes longer to cook

Metal vs. Wooden Spoon

• Both are at the same temperature in a hot kitchen
• The metal spoon feels hotter because metal conducts heat rapidly into your hand
• Wood conducts heat slowly, so your nerve endings don't receive energy as fast
• The temperature is the same; the rate of heat transfer differs

Cooking Pans

• Copper-bottom pans: high $k$ means heat spreads evenly across the bottom (no hot spots)
• Cast iron: lower $k$ but high heat capacity — holds heat well, cooks evenly once heated
• Wooden handles: low $k$ keeps the handle cool while the pan is hot

Microwave Ovens

• Microwaves directly excite water molecules (radiation at 2.45 GHz)
• Unlike conventional ovens, they heat food from within — no conduction through layers needed
• Dry materials (ceramics) barely absorb microwaves, so the plate stays relatively cool

The Greenhouse Effect

How It Works

1. Solar radiation (mostly visible light, short wavelength) passes through the atmosphere and heats Earth's surface

2. Earth re-radiates energy as infrared radiation (long wavelength) — because Earth is much cooler than the Sun

3. Greenhouse gases (CO₂, H₂O, CH₄, N₂O) absorb and re-emit infrared radiation in all directions

4. Some re-emitted IR goes back toward Earth's surface, trapping energy in the atmosphere

The Energy Balance

Without greenhouse gases, Earth's average temperature would be about $-18$°C (0°F). The natural greenhouse effect raises it to about $15$°C (59°F) — making the planet habitable.

Why It's Getting Stronger

Human activities (burning fossil fuels, deforestation) increase CO₂ concentration:

• Pre-industrial: ~280 ppm CO₂
• Current: ~420 ppm CO₂
• More CO₂ → more IR absorption → more energy trapped → higher temperatures

The Stefan-Boltzmann Connection

Earth must radiate energy at the same rate it absorbs solar energy (equilibrium). If more energy is trapped:

• Temperature rises until $P_{\text{out}} = \sigma \varepsilon A T^4$ matches the new energy input
• Since $P \propto T^4$, even small changes in trapped energy cause measurable temperature changes

Water's Role in Climate

Water's high specific heat means oceans absorb enormous amounts of energy with small temperature changes. Oceans act as a thermal buffer, slowing climate change but also storing vast amounts of energy.

Applications Concept Quiz 🧠

Match the Scenario to the Dominant Heat Transfer Mechanism 🎯

Exit Quiz — Real-World Applications

🎯 Synthesis & AP Review

Part 7 of 7 — Putting It All Together

You've learned the core concepts of heat and specific heat. Now it's time to integrate everything, avoid common AP mistakes, and tackle the kinds of problems you'll see on the exam.

Concept Map — How It All Connects

• Temperature → average molecular KE → measured in K
• Heat ($Q$) → energy transfer due to $\Delta T$ → measured in J
• Specific heat ($c$) → links $Q$, $m$, and $\Delta T$ via $Q = mc\Delta T$
• Calorimetry → conservation of energy: $\sum Q_i = 0$
• Conduction → $Q/t = kA\Delta T/L$ → through solids
• Convection → bulk fluid motion → natural or forced
• Radiation → $P = \sigma \varepsilon A T^4$ → no medium needed

Common AP Mistakes to Avoid

Mistake 1: Confusing Heat and Temperature

❌ "The object has a lot of heat" ✅ "The object has high internal energy" or "The object is at a high temperature"

Heat is a process (energy in transit), not a property of an object.

Mistake 2: Forgetting to Convert Units

• Temperature: use K for radiation problems ($P = \sigma \varepsilon A T^4$)
• Temperature: °C or K both work for $\Delta T$ in $Q = mc\Delta T$ (the change is the same)
• Length: convert cm or mm to m for conduction ($Q/t = kA\Delta T/L$)

Mistake 3: Sign Errors in Calorimetry

The equation $\sum Q_i = 0$ handles signs automatically IF you use $\Delta T = T_f - T_i$ consistently.

• Hot object: $T_f < T_i$ → $\Delta T < 0$ → $Q < 0$ ✅
• Cold object: $T_f > T_i$ → $\Delta T > 0$ → $Q > 0$ ✅

Mistake 4: Using °C in Stefan-Boltzmann

$P = \sigma \varepsilon A T^4$ requires absolute temperature in kelvins.

• 20°C → 293 K ✅
• Using $T = 20$ → catastrophically wrong answer ❌

Mistake 5: Thinking "Adding Heat Always Raises Temperature"

During a phase change (melting, boiling), temperature stays constant while heat is absorbed. The energy goes into breaking molecular bonds (latent heat), not increasing KE.

Mixed Concept Quiz 🧠

AP-Style FRQ Preview

Here's the type of multi-part free-response question you might see on the AP Physics 2 exam:

Scenario

A student performs a calorimetry experiment. She heats a 0.150 kg metal cylinder in boiling water (100°C) and then transfers it to an insulated cup containing 0.250 kg of water at 22.0°C. The final equilibrium temperature is 25.2°C.

Typical AP Questions

(a) Is the metal's specific heat greater than, less than, or equal to water's specific heat? Justify your answer using evidence from the experiment. (2 pts)

Answer: Less than water's. The final temperature (25.2°C) is much closer to water's initial temperature (22.0°C) than to the metal's (100.0°C). Since the masses are comparable, the metal must have a much smaller specific heat to account for this small final temperature shift.

(b) Calculate the specific heat of the metal. (3 pts)

$c_{\text{metal}} = \frac{-m_w c_w (T_f - T_{w,i})}{m_{\text{metal}}(T_f - T_{\text{metal},i})}$ $= \frac{-(0.250)(4{,}186)(25.2 - 22.0)}{(0.150)(25.2 - 100.0)}$ $= \frac{-(0.250)(4{,}186)(3.2)}{(0.150)(-74.8)}$ $= \frac{-3{,}348.8}{-11.22} = 298 \text{ J/(kg·°C)}$

(c) The student suspects some heat was lost to the environment. If this occurred, would the calculated specific heat be too high or too low? Explain. (2 pts)

Answer: Too low. If heat escaped to the environment, then the water didn't receive all the energy the metal lost. This means $T_f$ is lower than ideal, making $(T_f - T_{w,i})$ smaller and thus the numerator smaller. The calculated $c$ would underestimate the true value.

Synthesis Problem Drill 🔢

A 0.25 kg iron block ($c = 450$ J/(kg·°C)) at 350°C is dropped into 0.80 kg of water ($c = 4{,}186$ J/(kg·°C)) at 15°C in an insulated container.

1. Heat capacity (thermal mass) of the iron block, $mc$, in J/°C (round to nearest whole number)

2. Heat capacity (thermal mass) of the water, $mc$, in J/°C (round to nearest whole number)

3. Final equilibrium temperature in °C (round to 3 significant figures)

Final Mastery Quiz 🏆