AP Physics 2 Unit 1 Study Notes: Thermodynamics (Algebra-Based)

Temperature, Thermal Equilibrium, and Thermometers

What temperature really measures

Temperature is a measure of a system’s “hotness,” but in physics you should think of it more precisely: temperature is related to the average random kinetic energy of the microscopic particles (atoms and molecules) in a substance. When particles move faster on average, the temperature is higher.

This matters because many thermal phenomena—heat flow, phase changes, gas behavior—depend not on how much total energy an object has, but on how that energy is distributed among particles. A large bathtub of lukewarm water contains more total thermal energy than a small cup of boiling water, but the cup has the higher temperature because its particles have higher average kinetic energy.

A key subtlety: temperature is not the same as thermal energy (the total microscopic kinetic plus potential energy of the particles). Two objects can have the same temperature but very different thermal energies if they have different masses or are made of different materials.

Thermal equilibrium and the “direction” of heat flow

Thermal equilibrium occurs when two systems in contact stop exchanging energy due to temperature difference—meaning their temperatures are equal. The deep idea is the zeroth law of thermodynamics:

• If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other.

Why this matters: it’s what makes temperature a meaningful measurable property. It justifies the use of thermometers: if a thermometer comes to thermal equilibrium with an object, the thermometer’s reading tells you the object’s temperature.

A common misconception is to treat “heat” as something an object contains. In physics, heat is energy transferred due to temperature difference, not a substance stored inside an object. Once the transfer is complete, you talk about changes in internal energy, not “heat inside.”

Thermometer models (what makes a thermometer work)

A thermometer needs a property that changes predictably with temperature. Different thermometers use different “thermometric properties,” for example:

• Expansion of a liquid (mercury or alcohol)
• Electrical resistance (RTDs, thermistors)
• Thermoelectric voltage (thermocouples)
• Infrared radiation (non-contact IR thermometers)

The central requirement is calibration: assigning temperatures to specific states (such as the freezing and boiling points of water at 1 atm) and interpolating between them.

Temperature scales and why Kelvin matters

You’ll see three common temperature scales:

• Celsius: convenient for everyday water-based reference points
• Fahrenheit: common in the US but rare in physics
• Kelvin: the absolute temperature scale used in thermodynamics

The Kelvin scale is crucial because many thermodynamic relationships depend on absolute temperature. The conversion is:

T_K = T_C + 273.15

On AP problems you’ll often approximate 273.15 as 273 unless precision is emphasized.

Thermal expansion (a temperature effect you can see)

When temperature increases, most materials expand because particles vibrate with larger amplitude and, on average, take up more space.

For a solid rod expanding in length, the linear expansion model is:

\Delta L = \alpha L_0 \Delta T

• \Delta L is the change in length
• L_0 is the original length
• \Delta T is the temperature change (in Celsius or Kelvin, since changes are the same size)
• \alpha is the coefficient of linear expansion (material-dependent)

For volume expansion (often used for liquids and sometimes solids):

\Delta V = \beta V_0 \Delta T

A typical relationship for isotropic solids is \beta \approx 3\alpha (use this only if the problem indicates it’s appropriate).

Why it matters in thermodynamics: thermal expansion connects temperature to volume, which becomes important when you analyze gases and thermodynamic processes.

Example: expansion and engineering gaps

A steel bridge segment of length L_0 = 50\text{ m} experiences a temperature increase of \Delta T = 30\,^{\circ}\text{C}. If \alpha = 1.2\times 10^{-5}\,^{\circ}\text{C}^{-1}, then:

\Delta L = \alpha L_0 \Delta T
\Delta L = (1.2\times 10^{-5})(50)(30) = 1.8\times 10^{-2}\text{ m}

So the bridge length increases by about 1.8\text{ cm}. That’s why expansion joints exist.

Exam Focus

• Typical question patterns:
  • Conceptual: which object has higher temperature vs higher thermal energy; predicting heat flow direction.
  • Calculation: converting Celsius to Kelvin for gas-law or engine-efficiency problems.
  • Application: thermal expansion causing stress, gaps, or changes in volume.
• Common mistakes:
  • Using Celsius instead of Kelvin in formulas that require absolute temperature.
  • Saying “heat contained in an object” instead of “internal energy.”
  • Mixing up thermal equilibrium (equal temperature) with equal thermal energy.

Heat, Internal Energy, and Calorimetry

Heat vs internal energy (language that prevents confusion)

Internal energy is the total microscopic energy of a system: kinetic energy of random motion plus potential energy from intermolecular forces (especially important in liquids/solids and during phase changes).

Heat Q is energy that crosses the boundary of a system due to a temperature difference.

Work W is energy transferred by macroscopic forces (like a gas pushing a piston).

Thermodynamics is largely about tracking how internal energy changes when heat and work occur.

Specific heat: how “hard” it is to change temperature

Different materials require different amounts of energy to raise their temperature. The specific heat capacity c tells you how much energy is needed to raise the temperature of 1 kg of a substance by 1°C (or 1 K).

The temperature-change heat model is:

Q = mc\Delta T

• m is mass
• c is specific heat
• \Delta T is temperature change

Why it matters: this explains why water moderates climate (high specific heat) and why metals heat up quickly (often lower specific heat).

A common misconception is that “high specific heat means it gets hotter.” It’s the opposite: for the same energy input, a higher c gives a smaller \Delta T.

Calorimetry: energy conservation applied to heating/cooling

Calorimetry problems are usually conservation-of-energy problems. In an isolated setup (like a well-insulated coffee-cup calorimeter), energy lost by one part equals energy gained by another:

\sum Q = 0

The sign convention is important:

• If an object cools down, its \Delta T is negative, so its Q = mc\Delta T is negative (it releases energy).
• If an object warms up, Q is positive.

In many AP problems, you assume negligible heat exchange with the environment unless stated otherwise.

Example: mixing water at different temperatures

You mix 0.20\text{ kg} of water at 80^{\circ}\text{C} with 0.30\text{ kg} of water at 20^{\circ}\text{C} in an insulated container. Find the final temperature T_f (ignore the container’s heat capacity). Since both are water, they have the same c, which cancels.

Set energy gained + energy lost = 0:

m_1 c (T_f - T_1) + m_2 c (T_f - T_2) = 0

Let hot water be 1: m_1 = 0.20, T_1 = 80. Cold water be 2: m_2 = 0.30, T_2 = 20.

Cancel c:

0.20(T_f - 80) + 0.30(T_f - 20) = 0
0.20T_f - 16 + 0.30T_f - 6 = 0
0.50T_f = 22
T_f = 44^{\circ}\text{C}

The final temperature lands between the two initial values and closer to the larger mass (the cooler water), which is a good reasonableness check.

Phase changes and latent heat (temperature can stay constant)

When a substance melts, freezes, vaporizes, condenses, etc., energy is transferred but the temperature can remain constant during the phase change. That’s because the energy goes into changing the intermolecular potential energy, not the average kinetic energy.

The latent heat model is:

Q = mL

• L is the latent heat of fusion L_f (melting/freezing) or vaporization L_v (boiling/condensing)

A frequent error is to use Q = mc\Delta T during a phase change. During the phase change, \Delta T = 0 even though Q is not zero.

Heating curves (connecting all the pieces)

A heating curve graph (temperature vs energy added) usually shows slanted segments (temperature change: use Q = mc\Delta T) and flat segments (phase change: use Q = mL).

Conceptually, you can interpret the slope: a steeper slope means a smaller specific heat (temperature changes more per unit energy). Flat regions mean energy is being used to change phase.

Example: energy to heat and melt ice

Suppose you have m = 0.10\text{ kg} of ice at 0^{\circ}\text{C} and you add heat until it becomes liquid water at 0^{\circ}\text{C}. The energy required is only the melting energy:

Q = mL_f

If instead you started at -10^{\circ}\text{C}, you would need two stages:
1) Warm the ice: Q_1 = mc_{ice}\Delta T
2) Melt it: Q_2 = mL_f

The “two-stage” structure is exactly what AP questions test.

Exam Focus

• Typical question patterns:
  • Calorimetry: mixing, heating a substance with multiple stages, or finding final equilibrium temperature.
  • Graph interpretation: heating curves and identifying where phase changes occur.
  • Conceptual: distinguishing temperature change from phase change energy.
• Common mistakes:
  • Using Q = mc\Delta T during a phase change.
  • Forgetting sign conventions in calorimetry (energy lost should be negative).
  • Assuming equal final temperatures without applying conservation of energy (especially when different c values are involved).

Heat Transfer Mechanisms: Conduction, Convection, and Radiation

Why heat transfer mechanisms matter

Thermodynamics tracks energy, but you often also need the rate at which energy transfers. Mechanisms explain how quickly objects approach thermal equilibrium and which variables (area, thickness, temperature difference) control that process.

AP Physics 2 typically emphasizes:

• Understanding the physical mechanisms
• Using conduction and radiation equations when provided
• Reasoning about proportionalities and comparing situations

Conduction (microscopic collisions and energy diffusion)

Conduction is energy transfer through direct contact. In solids, it’s mostly driven by particle vibrations and, in metals, by mobile electrons transporting energy.

For a flat slab of thickness L and area A, with a temperature difference \Delta T across it, the steady-state conduction power (rate of heat transfer) is:

P = \frac{kA\Delta T}{L}

• P is power (watts)
• k is thermal conductivity (material property)
• A is cross-sectional area
• L is thickness
• \Delta T is temperature difference

Why it matters: this equation shows clear design principles for insulation.

• Larger L reduces heat loss.
• Smaller k (better insulator) reduces heat loss.
• Larger A increases heat loss.

A common conceptual mistake is thinking insulation “keeps cold out.” It actually reduces the rate of energy transfer, regardless of direction.

Example: comparing heat loss through two walls

Wall 1 has thickness L. Wall 2 has thickness 2L, same material and area, same temperature difference. From the formula, doubling L halves P. So the thicker wall loses heat at half the rate.

Convection (bulk fluid motion)

Convection transfers energy by the motion of a fluid (liquid or gas). Warmer fluid often becomes less dense and rises; cooler fluid sinks, creating convection currents.

In AP problems, convection is frequently treated qualitatively:

• Fans increase convection by forcing fluid motion.
• Convection is reduced in microgravity because buoyant rising and sinking is suppressed.

A common misconception is that convection requires a temperature difference only; it also requires a fluid that can move (so it’s not a major mechanism in rigid solids).

Radiation (electromagnetic energy transfer)

Thermal radiation is energy transfer by electromagnetic waves. It does not require matter, so it works through a vacuum (how the Sun warms Earth).

The net radiated power between an object and its environment is often modeled as:

P = \epsilon \sigma A (T^4 - T_{env}^4)

• \epsilon is emissivity (0 to 1)
• \sigma is the Stefan-Boltzmann constant
• A is surface area
• T and T_{env} are absolute temperatures in Kelvin

Even without heavy calculation, you should notice the strong T^4 dependence: small increases in absolute temperature can significantly increase radiated power.

A typical conceptual question: shiny surfaces (low emissivity) are good at reducing radiative heat loss; black matte surfaces (higher emissivity) radiate more effectively.

Exam Focus

• Typical question patterns:
  • Proportional reasoning with conduction: how changing k, A, L, or \Delta T affects P.
  • Conceptual identification: which mechanism dominates in a scenario (vacuum, solids, moving air).
  • Radiation comparisons: effect of doubling temperature (in Kelvin) on radiated power.
• Common mistakes:
  • Using Celsius instead of Kelvin in radiation formulas involving T^4.
  • Confusing convection and conduction (convection needs fluid motion).
  • Treating insulation as “blocking cold” rather than reducing heat transfer rate.

Ideal Gas Law and the Microscopic Picture of Gases

Modeling a gas: when “ideal” is a good approximation

An ideal gas is a simplified model where:

• Gas particles have negligible volume compared to the container.
• There are no intermolecular forces except during collisions.
• Collisions are perfectly elastic.

Real gases behave most ideally at low pressure (particles far apart) and high temperature (intermolecular attractions less important).

This matters because the ideal gas model connects macroscopic variables—pressure, volume, temperature—to a simple equation that underlies many thermodynamics problems.

The ideal gas law

The ideal gas law is:

PV = nRT

• P is absolute pressure (SI: pascals)
• V is volume (SI: cubic meters)
• n is number of moles
• R is the ideal gas constant
• T is absolute temperature (Kelvin)

Common unit awareness:

• 1\text{ atm} \approx 1.01\times 10^5\text{ Pa}
• Liter-to-cubic-meter conversion: 1\text{ L} = 1\times 10^{-3}\text{ m}^3

A common mistake is mixing units (like using liters with pascals) or using Celsius for T.

Combined gas law (a useful derived relationship)

If the amount of gas stays constant (fixed n), then from PV = nRT you can relate two states:

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

This is especially useful for “a gas changes from state 1 to state 2” problems when you’re not asked about heat or work.

Kinetic molecular theory connections (qualitative but powerful)

AP Physics 2 often expects you to connect the ideal gas law to a microscopic picture:

• Gas pressure comes from particles colliding with container walls.
• Higher temperature means higher average particle speed, leading to stronger/more frequent collisions and higher pressure (if volume fixed).

One quantitative relationship that is frequently emphasized is that average translational kinetic energy of an ideal gas particle is proportional to absolute temperature:

\langle K \rangle \propto T

You may also see the molar internal energy result for a monatomic ideal gas in later thermodynamics contexts:

U = \frac{3}{2}nRT

Use that only when the problem states or implies a monatomic ideal gas and ideal behavior.

Example: using the ideal gas law

A container holds n = 0.50\text{ mol} of an ideal gas at T = 300\text{ K} and V = 0.010\text{ m}^3. Find the pressure.

P = \frac{nRT}{V}

Using R = 8.31\text{ J/(mol\,K)}:

P = \frac{(0.50)(8.31)(300)}{0.010} \approx 1.25\times 10^5\text{ Pa}

That’s close to 1 atm, which is a good reasonableness check.

Exam Focus

• Typical question patterns:
  • State changes using PV = nRT or P_1V_1/T_1 = P_2V_2/T_2.
  • Conceptual particle-model questions: what happens to pressure if temperature increases at constant volume.
  • Unit conversion tasks embedded in multi-step problems.
• Common mistakes:
  • Forgetting to convert to Kelvin.
  • Treating pressure as gauge pressure rather than absolute pressure in gas-law contexts.
  • Plugging in inconsistent units (liters with pascals, for example).

Work Done by a Gas and PV Diagrams

What “work” means in thermodynamics

In mechanics you often calculate work with force and displacement. In thermodynamics, a common situation is a gas in a cylinder with a movable piston. If the gas expands, it pushes the piston outward—transferring energy to the surroundings as mechanical work.

For a gas at pressure P undergoing a small change in volume dV, the incremental work done by the gas is dW = P dV. For AP-level algebra-based problems, you typically use constant-pressure or geometric-area reasoning.

Sign convention (the one most commonly used in AP Physics):

• W is positive when the gas does work on the surroundings (expansion).
• W is negative when the surroundings do work on the gas (compression).

Work for constant pressure

If the pressure is constant during expansion/compression, then:

W = P\Delta V

• \Delta V = V_f - V_i

If \Delta V > 0 (expansion), work is positive.

PV diagrams: work as an area

A PV diagram plots pressure vs volume. The work done by the gas during a process is the area under the curve between V_i and V_f.

Why this matters: PV diagrams let you compare different processes between the same initial and final states. Even if \Delta V is the same, the work can be different because pressure can vary.

For a straight horizontal line (constant pressure), the area is a rectangle: W = P\Delta V.

For more complex paths, you may approximate the area using rectangles/triangles or reason qualitatively (a higher curve means larger pressure and thus larger work for the same volume change).

Cycles: net work is enclosed area

In a closed cycle, the gas returns to its initial state. The net work over a full cycle is the area enclosed by the loop on a PV diagram:

• Clockwise loop: positive net work done by the gas.
• Counterclockwise loop: negative net work (net work done on the gas).

This becomes central when you study heat engines.

Example: work from a PV rectangle

A gas expands at constant pressure P = 2.0\times 10^5\text{ Pa} from V_i = 1.0\times 10^{-3}\text{ m}^3 to V_f = 3.0\times 10^{-3}\text{ m}^3.

W = P\Delta V = (2.0\times 10^5)(2.0\times 10^{-3}) = 400\text{ J}

The positive value matches the idea that expansion means the gas is doing work.

Exam Focus

• Typical question patterns:
  • Compute work from a constant-pressure process or from the area under a PV curve.
  • Compare work for two different paths between the same states using PV diagrams.
  • Determine whether net work is positive/negative for a PV cycle based on loop direction.
• Common mistakes:
  • Mixing up work done by the gas vs work done on the gas (sign errors).
  • Using W = P\Delta V when pressure is not constant (instead of area reasoning).
  • Reading PV axes incorrectly (pressure is vertical, volume horizontal).

The First Law of Thermodynamics and Internal Energy Changes

The energy accounting rule for thermal systems

The first law of thermodynamics is energy conservation applied to thermodynamic systems:

\Delta U = Q - W

• \Delta U is the change in internal energy of the system
• Q is heat added to the system (positive if energy enters as heat)
• W is work done by the system (positive if the system does work on the surroundings)

This equation is the backbone of many AP Physics 2 thermodynamics problems. It tells you that internal energy can change because:

• energy enters/leaves as heat
• the system does work or has work done on it

A common mistake is using the wrong sign convention. If your course or teacher uses \Delta U = Q + W, then their W is defined as work done **on** the system. On the AP exam, you must stay consistent within the problem statement and your chosen convention. The most common AP classroom convention matches \Delta U = Q - W with W as work done by the gas.

What determines internal energy for an ideal gas

For an ideal gas, internal energy depends only on temperature (not on pressure or volume separately). That’s a powerful idea: if the temperature doesn’t change, then \Delta U = 0 for an ideal gas.

For a monatomic ideal gas, you may use:

U = \frac{3}{2}nRT

So:

\Delta U = \frac{3}{2}nR\Delta T

If the gas is not specified as monatomic, AP problems often keep it conceptual (internal energy proportional to temperature for an ideal gas) rather than requiring a specific coefficient.

Applying the first law step by step

When solving, it helps to follow a consistent process:

1) Identify the system (usually “the gas”).
2) Determine the process type (constant volume, constant pressure, etc.).
3) Decide what you know about \Delta U (often from temperature change).
4) Compute or infer W (from PV work).
5) Use \Delta U = Q - W to find the missing quantity.

This prevents a common trap: trying to memorize “special-case” shortcuts without understanding.

Example: constant-volume heating

A gas is heated in a rigid container (constant volume). Because \Delta V = 0, the boundary work is:

W = P\Delta V = 0

Then the first law becomes:

\Delta U = Q

So any heat added increases internal energy (and thus temperature for an ideal gas). Students often incorrectly assume “if volume is constant nothing changes”—but temperature and pressure can still change dramatically.

Example: expansion with known heat input

An ideal gas absorbs Q = 500\text{ J} of heat and does W = 200\text{ J} of work during expansion. Then:

\Delta U = Q - W = 500 - 200 = 300\text{ J}

Internal energy increases by 300\text{ J}.

Exam Focus

• Typical question patterns:
  • Use the first law to find Q, W, or \Delta U given the other two.
  • Identify when \Delta U = 0 (ideal gas with constant temperature) or when W = 0 (constant volume).
  • Multi-step processes: compute work from PV diagram then apply the first law.
• Common mistakes:
  • Sign confusion for W and Q.
  • Assuming internal energy depends on pressure/volume rather than temperature (for ideal gas contexts).
  • Forgetting that rigid containers can still have changing pressure and temperature.

Common Thermodynamic Processes (Isothermal, Isochoric, Isobaric, Adiabatic)

Why “process types” are useful

A thermodynamic process is a path the system takes from one state to another. Process labels are shortcuts that tell you which variables stay constant and what that implies for heat, work, and internal energy.

The crucial idea: in thermodynamics, the path matters. Even if the initial and final states are the same, Q and W can be different for different processes.

Isochoric (constant volume)

Isochoric means volume stays constant.

• \Delta V = 0, so W = 0
• First law becomes \Delta U = Q
• Heating increases temperature and pressure

On a PV diagram, an isochoric process is a vertical line.

Isobaric (constant pressure)

Isobaric means pressure stays constant.

• Work is W = P\Delta V
• Heat input typically increases both internal energy and does expansion work

On a PV diagram, an isobaric process is a horizontal line.

A typical conceptual point: for the same temperature increase, an isobaric heating generally requires more heat than isochoric heating because some energy goes into doing work as the gas expands.

Isothermal (constant temperature)

Isothermal means temperature stays constant.

For an ideal gas:

• \Delta T = 0 implies \Delta U = 0
• First law becomes 0 = Q - W, so Q = W

Interpretation: any heat added becomes work done by the gas (during expansion), because internal energy doesn’t change.

On a PV diagram, isotherms for ideal gases satisfy PV = \text{constant} and look like downward-curving hyperbolas.

Students often misread this as “no energy change happens” because temperature is constant. Energy transfer can still happen; it’s just balanced so that internal energy stays the same.

Adiabatic (no heat transfer)

Adiabatic means no heat is transferred:

Q = 0

Then the first law becomes:

\Delta U = -W

So if a gas expands adiabatically (positive W), its internal energy decreases, which means its temperature decreases for an ideal gas.

Adiabatic processes happen when:

• the system is well insulated, or
• the process happens so quickly that there isn’t time for significant heat exchange

On a PV diagram, adiabatic curves are typically steeper than isothermal curves (for the same starting point), but AP often tests this qualitatively rather than requiring a specific adiabatic equation.

Example: comparing isothermal vs adiabatic expansion

Start with the same ideal gas state and let it expand.

• Isothermal expansion: \Delta U = 0, so the gas must absorb heat to supply the work.
• Adiabatic expansion: Q = 0, so the work comes from internal energy, causing temperature to drop.

This comparison is a favorite conceptual question because it tests whether you understand the first law rather than just memorizing labels.

Exam Focus

• Typical question patterns:
  • Given a process type, infer which term is zero: W = 0 (isochoric), Q = 0 (adiabatic), \Delta U = 0 (ideal-gas isothermal).
  • Use PV diagrams to identify process types and compare work (area).
  • Conceptual comparisons: which process requires more heat for the same \Delta T.
• Common mistakes:
  • Thinking isothermal means Q = 0 (it usually does not).
  • Forgetting that adiabatic means no heat transfer, not necessarily “constant temperature.”
  • Assuming process labels automatically determine everything without using the first law.

The Second Law of Thermodynamics, Heat Engines, and Refrigerators

The “one-way” nature of thermal processes

The second law of thermodynamics captures the idea that even though energy is conserved (first law), not all energy transfers are equally useful. It explains why:

• Heat flows spontaneously from hot to cold, not cold to hot.
• You cannot convert heat entirely into work in a cyclic process.

This matters because it sets fundamental limits on engines, power plants, and refrigerators.

Heat engines: turning heat flow into work

A heat engine is a device that operates in a cycle and produces net work by absorbing heat from a hot reservoir and expelling some heat to a cold reservoir.

Key energy flows (over one full cycle):

• Q_H: heat absorbed from the hot reservoir
• Q_C: heat expelled to the cold reservoir
• W_{net}: net work output

Because the system returns to its initial state after a full cycle, internal energy returns to its original value:

\Delta U_{cycle} = 0

So from the first law over a complete cycle:

W_{net} = Q_H - Q_C

Efficiency

The thermal efficiency e of a heat engine is the fraction of input heat converted to net work:

e = \frac{W_{net}}{Q_H}

Using W_{net} = Q_H - Q_C:

e = 1 - \frac{Q_C}{Q_H}

The second law implies Q_C cannot be zero for a cyclic engine, so e < 1 always.

A common misconception is “a perfect engine is possible if we reduce friction.” Frictionless components don’t eliminate the need to dump waste heat; the limitation is deeper than engineering imperfections.

Carnot engine (the theoretical upper bound)

The Carnot engine is an idealized reversible engine that sets the maximum possible efficiency for any engine operating between two reservoirs at temperatures T_H and T_C (in Kelvin):

e_{Carnot} = 1 - \frac{T_C}{T_H}

This is not saying real engines achieve this; it’s an upper limit.

Two key insights AP likes to test:

• Increasing T_H increases max efficiency.
• Decreasing T_C increases max efficiency.

But you cannot make T_C = 0\text{ K} in practice, and engineering constraints limit how hot T_H can be.

Refrigerators and heat pumps (running an engine “backward”)

A refrigerator uses work input to move heat from a cold region to a hot region. Energy flow relationships:

Q_H = Q_C + W

• Q_C is heat removed from the cold compartment
• Q_H is heat expelled to the room
• W is electrical/mechanical work input

Instead of efficiency, refrigerators are rated by coefficient of performance (COP), because the goal is heat moved, not work produced.

Refrigerator COP:

COP_R = \frac{Q_C}{W}

Heat pump COP (goal is heating the warm space):

COP_{HP} = \frac{Q_H}{W}

A useful conceptual check: a good refrigerator can have COP_R > 1 because it is not creating energy; it is using work to move existing energy.

Example: engine energy accounting

A heat engine absorbs Q_H = 1000\text{ J} and expels Q_C = 600\text{ J}.

Net work:

W_{net} = Q_H - Q_C = 400\text{ J}

Efficiency:

e = \frac{400}{1000} = 0.40

So 40% of input heat becomes work; the rest must be exhausted as waste heat.

Example: refrigerator COP

A refrigerator removes Q_C = 500\text{ J} from the cold compartment while consuming W = 200\text{ J} of work.

COP_R = \frac{Q_C}{W} = \frac{500}{200} = 2.5

This means it moves 2.5 times as much heat as the work energy it consumes.

Exam Focus

• Typical question patterns:
  • Use W = Q_H - Q_C and e = W/Q_H to compute missing quantities.
  • Compare maximum possible efficiencies using e_{Carnot} = 1 - T_C/T_H.
  • Refrigerator/heat pump COP calculations from given heat flows and work.
• Common mistakes:
  • Using Celsius instead of Kelvin in Carnot efficiency.
  • Assuming 100% efficiency is possible if “no friction.”
  • Confusing Q_H and Q_C (remember: subscript matches the reservoir temperature).

Entropy and the Direction of Spontaneous Change

What entropy is trying to quantify

Entropy S is a measure that helps predict the natural direction of processes. At an introductory level, you can think of entropy as relating to:

• how spread out energy is among available microscopic states
• how dispersed energy becomes when systems interact

The second law can be stated in terms of entropy:

• For an isolated system, total entropy does not decrease.

This doesn’t mean entropy must always increase everywhere; it means that when you consider the whole isolated system (system plus surroundings), the net change is nonnegative.

Why this matters: entropy is the deeper reason heat engines have limits and why certain processes are irreversible.

Entropy change for heat transfer at constant temperature

For a reversible heat transfer at absolute temperature T, the entropy change is defined as:

\Delta S = \frac{Q_{rev}}{T}

• Q_{rev} is the heat transferred in a reversible manner
• T must be in Kelvin

Even if AP problems stay mostly conceptual, this formula is often used for reservoir entropy changes because reservoirs are modeled as staying at constant temperature.

Entropy and heat flow from hot to cold

Consider heat Q flowing from a hot reservoir at T_H to a cold reservoir at T_C.

Entropy change of the hot reservoir (loses heat):

\Delta S_H = -\frac{Q}{T_H}

Entropy change of the cold reservoir (gains heat):

\Delta S_C = \frac{Q}{T_C}

Total:

\Delta S_{total} = Q\left(\frac{1}{T_C} - \frac{1}{T_H}\right)

Since T_H > T_C, you have 1/T_C > 1/T_H, so \Delta S_{total} > 0. That matches the second law and explains why heat flows spontaneously from hot to cold.

A common misconception is that “entropy increase means everything gets more disordered in an everyday sense.” The “disorder” language can be helpful, but it can also mislead. A better AP-level view is: entropy tracks how energy disperses and whether a process is compatible with the second law.

Entropy and reversibility

A reversible process is an idealized process that can be reversed without leaving net changes in the system and surroundings. Reversible processes produce no net entropy.

An irreversible process increases total entropy. Examples:

• friction
• unrestrained expansion of a gas
• heat transfer across a finite temperature difference

Heat engines that achieve Carnot efficiency are reversible; real engines are irreversible and have lower efficiency.

Example: entropy change of two reservoirs

Heat Q = 200\text{ J} flows from T_H = 400\text{ K} to T_C = 300\text{ K}.

\Delta S_H = -\frac{200}{400} = -0.50\text{ J/K}

\Delta S_C = \frac{200}{300} \approx 0.667\text{ J/K}

\Delta S_{total} \approx 0.167\text{ J/K}

Positive total entropy confirms the spontaneity of the process.

Exam Focus

• Typical question patterns:
  • Conceptual: identify which processes are irreversible and thus increase total entropy.
  • Reservoir calculations using \Delta S = Q/T with Kelvin temperatures.
  • Linking entropy to engine limits and why waste heat is unavoidable.
• Common mistakes:
  • Using Celsius instead of Kelvin in entropy calculations.
  • Forgetting that the second law refers to total entropy of an isolated system (system plus surroundings).
  • Assuming entropy is a “force” that pushes things; it’s a state function used to evaluate directionality.