AP Physics C Interactive Thermodynamics Module

AP Physics C: Thermodynamics

Physics-centered review and inquiry covering thermodynamic laws, kinetic theory, ideal gases, work, entropy, p–V graphs, and absolute zero.

What You Should Be Able To Do

Glossary of Thermodynamics Terms

System

The part of the universe selected for analysis.

Surroundings

Everything outside the system that can exchange energy with it.

State Variable

A quantity determined only by the current state, such as \(P\), \(V\), \(T\), \(N\), and \(U\).

Path Variable

A quantity that depends on the process, not just the endpoints. Heat \(Q\) and work \(W\) are path variables.

Thermal Equilibrium

A condition where systems have the same temperature and no net heat flows.

Pressure

Force per unit area. In a gas, pressure comes from molecular collisions with container walls.

Temperature

A macroscopic measure related to average molecular kinetic energy.

Internal Energy

The total microscopic energy stored in the system, including particle kinetic and potential energies.

Heat

Energy transferred because of a temperature difference.

Work

Energy transferred by mechanical interaction, such as compression or expansion.

Entropy

A state function related to energy dispersal and the number of possible microscopic arrangements.

Isothermal

A constant-temperature process.

Adiabatic

A process with no heat transfer: \(Q=0\).

Isochoric

A constant-volume process. No \(P\,dV\) work occurs.

Isobaric

A constant-pressure process.

Quasi-static

A slow process where the system remains close to equilibrium at each step.

Reversible Process

An ideal process that can be reversed without increasing total entropy.

Absolute Zero

\(0\ \text{K}\), or \(-273.15^\circ\text{C}\), the theoretical lower limit of temperature.

Boltzmann Constant

\(k_B=1.38\times10^{-23}\ \text{J/K}\), connecting temperature to microscopic particle energy.

Conduction

Heat transfer through direct molecular contact within a material or between materials in contact.

Convection

Heat transfer by the bulk movement of a fluid (liquid or gas) carrying thermal energy.

Radiation

Heat transfer via electromagnetic waves. No medium is required.

Specific Heat Capacity

The energy required to raise \(1\ \text{kg}\) of a substance by \(1\ \text{K}\). Symbol: \(c\).

Latent Heat

The energy per unit mass absorbed or released during a phase change at constant temperature. Symbol: \(L\).

Heat of Fusion

The latent heat for the solid–liquid transition. Symbol: \(L_f\).

Heat of Vaporization

The latent heat for the liquid–gas transition. Symbol: \(L_v\).

Phase Change

A transition between solid, liquid, and gas states. Temperature remains constant during phase changes.

Thermal Expansion

The tendency of matter to increase in size when heated, due to increased molecular motion.

Coefficient of Linear Expansion

The fractional change in length per degree of temperature change. Symbol: \(\alpha\).

The Laws of Thermodynamics

Zeroth Law

If system A is in thermal equilibrium with system B, and B is in thermal equilibrium with system C, then A and C are in thermal equilibrium with each other. This is why thermometers work.

First Law

\[\Delta U=Q+W_{\text{on}}\]

This version uses \(W_{\text{on}}\), the work done on the system. If work is done by the system, then \(W_{\text{on}}\) is negative.

Sign convention warning: Many physics books write \(\Delta U=Q-W_{\text{by}}\), where \(W_{\text{by}}\) is work done by the gas. Both are correct if used consistently.

Second Law

For any real process, the total entropy of the universe does not decrease:

\[\Delta S_{\text{universe}}\ge 0\]

Third Law

As a perfect crystal approaches absolute zero, its entropy approaches a minimum value. In practice, no physical process can cool a system all the way to \(0\ \text{K}\).

The Ideal Gas Law from a Physics Perspective

The chemistry form of the ideal gas law is:

\[PV=nRT\]

The physics form is often more useful because it describes particles directly:

\[PV=Nk_BT\]

Macroscopic Meaning

  • \(P\): pressure in pascals \((\text{N/m}^2)\)
  • \(V\): volume in cubic meters \((\text{m}^3)\)
  • \(T\): absolute temperature in kelvin
  • \(n\): moles of gas
  • \(N\): number of gas particles

Microscopic Meaning

  • Pressure comes from particles colliding with walls.
  • Temperature measures average translational kinetic energy.
  • Volume gives particles more or less space to move.
  • More particles means more wall collisions.

Connecting Moles and Particles

\[N=nN_A\]

where \(N_A=6.022\times10^{23}\ \text{particles/mol}\). Since \(R=N_Ak_B\), the two forms of the ideal gas law are the same physics written at different scales.

Common Physics Mistakes

Kinetic Theory of Gases

Kinetic theory explains thermodynamics using Newtonian mechanics and statistics. For an ideal monatomic gas:

\[\overline{K}_{\text{trans}}=\frac{3}{2}k_BT\]

The total internal energy is:

\[U=\frac{3}{2}Nk_BT=\frac{3}{2}nRT\]

A deeper kinetic theory result connects pressure to molecular speed:

\[PV=\frac{1}{3}Nm\overline{v^2}\]

This says pressure increases when particles are more massive, more numerous, moving faster, or confined to a smaller volume.

Quick Kinetic Theory Check

A sealed rigid container of ideal gas is heated from \(300\ \text{K}\) to \(600\ \text{K}\). What happens to pressure and average kinetic energy?

Show Solution
Since \(V\) and \(N\) are constant, \(P\propto T\), so pressure doubles. Since \(\overline{K}_{\text{trans}}=\frac{3}{2}k_BT\), average translational kinetic energy also doubles.

Calculus-Based Work in Thermodynamics

For a quasi-static expansion or compression, the work done by a gas is the area under the \(P\)-versus-\(V\) graph:

\[W_{\text{by}}=\int_{V_i}^{V_f}P(V)\,dV\]

Using work done on the gas:

\[W_{\text{on}}=-\int_{V_i}^{V_f}P(V)\,dV\]

Example: Variable Pressure Work

A gas expands according to \(P(V)=\frac{5000}{V}\), where \(P\) is in pascals and \(V\) is in cubic meters. Find the work done by the gas as it expands from \(1.0\ \text{m}^3\) to \(4.0\ \text{m}^3\).

Show Solution
\[ W_{\text{by}}=\int_1^4 \frac{5000}{V}\,dV =5000\ln(V)\Big|_1^4 =5000\ln(4) \approx 6.93\times10^3\ \text{J} \]

Example: Work from a Linear p–V Path

A gas expands from \(V=2.0\ \text{m}^3\) to \(V=6.0\ \text{m}^3\). Its pressure decreases linearly from \(500\ \text{kPa}\) to \(100\ \text{kPa}\). Find the work done by the gas.

Show Solution
The area under a straight-line p–V graph is a trapezoid:
\[ W_{\text{by}}=\frac{P_i+P_f}{2}(V_f-V_i) =\frac{500000+100000}{2}(6.0-2.0) =1.20\times10^6\ \text{J} \]

Entropy and the Second Law

Entropy change for a reversible process is calculated using:

\[\Delta S=\int\frac{dQ_{\text{rev}}}{T}\]

For a constant-temperature reversible process:

\[\Delta S=\frac{Q_{\text{rev}}}{T}\]

For heating or cooling a substance with approximately constant specific heat:

\[\Delta S=mc\ln\left(\frac{T_f}{T_i}\right)\]

Example: Entropy Change During Cooling

A \(2.0\ \text{kg}\) aluminum block cools from \(400\ \text{K}\) to \(300\ \text{K}\). Use \(c=900\ \text{J/(kg\cdot K)}\). Find the entropy change of the block.

Show Solution
\[ \Delta S=mc\ln\left(\frac{T_f}{T_i}\right) =(2.0)(900)\ln\left(\frac{300}{400}\right) \approx -5.18\times10^2\ \text{J/K} \]
The block's entropy decreases because it cools. The surroundings gain entropy, and for a real process the total entropy increases.

Example: Isothermal Expansion Entropy

One mole of ideal gas expands isothermally at \(300\ \text{K}\) from \(2.0\ \text{L}\) to \(8.0\ \text{L}\). Find \(\Delta S\).

Show Solution
For an isothermal ideal gas expansion:
\[ \Delta S=nR\ln\left(\frac{V_f}{V_i}\right) =(1)(8.314)\ln\left(\frac{8.0}{2.0}\right) \approx 11.5\ \text{J/K} \]

Heat Transfer

Heat flows from hotter objects to cooler objects through three mechanisms: conduction, convection, and radiation. The quantity of heat transferred to change the temperature of a substance (without a phase change) is:

\[Q=mc\Delta T\]

where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T = T_f - T_i\).

Conduction

Conduction is heat transfer through direct molecular contact. Energy passes from faster-vibrating molecules to slower ones. Fourier’s law gives the rate of heat flow through a material:

\[\frac{Q}{t}=\frac{kA\,\Delta T}{L}\]

where \(k\) is the thermal conductivity of the material (W/(m·K)), \(A\) is the cross-sectional area, \(\Delta T\) is the temperature difference across the material, and \(L\) is the thickness.

Physical intuition: Metals feel cold because they conduct heat away from your hand quickly. Wood at the same temperature feels warmer because its thermal conductivity is much lower.

Convection

Convection is heat transfer by the bulk motion of a fluid. Warmer fluid is less dense and rises, while cooler fluid sinks, setting up convection currents. Convection can be natural (driven by buoyancy) or forced (driven by a fan or pump).

Radiation

All objects with temperature above absolute zero emit electromagnetic radiation. The power radiated by an object is given by the Stefan–Boltzmann law:

\[P=\varepsilon\sigma A T^4\]

where \(\varepsilon\) is emissivity (0 to 1), \(\sigma = 5.67\times10^{-8}\ \text{W/(m}^2\text{K}^4)\) is the Stefan–Boltzmann constant, \(A\) is the surface area, and \(T\) is the absolute temperature in kelvin. Dark, rough surfaces have higher emissivity than shiny, polished surfaces.

Problem: Conduction Through a Window

A glass window is \(1.2\ \text{m}\) wide, \(1.5\ \text{m}\) tall, and \(0.005\ \text{m}\) thick. The inside surface is at \(20^\circ\text{C}\) and the outside surface is at \(-5^\circ\text{C}\). The thermal conductivity of glass is \(k=0.80\ \text{W/(m\cdot K)}\). Find the rate of heat loss through the window.

Show Solution
\[ \frac{Q}{t}=\frac{kA\,\Delta T}{L} =\frac{(0.80)(1.2\times 1.5)(25)}{0.005} =\frac{(0.80)(1.80)(25)}{0.005} =7200\ \text{W} \]
That is \(7.2\ \text{kW}\) of heat loss, which illustrates why single-pane windows are poor insulators.

Problem: Heating Water

How much heat is required to raise the temperature of \(0.500\ \text{kg}\) of water from \(20.0^\circ\text{C}\) to \(85.0^\circ\text{C}\)? Use \(c_{\text{water}}=4186\ \text{J/(kg\cdot K)}\).

Show Solution
\[ Q=mc\Delta T=(0.500)(4186)(85.0-20.0)=(0.500)(4186)(65.0)\approx 1.36\times10^5\ \text{J} \]
About \(136\ \text{kJ}\) of heat energy is needed.

Phase Changes and Latent Heat

Matter exists in three common phases: solid, liquid, and gas. Transitions between these phases are called phase changes. During a phase change, energy is absorbed or released while the temperature remains constant. The energy goes into breaking or forming intermolecular bonds rather than changing molecular kinetic energy.

Key Phase Transitions

Melting

Solid → Liquid. Absorbs energy (endothermic).

Freezing

Liquid → Solid. Releases energy (exothermic).

Vaporization

Liquid → Gas. Absorbs energy (endothermic).

Condensation

Gas → Liquid. Releases energy (exothermic).

Sublimation

Solid → Gas directly. Absorbs energy.

Deposition

Gas → Solid directly. Releases energy.

Latent Heat

The energy required for a phase change is calculated using:

\[Q=mL\]

where \(m\) is the mass and \(L\) is the latent heat (specific to the substance and the transition). The temperature does not change during this process.

Heat of Fusion (\(L_f\))

The latent heat of fusion is the energy per unit mass required to change a substance from solid to liquid (or released going from liquid to solid) at its melting point:

\[Q_f=mL_f\]

For water, \(L_f=3.34\times10^5\ \text{J/kg}\) at \(0^\circ\text{C}\). This means it takes \(334\ \text{kJ}\) to melt \(1\ \text{kg}\) of ice at \(0^\circ\text{C}\) into water at \(0^\circ\text{C}\).

Heat of Vaporization (\(L_v\))

The latent heat of vaporization is the energy per unit mass required to change a substance from liquid to gas (or released going from gas to liquid) at its boiling point:

\[Q_v=mL_v\]

For water, \(L_v=2.26\times10^6\ \text{J/kg}\) at \(100^\circ\text{C}\). This is nearly seven times larger than \(L_f\), because converting liquid to gas requires completely separating molecules from their neighbors.

Common mistake: Forgetting that the heat of vaporization is much larger than the heat of fusion. It takes far more energy to boil water than to melt ice of the same mass.

Heating Curves

A heating curve plots temperature versus heat added for a substance. It has five distinct regions for a substance going from solid to gas:

  1. Solid heating: Temperature rises. Use \(Q=mc_{\text{solid}}\Delta T\).
  2. Melting plateau: Temperature stays at the melting point. Use \(Q=mL_f\).
  3. Liquid heating: Temperature rises. Use \(Q=mc_{\text{liquid}}\Delta T\).
  4. Boiling plateau: Temperature stays at the boiling point. Use \(Q=mL_v\).
  5. Gas heating: Temperature rises. Use \(Q=mc_{\text{gas}}\Delta T\).

Problem: Melting Ice

How much energy is needed to melt \(0.200\ \text{kg}\) of ice at \(0^\circ\text{C}\) into water at \(0^\circ\text{C}\)?

Show Solution
\[ Q=mL_f=(0.200)(3.34\times10^5)=6.68\times10^4\ \text{J}\approx 66.8\ \text{kJ} \]

Problem: Boiling Water

How much energy is needed to completely vaporize \(0.150\ \text{kg}\) of water already at \(100^\circ\text{C}\)?

Show Solution
\[ Q=mL_v=(0.150)(2.26\times10^6)=3.39\times10^5\ \text{J}\approx 339\ \text{kJ} \]

Problem: Full Heating Curve Calculation

Calculate the total energy needed to convert \(0.100\ \text{kg}\) of ice at \(-20^\circ\text{C}\) to steam at \(120^\circ\text{C}\). Use: \(c_{\text{ice}}=2090\ \text{J/(kg\cdot K)}\), \(L_f=3.34\times10^5\ \text{J/kg}\), \(c_{\text{water}}=4186\ \text{J/(kg\cdot K)}\), \(L_v=2.26\times10^6\ \text{J/kg}\), \(c_{\text{steam}}=2010\ \text{J/(kg\cdot K)}\).

Show Solution
Step 1 — Heat ice from \(-20^\circ\text{C}\) to \(0^\circ\text{C}\):
\[Q_1=mc_{\text{ice}}\Delta T=(0.100)(2090)(20)=4180\ \text{J}\]
Step 2 — Melt ice at \(0^\circ\text{C}\):
\[Q_2=mL_f=(0.100)(3.34\times10^5)=33{,}400\ \text{J}\]
Step 3 — Heat water from \(0^\circ\text{C}\) to \(100^\circ\text{C}\):
\[Q_3=mc_{\text{water}}\Delta T=(0.100)(4186)(100)=41{,}860\ \text{J}\]
Step 4 — Vaporize water at \(100^\circ\text{C}\):
\[Q_4=mL_v=(0.100)(2.26\times10^6)=226{,}000\ \text{J}\]
Step 5 — Heat steam from \(100^\circ\text{C}\) to \(120^\circ\text{C}\):
\[Q_5=mc_{\text{steam}}\Delta T=(0.100)(2010)(20)=4020\ \text{J}\]
Total:
\[Q_{\text{total}}=4180+33{,}400+41{,}860+226{,}000+4020\approx 3.09\times10^5\ \text{J}\]
Notice that vaporization accounts for the largest share of the total energy.

Thermal Expansion

Most materials expand when heated because increased thermal energy causes atoms to vibrate with larger amplitude around their equilibrium positions. Thermal expansion is important in engineering: bridges have expansion joints, railroad tracks are laid with gaps, and concrete highways have relief joints.

Linear Expansion (One Dimension)

For a solid object with initial length \(L_0\), the change in length due to a temperature change \(\Delta T\) is:

\[\Delta L=\alpha L_0\,\Delta T\]

where \(\alpha\) is the coefficient of linear expansion in units of \(1/\text{K}\) or \(1/{}^\circ\text{C}\). The new length is:

\[L=L_0(1+\alpha\,\Delta T)\]

Area Expansion (Two Dimensions)

For a flat surface or sheet with initial area \(A_0\), the change in area is approximately:

\[\Delta A\approx 2\alpha A_0\,\Delta T\]

This approximation comes from expanding both length and width: if \(L\) and \(W\) each grow by a factor of \((1+\alpha\,\Delta T)\), then:

\[A=LW=L_0W_0(1+\alpha\,\Delta T)^2\approx A_0(1+2\alpha\,\Delta T)\]

The \(\alpha^2(\Delta T)^2\) term is negligibly small for typical temperature changes.

Volume expansion: For three-dimensional expansion, \(\Delta V\approx 3\alpha V_0\,\Delta T\) by the same reasoning. For liquids, a volumetric coefficient \(\beta\) is used directly: \(\Delta V=\beta V_0\,\Delta T\).

Problem: Expanding Steel Bridge

A steel bridge span is \(200.0\ \text{m}\) long at \(15^\circ\text{C}\). How much longer is it on a \(40^\circ\text{C}\) summer day? Use \(\alpha_{\text{steel}}=12\times10^{-6}\ /{}^\circ\text{C}\).

Show Solution
\[ \Delta L=\alpha L_0\,\Delta T=(12\times10^{-6})(200.0)(40-15)=(12\times10^{-6})(200.0)(25)=0.060\ \text{m} \]
The bridge grows by \(6.0\ \text{cm}\). This is why expansion joints are essential.

Problem: Area Expansion of an Aluminum Sheet

A square aluminum sheet has sides of \(1.00\ \text{m}\) at \(20^\circ\text{C}\). By how much does its area increase when heated to \(120^\circ\text{C}\)? Use \(\alpha_{\text{Al}}=23\times10^{-6}\ /{}^\circ\text{C}\).

Show Solution
Initial area:
\[A_0=(1.00)^2=1.00\ \text{m}^2\]
Area change:
\[ \Delta A=2\alpha A_0\,\Delta T=2(23\times10^{-6})(1.00)(120-20)=2(23\times10^{-6})(1.00)(100)=4.6\times10^{-3}\ \text{m}^2 \]
The area increases by about \(46\ \text{cm}^2\).

Constants and Known Values

Reference tables for thermodynamics calculations. Values are approximate and at standard conditions unless noted.

Fundamental Constants

ConstantSymbolValue
Boltzmann constant\(k_B\)\(1.38\times10^{-23}\ \text{J/K}\)
Universal gas constant\(R\)\(8.314\ \text{J/(mol\cdot K)}\)
Avogadro’s number\(N_A\)\(6.022\times10^{23}\ \text{mol}^{-1}\)
Stefan–Boltzmann constant\(\sigma\)\(5.67\times10^{-8}\ \text{W/(m}^2\text{K}^4)\)
Standard atmospheric pressure\(P_0\)\(1.013\times10^5\ \text{Pa}\)

Specific Heat Capacities

Substance\(c\) (J/(kg·K))
Water (liquid)4186
Ice2090
Steam2010
Aluminum900
Copper385
Iron / Steel450
Lead128
Glass840
Ethanol2440
Air (at constant pressure)1005

Latent Heats

SubstanceMelting Point (°C)\(L_f\) (J/kg)Boiling Point (°C)\(L_v\) (J/kg)
Water0\(3.34\times10^5\)100\(2.26\times10^6\)
Ethanol−114\(1.09\times10^5\)78\(8.46\times10^5\)
Lead327\(2.45\times10^4\)1750\(8.70\times10^5\)
Aluminum660\(3.97\times10^5\)2519\(1.07\times10^7\)
Copper1085\(2.07\times10^5\)2562\(4.73\times10^6\)
Nitrogen−210\(2.56\times10^4\)−196\(2.01\times10^5\)
Oxygen−219\(1.39\times10^4\)−183\(2.13\times10^5\)

Thermal Conductivities

Material\(k\) (W/(m·K))
Silver429
Copper401
Aluminum237
Iron / Steel80
Glass0.80
Concrete0.80
Wood0.08–0.16
Styrofoam0.033
Air0.026

Coefficients of Linear Expansion

Material\(\alpha\) (\(\times10^{-6}\)/°C)
Aluminum23
Brass19
Copper17
Iron / Steel12
Glass (ordinary)9
Glass (Pyrex)3.2
Concrete12
Lead29
Invar (nickel-iron alloy)0.9

Thermal Equilibrium Problems

When two or more objects at different temperatures are placed in thermal contact (and insulated from the surroundings), heat flows from the hotter object to the cooler one until they reach a common final temperature \(T_f\). By conservation of energy:

\[Q_{\text{lost}}+Q_{\text{gained}}=0\]

or equivalently, the total heat gained by cooler objects equals the total heat lost by warmer objects:

\[\sum m_i c_i (T_f-T_i)=0\]
Strategy: Write \(Q=mc(T_f-T_i)\) for each object. Objects that cool down will naturally produce a negative \(Q\) (since \(T_f < T_i\)), and objects that warm up will produce a positive \(Q\). Set the sum equal to zero and solve for \(T_f\).

Problem: Two Metals Reaching Equilibrium

A \(0.300\ \text{kg}\) copper block at \(250^\circ\text{C}\) is dropped into \(0.500\ \text{kg}\) of water at \(20.0^\circ\text{C}\) in an insulated container. Find the final equilibrium temperature. Use \(c_{\text{Cu}}=385\ \text{J/(kg\cdot K)}\) and \(c_{\text{water}}=4186\ \text{J/(kg\cdot K)}\).

Show Solution
Set heat lost by copper equal to heat gained by water:
\[ m_{\text{Cu}}c_{\text{Cu}}(T_f-250)+m_{\text{w}}c_{\text{w}}(T_f-20.0)=0 \]
Substitute:
\[ (0.300)(385)(T_f-250)+(0.500)(4186)(T_f-20.0)=0 \]
\[ 115.5(T_f-250)+2093(T_f-20.0)=0 \]
\[ 115.5\,T_f-28{,}875+2093\,T_f-41{,}860=0 \]
\[ 2208.5\,T_f=70{,}735 \]
\[ T_f\approx 32.0^\circ\text{C} \]
The final temperature is close to the initial water temperature because water has a much larger heat capacity than copper.

Problem: Ice Added to Warm Water

\(0.050\ \text{kg}\) of ice at \(0^\circ\text{C}\) is added to \(0.400\ \text{kg}\) of water at \(35.0^\circ\text{C}\) in an insulated cup. Find the final temperature. Assume all ice melts. Use \(L_f=3.34\times10^5\ \text{J/kg}\), \(c_{\text{water}}=4186\ \text{J/(kg\cdot K)}\).

Show Solution
The ice must first melt (absorbing \(Q=mL_f\)), then the resulting meltwater warms to \(T_f\), while the original warm water cools to \(T_f\):
\[ m_{\text{ice}}L_f+m_{\text{ice}}c_{\text{w}}(T_f-0)+m_{\text{w}}c_{\text{w}}(T_f-35.0)=0 \]
\[ (0.050)(3.34\times10^5)+(0.050)(4186)(T_f)+(0.400)(4186)(T_f-35.0)=0 \]
\[ 16{,}700+209.3\,T_f+1674.4\,T_f-58{,}604=0 \]
\[ 1883.7\,T_f=41{,}904 \]
\[ T_f\approx 22.2^\circ\text{C} \]
Verification: \(T_f\) is between \(0^\circ\text{C}\) and \(35^\circ\text{C}\), and all ice melts since the warm water provides enough energy. The answer is reasonable.

Problem: Three-Object Equilibrium

An insulated container holds \(0.200\ \text{kg}\) of water at \(25.0^\circ\text{C}\). A \(0.150\ \text{kg}\) aluminum block at \(90.0^\circ\text{C}\) and a \(0.100\ \text{kg}\) iron block at \(200^\circ\text{C}\) are both dropped in simultaneously. Find the final equilibrium temperature. Use \(c_{\text{water}}=4186\ \text{J/(kg\cdot K)}\), \(c_{\text{Al}}=900\ \text{J/(kg\cdot K)}\), \(c_{\text{Fe}}=450\ \text{J/(kg\cdot K)}\).

Show Solution
Write the energy balance for all three objects:
\[ m_{\text{w}}c_{\text{w}}(T_f-25.0)+m_{\text{Al}}c_{\text{Al}}(T_f-90.0)+m_{\text{Fe}}c_{\text{Fe}}(T_f-200)=0 \]
\[ (0.200)(4186)(T_f-25.0)+(0.150)(900)(T_f-90.0)+(0.100)(450)(T_f-200)=0 \]
\[ 837.2(T_f-25.0)+135(T_f-90.0)+45(T_f-200)=0 \]
\[ 837.2\,T_f-20{,}930+135\,T_f-12{,}150+45\,T_f-9000=0 \]
\[ 1017.2\,T_f=42{,}080 \]
\[ T_f\approx 41.4^\circ\text{C} \]
The water dominates because of its large specific heat, pulling \(T_f\) closer to its initial temperature.

Graph Interpretation

p–V Graphs

On a p–V diagram, what does the area under a process curve represent?

Show Answer
The area represents work done by the gas: \[W_{\text{by}}=\int P\,dV\] Expansion gives positive work by the gas. Compression gives negative work by the gas.

Cycles

A closed loop on a p–V diagram represents a heat engine cycle. What does the area inside the loop represent?

Show Answer
The enclosed area represents the net work done over one complete cycle. Clockwise cycles usually represent positive net work output by the gas.

Heating Curves

A temperature-versus-time graph has a flat region while heat is still being added. What does this mean?

Show Answer
The added energy is going into a phase change rather than increasing temperature. During a phase change, energy changes the arrangement of particles rather than their average kinetic energy.

Inquiry Lab: Determining Absolute Zero

Purpose

Use pressure-temperature data for a fixed-volume gas to estimate absolute zero. You will generate your own graph, determine a best-fit line, and extrapolate to the temperature where pressure would become zero.

Physics Background

For a fixed amount of gas at constant volume:

\[\frac{P}{T}=\text{constant}\]

Since kelvin temperature is related to Celsius temperature by:

\[T_K=T_C+273.15\]

a graph of pressure \(P\) versus Celsius temperature \(T_C\) should be approximately linear. If the line is extended until \(P=0\), the x-intercept estimates absolute zero.

Important: The graph is intentionally not provided. You must generate the graph yourself using graph paper, a spreadsheet, Desmos, Python, or another graphing tool.

Provided Data Set

Temperature, \(T_C\) (°C)Pressure, \(P\) (kPa)
−5081.2
−2589.0
098.3
25107.2
50116.0
75125.1
100133.9

Student Tasks

  1. Plot pressure \(P\) on the vertical axis and Celsius temperature \(T_C\) on the horizontal axis.
  2. Draw or calculate a best-fit line.
  3. Write the equation of your line in the form \(P=mT_C+b\).
  4. Set \(P=0\) and solve for the x-intercept.
  5. Use the x-intercept as your experimental estimate of absolute zero.
  6. Calculate percent error using \(-273.15^\circ\text{C}\) as the accepted value.
  7. Explain why this experiment supports using the kelvin temperature scale.

Blank Graph Space

Use this space if printing, or recreate it in your notebook or spreadsheet.

Analysis Questions

  1. What physical variable was held constant in this experiment?
  2. Why should pressure decrease as temperature decreases?
  3. Why is it reasonable to extrapolate the trend even though no data was collected near \(-273^\circ\text{C}\)?
  4. Why would a real gas stop behaving ideally before reaching absolute zero?
  5. How would random measurement error affect the x-intercept?
Teacher Check: Expected Result
Using a linear fit, students should get an x-intercept close to \(-273^\circ\text{C}\), though the exact result depends on their line of best fit. A reasonable student result might fall within a few degrees of the accepted value.

Practice Problems with Hidden Solutions

Problem 1: Ideal Gas Calculation

A container holds \(0.50\ \text{mol}\) of ideal gas at \(300\ \text{K}\) in a volume of \(0.012\ \text{m}^3\). Find the pressure.

Show Solution
\[ P=\frac{nRT}{V} =\frac{(0.50)(8.314)(300)}{0.012} \approx 1.04\times10^5\ \text{Pa} \]

Problem 2: First Law

A gas absorbs \(500\ \text{J}\) of heat and does \(200\ \text{J}\) of work on its surroundings. Find \(\Delta U\).

Show Solution
Using \(\Delta U=Q-W_{\text{by}}\):
\[ \Delta U=500-200=300\ \text{J} \]

Problem 3: RMS Speed

For a gas molecule of mass \(4.65\times10^{-26}\ \text{kg}\) at \(300\ \text{K}\), find the rms speed.

Show Solution
\[ v_{\text{rms}}=\sqrt{\frac{3k_BT}{m}} =\sqrt{\frac{3(1.38\times10^{-23})(300)}{4.65\times10^{-26}}} \approx 517\ \text{m/s} \]

Problem 4: Entropy of Heat Transfer

A reservoir at \(400\ \text{K}\) loses \(1200\ \text{J}\) of heat. What is its entropy change?

Show Solution
\[ \Delta S=\frac{Q}{T}=\frac{-1200}{400}=-3.0\ \text{J/K} \]

Problem 5: Conduction Rate

A copper rod is \(0.50\ \text{m}\) long and has a cross-sectional area of \(1.0\times10^{-4}\ \text{m}^2\). One end is held at \(100^\circ\text{C}\) and the other at \(0^\circ\text{C}\). Find the rate of heat transfer. Use \(k_{\text{Cu}}=401\ \text{W/(m\cdot K)}\).

Show Solution
\[ \frac{Q}{t}=\frac{kA\,\Delta T}{L}=\frac{(401)(1.0\times10^{-4})(100)}{0.50}=8.0\ \text{W} \]

Problem 6: Latent Heat of Fusion

A \(0.500\ \text{kg}\) block of lead at its melting point (\(327^\circ\text{C}\)) is completely melted. How much energy was absorbed? Use \(L_f=2.45\times10^4\ \text{J/kg}\).

Show Solution
\[ Q=mL_f=(0.500)(2.45\times10^4)=1.23\times10^4\ \text{J}\approx 12.3\ \text{kJ} \]

Problem 7: Linear Thermal Expansion

A brass rod is \(1.500\ \text{m}\) at \(20^\circ\text{C}\). What is its length at \(180^\circ\text{C}\)? Use \(\alpha_{\text{brass}}=19\times10^{-6}\ /{}^\circ\text{C}\).

Show Solution
\[ \Delta L=\alpha L_0\,\Delta T=(19\times10^{-6})(1.500)(160)=4.56\times10^{-3}\ \text{m} \]
\[ L=L_0+\Delta L=1.500+0.00456=1.505\ \text{m} \]

Problem 8: Thermal Equilibrium

A \(0.250\ \text{kg}\) iron horseshoe at \(500^\circ\text{C}\) is plunged into \(2.00\ \text{kg}\) of water at \(22.0^\circ\text{C}\) in an insulated bucket. Find the final temperature. Use \(c_{\text{Fe}}=450\ \text{J/(kg\cdot K)}\) and \(c_{\text{water}}=4186\ \text{J/(kg\cdot K)}\).

Show Solution
\[ m_{\text{Fe}}c_{\text{Fe}}(T_f-500)+m_{\text{w}}c_{\text{w}}(T_f-22.0)=0 \]
\[ (0.250)(450)(T_f-500)+(2.00)(4186)(T_f-22.0)=0 \]
\[ 112.5\,T_f-56{,}250+8372\,T_f-184{,}184=0 \]
\[ 8484.5\,T_f=240{,}434 \]
\[ T_f\approx 28.3^\circ\text{C} \]
The large mass and specific heat of the water keep the final temperature close to the water’s initial temperature.

Problem 9: Phase Change Energy

How much total energy is needed to convert \(0.300\ \text{kg}\) of ice at \(-10^\circ\text{C}\) into water at \(25^\circ\text{C}\)? Use \(c_{\text{ice}}=2090\ \text{J/(kg\cdot K)}\), \(L_f=3.34\times10^5\ \text{J/kg}\), \(c_{\text{water}}=4186\ \text{J/(kg\cdot K)}\).

Show Solution
Step 1 — Heat ice to \(0^\circ\text{C}\):
\[Q_1=(0.300)(2090)(10)=6270\ \text{J}\]
Step 2 — Melt ice:
\[Q_2=(0.300)(3.34\times10^5)=100{,}200\ \text{J}\]
Step 3 — Heat water to \(25^\circ\text{C}\):
\[Q_3=(0.300)(4186)(25)=31{,}395\ \text{J}\]
Total:
\[Q=6270+100{,}200+31{,}395\approx 1.38\times10^5\ \text{J}\]

Check Understanding

Use these quick checks to test the main ideas. Your score is not saved.

1. In the ideal gas law, which temperature scale must be used?

2. On a p–V graph, the area under the curve represents:

3. If an ideal gas is held at constant volume and temperature doubles in kelvin, pressure:

4. For an ideal monatomic gas, internal energy depends directly on:

5. A real irreversible process has:

6. Which mechanism of heat transfer does NOT require a medium?

7. During a phase change, the temperature of the substance:

8. For water, the heat of vaporization compared to the heat of fusion is:

9. A steel beam expands when heated. If the temperature change doubles, the expansion:

10. For area expansion of a flat sheet, the coefficient is approximately:

11. When a hot metal block is placed in cool water in an insulated container, at equilibrium:

12. In Fourier’s law of conduction, doubling the thickness of a wall while keeping everything else the same will: