Chapter 11 · Thermodynamics

11.1 Thermal Equilibrium, the Zeroth Law and the First Law

Bodies in thermal contact settle to a common temperature — the zeroth law makes that a measurable property. Energy crosses the boundary of a system as heat $Q$ or as work $W$, and the first law balances the account: $\Delta U=Q-W$.

Leave a cup of hot tea and a glass of cold juice on the same table and, an hour later, both are at room temperature. Put a thermometer under your tongue and you are told to wait a minute before reading it. Pump a bicycle tyre and the pump gets hot; let the air out of the tyre and the valve feels cold. Each of these is a small piece of thermodynamics: bodies exchanging energy until they stop changing, and energy entering or leaving a body in two quite different ways — as heat, because of a temperature difference, or as work, because something was pushed through a distance.

This section sets up the language of thermodynamics. You will meet systems, their surroundings and the walls between them, the idea of thermal equilibrium, and the zeroth law, which is what makes a thermometer reading mean something. Then you will separate heat, work and internal energy, learn to read the work done by a gas as an area on a $p$–$V$ diagram, and put everything together in the first law of thermodynamics — the law of conservation of energy for systems that can be heated.

Systems, walls and thermal equilibrium

In mechanics we followed a few bodies, each with a position and a velocity. A litre of air contains about $2.5\times10^{22}$ molecules, and following each of them is hopeless — and unnecessary. Thermodynamics describes a large collection of particles by a few bulk (macroscopic) quantities that you can measure directly: pressure $p$, volume $V$, temperature $T$, mass, internal energy. It does not ask what any single molecule is doing.

📘 Definition 11.1 — System, surroundings and walls

A thermodynamic system is the part of the universe we choose to study — a gas in a cylinder, the water in a kettle, a block of ice. Everything outside it that can exchange energy with it is its surroundings. The system is separated from its surroundings by a boundary, or wall:

an adiabatic wall is a perfect insulator: no heat can flow through it (the double wall of a vacuum flask is a good approximation);
a diathermic wall lets heat flow freely (a thin sheet of copper).

A wall may be fixed or movable (a piston): a movable wall lets the system exchange energy by doing work, even if the wall is adiabatic.

Put two systems A and B in contact through a diathermic wall. If one was hotter, energy flows from it to the other, and the pressures and volumes of both change. After a while every change stops: the bulk quantities of A and of B have settled to constant values. Separate A and B with an adiabatic wall instead, and each keeps whatever state it had — they cannot influence each other thermally.

📘 Definition 11.2 — Thermal equilibrium

A system is in equilibrium when its macroscopic variables (pressure, volume, temperature, composition) do not change with time. Two systems are in thermal equilibrium with each other if, when they are placed in contact through a diathermic wall, there is no net flow of heat between them and neither state changes. Systems in thermal equilibrium have the same temperature.

Thermal equilibrium needs a diathermic contact; mechanical equilibrium (no net force on a movable wall — equal pressures on both sides) and chemical equilibrium (no change of composition) are separate conditions. In this chapter a system "in equilibrium" is in all three at once.

Now take three systems. A and B are separated by an adiabatic wall, and each is in contact with a third system C through diathermic walls. Wait until A and C are in thermal equilibrium, and B and C are in thermal equilibrium. Then replace the wall between A and B by a diathermic one. Experiment shows that nothing happens: no heat flows from A to B. This is so basic that it was formulated as a law only after the first and second laws had been numbered — hence its name.

📐 Law 11.1 — Zeroth law of thermodynamics

If two systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

So thermal equilibrium is transitive, like equality of numbers. It follows that there is a property that has the same value for all systems in thermal equilibrium with one another. This property is the temperature: two systems are in thermal equilibrium if and only if their temperatures are equal.

The zeroth law is the reason a thermometer works. The thermometer is system C. When it is in thermal equilibrium with your body, its reading (the length of a mercury column, the resistance of a wire, the pressure of a gas — Chapter 10) has the value that belongs to your temperature; any other body that gives the same reading is at the same temperature as you, even though you have never touched it. Notice one practical point: a thermometer always shows its own temperature. It shows the temperature of the body only after the two have reached equilibrium — and the heat it takes from the body on the way changes that body a little.

✏️ Example 1 — A thermometer measures its own temperature

A thermometer at room temperature, $25.0\ ^\circ\text{C}$, has a heat capacity of $2.0\ \text{J/K}$. It is dipped into a small insulated sample of $5.0\ \text{g}$ of water at $60.0\ ^\circ\text{C}$ ($c=4186\ \text{J kg}^{-1}\text{K}^{-1}$). (a) Why must you wait before reading it? (b) What does it read once equilibrium is reached? (c) How much heat did the thermometer take from the water?

💡 Show solution▶

(a) The thermometer shows its own temperature. Heat must flow from the water into it until the two are in thermal equilibrium; only then (zeroth law) is its reading the temperature of the water.

(b) Heat lost by water = heat gained by thermometer (insulated): $0.0050\times4186\,(60.0-T)=2.0\,(T-25.0)$. With $mc=20.93\ \text{J/K}$: $T=\dfrac{20.93\times60.0+2.0\times25.0}{20.93+2.0}=\dfrac{1305.8}{22.93}=56.9\ ^\circ\text{C}$.

Check The water lost $20.93\times(60.0-56.95)=64\ \text{J}$ ✓. A good thermometer has a heat capacity much smaller than that of the body it measures; here it does not, so the reading is $3.1\ \text{K}$ below the original temperature of the water.

Three blocks with heat capacities $C_A=3.0$, $C_B=1.0$ and $C_C=2.0\ \text{kJ/K}$ sit in an insulating box. Choose the starting temperatures of A and C (B starts at $20\ ^\circ\text{C}$) and which walls let heat through. Press Play: heat flows from hot to cold through every diathermic wall (the arrows), at a rate proportional to the temperature difference, and the graph records the temperatures. Start with A and C touching only B; when both are in equilibrium with B, switch the A–C wall to diathermic and see whether anything happens.

$T_A$:80 °C$T_C$:5 °Cwalls:

$T_A$—

$T_B$—

$T_C$—

common final $T$—

Heat, work and internal energy

A gas in a cylinder can be warmed in two ways. You can stand it on a hot plate: energy flows in because the plate is hotter. Or you can push the piston in quickly: you do work on the gas, and it warms up just the same. Once the energy is inside, the gas cannot tell which route it came by — that was the point of the teaser on the chapter page. We need a name for the energy that is in the system, and separate names for the two ways of getting it there.

📘 Definition 11.3 — Internal energy

The internal energy $U$ of a system is the total energy of its molecules measured in the frame in which the system as a whole is at rest: the sum of the kinetic energies of their random motion (translation, rotation, vibration) and the potential energies of the forces between them. It does not include the kinetic energy of the system moving as a whole or its potential energy in an external field.

Internal energy depends only on the state of the system — on its $p$, $V$, $T$ — and not on how it got there. For an ideal gas the molecules exert no forces on each other except in collisions, so $U$ depends on the temperature only: $U=U(T)$. For a monatomic ideal gas (helium, argon) $U=\tfrac32nRT$ (Chapter 12 — Kinetic Theory explains why).

📘 Definition 11.4 — Heat and work

Heat $Q$ is energy transferred between a system and its surroundings because of a temperature difference between them. Work $W$ is energy transferred by any other means — by a force acting through a displacement, such as a gas pushing a piston, a paddle stirring a liquid or an electric current through a resistor.

Heat and work are energy in transit. A body contains internal energy; it does not contain "heat" or "work". Both are measured in joules.

The distinction is between a quantity of the state and quantities of the process. A bank balance is like $U$: it has a definite value at any moment. Deposits by cash and deposits by cheque are like $Q$ and $W$: they tell you how the balance changed, and the same final balance can be reached by many different mixtures of them. Asking "how much heat is in this gas?" is like asking how much of your balance is cash.

Work done by a gas. A gas at pressure $p$ in a cylinder pushes on a piston of area $A$ with a force $pA$. If the piston moves out a small distance $dx$, the gas does work $dW=pA\,dx=p\,dV$, where $dV=A\,dx$ is the increase in volume. While the gas expands $dV$ is positive and the gas does positive work on the surroundings; when it is compressed $dV$ is negative and $W$ is negative — the surroundings do work on the gas. Adding up the small pieces (the integral as an area, Section 2.2):

📐 Result 11.2 — Work done by a gas

When a gas changes volume slowly from $V_1$ to $V_2$, keeping a well-defined pressure $p$ at every stage, the work done by the gas is

$$W=\int_{V_1}^{V_2}p\,dV,$$

the area under the curve on a $p$–$V$ diagram — positive for an expansion, negative for a compression. At constant pressure it is simply

$$W=p\,(V_2-V_1)=p\,\Delta V .$$

The work depends on the path between the initial and final states, not just on the end points.

Units: $1\ \text{Pa}\times1\ \text{m}^3=1\ \text{N m}=1\ \text{J}$. Volumes are often given in litres ($1\ \text{L}=10^{-3}\ \text{m}^3$) and pressures in units of $10^5\ \text{Pa}$, so a handy rule is $(10^5\ \text{Pa})\times(1\ \text{L})=100\ \text{J}$. A graph of $p$ against $V$ that records every state of a gas during a process is called a $p$–$V$ diagram (or indicator diagram); engineers used them to measure the work of steam engines long before thermodynamics had a theory.

✏️ Example 2 — Expansion and compression at constant pressure

(a) A gas at a constant pressure of $1.0\times10^5\ \text{Pa}$ expands from $2.0\ \text{L}$ to $5.0\ \text{L}$. Find the work done by the gas. (b) The gas is then compressed at the same pressure from $3.0\ \text{L}$ to $1.0\ \text{L}$. Find the work done by the gas and the work done on it.

💡 Show solution▶

(a) $W=p\,\Delta V=1.0\times10^5\times(5.0-2.0)\times10^{-3}=3.0\times10^2\ \text{J}$, done by the gas on the piston.

(b) $W=1.0\times10^5\times(1.0-3.0)\times10^{-3}=-2.0\times10^2\ \text{J}$. The negative sign means the surroundings did $+200\ \text{J}$ of work on the gas.

Check Rule of thumb: $(1.0\times10^5\ \text{Pa})(3.0\ \text{L})=3.0\times100\ \text{J}=300\ \text{J}$ ✓.

✏️ Example 3 — Three paths, three amounts of work

A gas goes from state A ($V_A=1.0\ \text{L}$, $p_A=1.0\times10^5\ \text{Pa}$) to state B ($V_B=3.0\ \text{L}$, $p_B=3.0\times10^5\ \text{Pa}$). Find the work done by the gas if (a) the state point moves along a straight line from A to B on the $p$–$V$ diagram; (b) the pressure is first raised to $p_B$ at constant volume, then the gas expands at constant pressure; (c) the gas first expands at constant pressure $p_A$, then the pressure is raised at constant volume.

💡 Show solution▶

(a) The area under the straight line is a trapezium: $W=\tfrac12(p_A+p_B)(V_B-V_A)=\tfrac12(4.0\times10^5)(2.0\times10^{-3})=4.0\times10^2\ \text{J}$.

(b) Constant volume: no work. Then $W=p_B\,\Delta V=3.0\times10^5\times2.0\times10^{-3}=6.0\times10^2\ \text{J}$.

Check Same end points, three different works ($200$, $400$, $600\ \text{J}$): work is a property of the path. Interactive 2 draws all three paths.

The first law of thermodynamics

Suppose heat $Q$ flows into a system and the system does work $W$ on its surroundings. Energy is conserved, so whatever came in and did not go out as work must still be inside, as an increase in internal energy. Joule's experiments in the 1840s — warming water by stirring it with paddles driven by falling weights, and by heating it — showed that a given rise in temperature always needs the same number of joules, whatever mixture of work and heat supplies them. This is the experimental basis of the first law.

📐 Law 11.3 — First law of thermodynamics

For any process in which a system goes from one equilibrium state to another,

$$\Delta U=Q-W\qquad\text{or}\qquad Q=\Delta U+W,$$

where $Q$ is the heat supplied to the system and $W$ is the work done by the system. The change in internal energy $\Delta U=U_2-U_1$ depends only on the initial and final states, although $Q$ and $W$ separately depend on the process. For a small step, $dQ=dU+p\,dV$.

Sign convention (used throughout this course). $Q\gt0$: heat flows into the system; $Q\lt0$: heat flows out. $W\gt0$: the system does work on the surroundings (it expands); $W\lt0$: work is done on the system (it is compressed). $\Delta U\gt0$: the internal energy increases (for an ideal gas, the temperature rises). Some books, and chemistry courses, write $\Delta U=Q+W$ with $W$ the work done on the system; the physics is the same, only the sign of $W$ is flipped.

🧭 Problem-solving strategy — the first law

Say clearly what the system is (usually the gas).
Write each energy transfer with its sign: heat in $+$, heat out $-$; work by the system $+$, work on the system $-$.
Find what you can directly: $W$ from the area under the $p$–$V$ path; $W=0$ at constant volume; $Q=0$ for an insulated system or a very fast change; $\Delta U=0$ for an ideal gas at constant temperature or for a complete cycle.
Use $\Delta U=Q-W$ for the remaining quantity. Since $\Delta U$ is the same for every path between two states, find it on the easiest path and reuse it on the hard one.

✏️ Example 4 — Keeping the signs straight

(a) A gas absorbs $1500\ \text{J}$ of heat and does $600\ \text{J}$ of work pushing out a piston. Find the change in its internal energy. (b) Another gas is compressed: $400\ \text{J}$ of work is done on it, and at the same time it gives out $150\ \text{J}$ of heat to its surroundings. Find $\Delta U$. Does its temperature rise or fall?

💡 Show solution▶

(a) $Q=+1500\ \text{J}$, $W=+600\ \text{J}$: $\Delta U=1500-600=+900\ \text{J}$.

(b) $Q=-150\ \text{J}$ (heat out), $W=-400\ \text{J}$ (work on the gas): $\Delta U=-150-(-400)=+250\ \text{J}$. The internal energy rises, so (for an ideal gas) the temperature rises.

Check In (b), $400\ \text{J}$ came in as work and only $150\ \text{J}$ left as heat, so $250\ \text{J}$ stayed inside ✓.

✏️ Example 5 — Where does the latent heat go?

$1.00\ \text{g}$ of water at $100\ ^\circ\text{C}$ is boiled at atmospheric pressure, $1.013\times10^5\ \text{Pa}$, turning into $1671\ \text{cm}^3$ of steam at $100\ ^\circ\text{C}$. The volume of the liquid was $1.00\ \text{cm}^3$ and the latent heat of vaporisation is $2256\ \text{J/g}$. Find (a) the work done by the water as it expands into steam, (b) the change in its internal energy.

💡 Show solution▶

(a) The steam pushes back the atmosphere at constant pressure: $W=p\,\Delta V=1.013\times10^5\times(1671-1.00)\times10^{-6}=169\ \text{J}$.

(b) $Q=2256\ \text{J}$ (absorbed). $\Delta U=Q-W=2256-169=2087\ \text{J}\approx2.09\times10^3\ \text{J}$.

Check The temperature does not change, yet $U$ rises by about $93\%$ of the heat supplied: that energy goes into pulling the molecules apart against their attractions (potential energy). Only about $7\%$ is spent pushing back the air. Water and steam are not ideal gases, so "$U$ depends only on $T$" does not apply here.

✏️ Example 6 — One change of internal energy, three heats

The gas of Example 3 is a monatomic ideal gas, for which $U=\tfrac32nRT=\tfrac32pV$. (a) Find $\Delta U$ from A to B. (b) Find the heat absorbed along each of the three paths of Example 3. (c) By what factor does its temperature rise?

💡 Show solution▶

(a) $\Delta U=\tfrac32(p_BV_B-p_AV_A)=\tfrac32(900-100)\ \text{J}=1.2\times10^3\ \text{J}$, using $pV$ in joules: $p_BV_B=3.0\times10^5\times3.0\times10^{-3}=900\ \text{J}$, $p_AV_A=100\ \text{J}$.

(b) $Q=\Delta U+W$: straight line $1200+400=1.6\times10^3\ \text{J}$; constant volume first $1200+600=1.8\times10^3\ \text{J}$; constant pressure first $1200+200=1.4\times10^3\ \text{J}$.

Check $Q-W$ is $1200\ \text{J}$ on every path ✓ — that is the first law. Interactive 2 starts from these values.

A monatomic ideal gas ($U=\tfrac32pV$) goes from the fixed state A to a state B that you choose. Pick a path: the state point travels along it and the area under the path — the work done by the gas — fills in. Compare the three paths: $W$ and $Q$ change, $\Delta U=Q-W$ does not. Make $V_B\lt V_A$ to see compressions, where the area counts as negative.

$V_B$:3.0 L$p_B$:3.0 × 10⁵ Papath:

work by gas $W$—

$\Delta U=\tfrac32(p_BV_B-p_AV_A)$—

heat in $Q=\Delta U+W$—

$T_B/T_A$—

⚠️ Common mistakes

1. Sign of $W$. In $\Delta U=Q-W$, $W$ is the work done by the gas. Work done on the gas is $-W$. Decide the sign of each transfer before you substitute.

2. "The gas contains a lot of heat." A system contains internal energy. Heat and work only describe energy crossing the boundary.

3. Heat and temperature. Supplying heat does not always raise the temperature (boiling water, an isothermal expansion), and the temperature can rise with no heat at all (a fast compression).

4. Work without a path. You cannot find $W$ (or $Q$) from the end states alone; you need the path. $\Delta U$, on the other hand, needs only the end states.

5. Units. Convert litres to $\text{m}^3$ and kPa or atm to Pa before multiplying: $(10^5\ \text{Pa})(1\ \text{L})=100\ \text{J}$, not $10^5\ \text{J}$.

6. Thermal equilibrium means equal temperature, not equal internal energy. A bathtub and a cup of water at the same temperature are in equilibrium although the bath has far more internal energy.

Exercises

Take $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$, $g=9.8\ \text{m/s}^2$, $c_\text{water}=4186\ \text{J kg}^{-1}\text{K}^{-1}$ and $1\ \text{atm}=1.013\times10^5\ \text{Pa}$. $W$ is always the work done by the system, so $\Delta U=Q-W$.

Thermal equilibrium and the zeroth law.

1 What is meant by thermal equilibrium between two systems? Which quantity must be the same for both?
2 State the zeroth law of thermodynamics and explain how it justifies the use of a thermometer.
3 Distinguish between an adiabatic wall and a diathermic wall, giving one practical example of each.
4 System P is in thermal equilibrium with Q, but Q is not in thermal equilibrium with R. Can P be in thermal equilibrium with R? Explain.
5 Why must a clinical thermometer be held under the tongue for about a minute before it is read?
6 A thermometer of heat capacity $3.0\ \text{J/K}$, initially at $20.0\ ^\circ\text{C}$, is placed in $10.0\ \text{g}$ of water at $50.0\ ^\circ\text{C}$ in an insulated cup. What does it read at equilibrium?
7 A large block of iron and a small iron nail, both at $40\ ^\circ\text{C}$, are placed in contact. Does heat flow between them? Which has more internal energy?

💡 Show solutions▶

1 When the two systems are in contact through a diathermic wall and there is no net heat flow, so neither state changes. Their temperatures are equal.

2 If A and B are each in thermal equilibrium with C, they are in thermal equilibrium with each other. The thermometer is C: bodies that give it the same reading are in equilibrium with it, hence with each other, and so have the same temperature — without ever touching each other.

3 An adiabatic wall allows no heat flow (the evacuated double wall of a vacuum flask, thick polystyrene); a diathermic wall allows heat to flow freely (the thin metal base of a cooking pot, a copper sheet).

4 No. If P were in equilibrium with R as well, then Q and R would both be in equilibrium with P, and by the zeroth law with each other — which contradicts the data.

5 The thermometer shows its own temperature. Heat must flow from the mouth into the bulb until they reach thermal equilibrium; only then does the reading equal body temperature.

6 $0.0100\times4186\,(50.0-T)=3.0\,(T-20.0)$, so $T=\dfrac{41.86\times50.0+3.0\times20.0}{44.86}=48.0\ ^\circ\text{C}$.

7 No heat flows — they are at the same temperature, so already in thermal equilibrium. The block has far more internal energy (more molecules); equilibrium needs equal temperature, not equal energy.

Heat, work and internal energy.

1 A gas at a constant pressure of $2.0\times10^5\ \text{Pa}$ expands from $3.0\ \text{L}$ to $7.5\ \text{L}$. Find the work done by the gas.
2 A gas is compressed at a constant pressure of $1.5\times10^5\ \text{Pa}$ from $6.0\ \text{L}$ to $2.0\ \text{L}$. Find the work done by the gas and the work done on it.
3 A gas is heated in a closed rigid steel container. How much work does it do? Where does all the heat go?
4 On a $p$–$V$ diagram a gas goes along a straight line from $(2.0\ \text{L},\ 4.0\times10^5\ \text{Pa})$ to $(6.0\ \text{L},\ 1.0\times10^5\ \text{Pa})$. Find the work done by the gas.
5 Explain why the statement "a hot body contains more heat than a cold one" is wrong. What should be said instead?
6 Explain, with an example, why $\Delta U$ depends only on the initial and final states while $Q$ and $W$ depend on the path.
7 Gas in a cylinder with a piston of area $0.010\ \text{m}^2$ pushes the piston out by $5.0\ \text{cm}$ at a constant pressure of $1.2\times10^5\ \text{Pa}$. Find the work done.

💡 Show solutions▶

1 $W=2.0\times10^5\times4.5\times10^{-3}=9.0\times10^2\ \text{J}$.

2 $W=1.5\times10^5\times(-4.0\times10^{-3})=-6.0\times10^2\ \text{J}$ by the gas; $+600\ \text{J}$ of work is done on it.

3 $\Delta V=0$, so $W=0$; all the heat goes into internal energy, $\Delta U=Q$, and the temperature and pressure rise.

4 Trapezium: $W=\tfrac12(4.0+1.0)\times10^5\times4.0\times10^{-3}=1.0\times10^3\ \text{J}$ (positive — the gas expands, even though its pressure falls).

5 Heat is energy in transit because of a temperature difference; a body does not contain it. A hotter body (of the same material and mass) has more internal energy, and heat flows from it to a colder body in contact.

6 $U$ is fixed by the state ($p$, $V$, $T$), so $\Delta U=U_2-U_1$ is fixed by the end states. $W$ is the area under the path, which differs from path to path (Example 3: $200$, $400$, $600\ \text{J}$), and then $Q=\Delta U+W$ differs too.

7 $\Delta V=0.010\times0.050=5.0\times10^{-4}\ \text{m}^3$; $W=1.2\times10^5\times5.0\times10^{-4}=60\ \text{J}$.

III

The first law.

1 A gas absorbs $800\ \text{J}$ of heat and does $300\ \text{J}$ of work. Find $\Delta U$.
2 $250\ \text{J}$ of work is done on a gas while it gives out $100\ \text{J}$ of heat. Find $\Delta U$.
3 A gas in a perfectly insulated cylinder expands and does $450\ \text{J}$ of work. Find $\Delta U$. Does the gas get hotter or colder?
4 A gas absorbs $600\ \text{J}$ of heat at constant volume. Find $W$ and $\Delta U$.
5 $2.0\ \text{mol}$ of a monatomic ideal gas ($U=\tfrac32nRT$) is heated from $300\ \text{K}$ to $400\ \text{K}$ in a rigid container. Find $\Delta U$ and $Q$.
6 The same gas is heated from $300\ \text{K}$ to $400\ \text{K}$ at constant pressure. Using $pV=nRT$, find the work done by the gas, then $Q$.
7 An ideal gas expands at constant temperature, absorbing $1000\ \text{J}$ of heat. Find $\Delta U$ and $W$.
8 $2.0\ \text{g}$ of water at $100\ ^\circ\text{C}$ is boiled at $1\ \text{atm}$. Each gram of liquid ($1.00\ \text{cm}^3$) becomes $1671\ \text{cm}^3$ of steam; the latent heat is $2256\ \text{J/g}$. Find $Q$, $W$ and $\Delta U$.

💡 Show solutions▶

1 $\Delta U=800-300=+500\ \text{J}$.

2 $Q=-100\ \text{J}$, $W=-250\ \text{J}$: $\Delta U=-100+250=+150\ \text{J}$.

3 $Q=0$: $\Delta U=-450\ \text{J}$. The internal energy falls, so the gas gets colder — the energy for the work came from inside it.

4 $W=0$; $\Delta U=Q=+600\ \text{J}$.

5 $\Delta U=\tfrac32(2.0)(8.31)(100)=2.49\times10^3\ \text{J}$; $W=0$, so $Q=2.49\times10^3\ \text{J}$.

6 $W=p\,\Delta V=nR\,\Delta T=2.0\times8.31\times100=1.66\times10^3\ \text{J}$. $\Delta U$ is the same as in part 5 (same $\Delta T$): $Q=2493+1662=4.16\times10^3\ \text{J}$ — more heat is needed at constant pressure (Section 11.2).

7 $U$ of an ideal gas depends only on $T$: $\Delta U=0$, so $W=Q=1000\ \text{J}$ — all the heat is turned into work.

8 $Q=2.0\times2256=4.51\times10^3\ \text{J}$; $W=1.013\times10^5\times2\times1670\times10^{-6}=338\ \text{J}$; $\Delta U=4512-338=4.17\times10^3\ \text{J}$.

Applications and harder problems.

1 (Modelling) In one quick stroke of a bicycle pump, $24\ \text{J}$ of work is done on $0.020\ \text{mol}$ of air, and $6.0\ \text{J}$ of heat leaks into the barrel. Treating air as an ideal gas with $U=\tfrac52nRT$, find $\Delta U$ and the rise in temperature of the air.
2 (Modelling) During a long climb, a hiker does $180\ \text{kJ}$ of work against gravity and the internal energy of her body (stored food energy) falls by $650\ \text{kJ}$. How much heat does her body give out?
3 A gas goes from state A to state B along path 1, absorbing $80\ \text{J}$ of heat and doing $30\ \text{J}$ of work. (a) Along path 2 it does $10\ \text{J}$ of work; how much heat does it absorb? (b) It returns from B to A along path 3, with $20\ \text{J}$ of work done on it. Find the heat absorbed or released.
4 A $10\ \text{g}$ lead bullet moving at $300\ \text{m/s}$ stops in a wooden block. Half of its kinetic energy goes into the internal energy of the bullet ($c_\text{lead}=128\ \text{J kg}^{-1}\text{K}^{-1}$). Find the rise in its temperature.
5 Water falls $50\ \text{m}$ down a waterfall. If all its potential energy became internal energy of the water, by how much would its temperature rise?
6 A gas expands at a constant pressure of $1.0\times10^5\ \text{Pa}$ from $2.0\ \text{L}$ to $6.0\ \text{L}$ and absorbs $1000\ \text{J}$ of heat. Find $W$ and $\Delta U$. Show that this is consistent with the gas being a monatomic ideal gas, $U=\tfrac32pV$.

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1 $Q=-6.0\ \text{J}$, $W=-24\ \text{J}$: $\Delta U=-6.0+24=+18\ \text{J}$. $\Delta T=\dfrac{\Delta U}{\tfrac52nR}=\dfrac{18}{2.5\times0.020\times8.31}=43\ \text{K}$ — which is why the pump barrel warms up.

2 $\Delta U=-650\ \text{kJ}$, $W=+180\ \text{kJ}$: $Q=\Delta U+W=-470\ \text{kJ}$, i.e. $470\ \text{kJ}$ of heat is given out (sweat and warm breath carry it away).

3 $\Delta U_{AB}=80-30=50\ \text{J}$. (a) $Q_2=50+10=60\ \text{J}$ absorbed. (b) $\Delta U_{BA}=-50\ \text{J}$, $W=-20\ \text{J}$: $Q_3=-50-20=-70\ \text{J}$, so $70\ \text{J}$ is released.

4 $K=\tfrac12(0.010)(300)^2=450\ \text{J}$; $\Delta U=225\ \text{J}$; $\Delta T=\dfrac{225}{0.010\times128}=1.8\times10^2\ \text{K}$ ($176\ \text{K}$) — enough to soften lead.

5 $mgh=mc\,\Delta T$: $\Delta T=\dfrac{gh}{c}=\dfrac{9.8\times50}{4186}=0.12\ \text{K}$, independent of the mass.

6 $W=1.0\times10^5\times4.0\times10^{-3}=400\ \text{J}$; $\Delta U=1000-400=600\ \text{J}$. For a monatomic ideal gas at constant $p$: $\Delta U=\tfrac32p\,\Delta V=\tfrac32\times400=600\ \text{J}$ ✓.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks State the zeroth law of thermodynamics. Explain how it leads to the concept of temperature.

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Statement If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other. [1]

Temperature Thermal equilibrium is therefore transitive, so all systems in mutual equilibrium share some property with a common value; this property is temperature. Two systems are in thermal equilibrium exactly when their temperatures are equal, and a thermometer (the third system) can compare bodies that never touch. [1]

Exam tip Mention "diathermic contact" and "no net heat flow" — the words examiners look for in the definition of thermal equilibrium.

E2. NEET A system absorbs $2.0\ \text{kJ}$ of heat while $500\ \text{J}$ of work is done on it. The change in its internal energy is (a) $1.5\ \text{kJ}$ (b) $2.5\ \text{kJ}$ (c) $-2.5\ \text{kJ}$ (d) $2.0\ \text{kJ}$

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Idea $Q=+2000\ \text{J}$; work done on the system means $W=-500\ \text{J}$. $\Delta U=Q-W=2000+500=2500\ \text{J}$.

Answer (b) $2.5\ \text{kJ}$.

Exam tip (a) treats the $500\ \text{J}$ as work done by the system. Both transfers here bring energy in, so $\Delta U$ must exceed each of them.

E3. JEE Main (Numerical value) A gas is taken from state A to state B along path 1, absorbing $500\ \text{J}$ of heat and doing $200\ \text{J}$ of work. Along path 2 it absorbs $420\ \text{J}$. It is then brought back from B to A along path 3, during which $150\ \text{J}$ of work is done on it. Find the heat, in joules, released by the gas along path 3.

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Path 1 $\Delta U_{AB}=Q-W=500-200=300\ \text{J}$, the same for every path from A to B (so along path 2 the gas does $420-300=120\ \text{J}$ of work).

Path 3 $\Delta U_{BA}=-300\ \text{J}$, $W=-150\ \text{J}$: $Q=\Delta U+W=-300-150=-450\ \text{J}$.

Answer $450$ (J released).

Exam tip Over the full cycle A → B (path 1) → A: net $Q=500-450=50\ \text{J}$ and net $W=200-150=50\ \text{J}$ ✓, $\Delta U=0$.

E4. JEE Advanced (One or more correct) A monatomic ideal gas ($U=\tfrac32pV$) is taken from A $(p_0,V_0)$ to B $(2p_0,2V_0)$ by three paths: I, a straight line on the $p$–$V$ diagram; II, constant volume then constant pressure; III, constant pressure then constant volume. Which statements are correct? (A) $\Delta U=4.5\,p_0V_0$ on every path. (B) The work done along I is $1.5\,p_0V_0$. (C) The heat absorbed along II is $5.5\,p_0V_0$. (D) Less heat is absorbed along III than along either of the other two paths.

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(A) $\Delta U=\tfrac32(4p_0V_0-p_0V_0)=4.5\,p_0V_0$, path-independent ✓.

(B) Trapezium: $W_I=\tfrac12(p_0+2p_0)V_0=1.5\,p_0V_0$ ✓.

(D) $W_{III}=p_0V_0$: $Q_{III}=5.5\,p_0V_0\lt Q_I=6.0\,p_0V_0\lt Q_{II}=6.5\,p_0V_0$ ✓.

Answer (A), (B), (D).

Exam tip With $\Delta U$ fixed, the path with the least area under it needs the least heat — compare areas instead of computing every $Q$.

E5. NEET JEE Main Water falls from a height of $84\ \text{m}$. If all of its gravitational potential energy is converted into its internal energy, the rise in its temperature is about ($g=9.8\ \text{m/s}^2$, $c=4200\ \text{J kg}^{-1}\text{K}^{-1}$) (a) $0.020\ \text{K}$ (b) $0.20\ \text{K}$ (c) $2.0\ \text{K}$ (d) $0.82\ \text{K}$

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Idea $mgh=mc\,\Delta T$, so $\Delta T=\dfrac{gh}{c}=\dfrac{9.8\times84}{4200}=0.196\ \text{K}$.

Answer (b) $0.20\ \text{K}$.

Exam tip The mass cancels. (d) uses $c=1000\ \text{J kg}^{-1}\text{K}^{-1}$, a slip from "$1\ \text{kcal}$".

E6. Board · Case study A cylinder fitted with a frictionless piston of area $0.020\ \text{m}^2$ contains an ideal gas. The gas is heated slowly; it absorbs $1200\ \text{J}$ and pushes the piston out by $10\ \text{cm}$ at a constant pressure of $1.5\times10^5\ \text{Pa}$. (i) Find the work done by the gas. [1] (ii) Find the change in its internal energy. [1] (iii) The experiment is repeated from the same start with the piston clamped, supplying the same $1200\ \text{J}$. Find $\Delta U$ and say whether the temperature rises more or less than before, and by what factor. [2]

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(i) $\Delta V=0.020\times0.10=2.0\times10^{-3}\ \text{m}^3$; $W=p\,\Delta V=1.5\times10^5\times2.0\times10^{-3}=300\ \text{J}$. [1]

(ii) $\Delta U=Q-W=1200-300=900\ \text{J}$. [1]

(iii) Clamped: $W=0$, so $\Delta U=1200\ \text{J}$. The internal energy of an ideal gas depends only on $T$ and $\Delta U\propto\Delta T$, so the temperature rises more, by a factor $\dfrac{1200}{900}=\dfrac43$. [2]

Exam tip State the reason "$U$ of an ideal gas depends only on $T$" — without it the comparison of temperature rises earns no credit.

🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. Can you warm water by shaking it — and does the temperature rise match the work you put in, as the first law predicts?

Why it matters. Mechanical work turning into internal energy is everywhere in engineering: food processors and industrial mixers heat what they stir, pumps warm the fluid they move, the brakes of a lorry descending a mountain road can overheat, and the drilling of rock or bone must be cooled. Joule's version of this experiment, with paddles and falling weights, established that heat and work are two forms of the same energy.

What to do. Use a vacuum flask (a good adiabatic wall) about one-third full of water at room temperature, and a digital thermometer that reads to $0.1\ ^\circ\text{C}$.

Weigh the water ($m$). Record its starting temperature after it has stood for ten minutes.
Close the flask and shake it vigorously up and down through a measured height $h$ (about $0.3\ \text{m}$), counting the shakes. Each shake lifts the water and lets it crash down; estimate the work per shake as $mgh$.
After every 200 shakes, open the flask briefly and read the temperature.
Plot the temperature rise $\Delta T$ against the number of shakes $N$. Compare the slope with the prediction $\dfrac{gh}{c}$ per shake.
Leave the flask for ten minutes without shaking and record how fast it cools; use this to correct your readings for heat loss.

Tools: Vacuum flask, kitchen scale, digital thermometer, ruler, a friend to count, spreadsheet for the graph.

Hand in: Your table and graph, the measured and predicted slopes, a value of $c$ deduced from your data, and a paragraph on the main sources of error (heat from your hands, the uncertain fall height, heat leaking out).

Going further. Repeat with a bag of lead shot or sand dropped repeatedly inside a cardboard tube (a classic school version of Joule's experiment). Why does the temperature rise come out much larger for the same number of drops?

Instructor note: with $h=0.30\ \text{m}$, each shake gives $\Delta T=\dfrac{gh}{c}=\dfrac{9.8\times0.30}{4186}=7.0\times10^{-4}\ \text{K}$, so 1000 shakes give about $0.70\ \text{K}$ — clearly measurable, independent of the mass of water. In practice students obtain 50–80% of this: not every shake lifts the water the full height, and the flask itself and the air inside absorb some energy. For lead shot ($c=128\ \text{J kg}^{-1}\text{K}^{-1}$) falling $1.0\ \text{m}$, each drop gives $0.077\ \text{K}$, so 50 drops give nearly $4\ \text{K}$.

🧪 Self-check quiz

0/5 correct

Q1. The zeroth law of thermodynamics is what allows us to define

Q2. A gas absorbs $500\ \text{J}$ of heat and does $200\ \text{J}$ of work. Its internal energy

Q3. Which of these depends only on the state of a system, not on how it reached that state?

Q4. An adiabatic wall

Q5. A gas is compressed at a constant pressure of $1.0\times10^5\ \text{Pa}$ from $4.0\ \text{L}$ to $1.0\ \text{L}$. The work done by the gas is

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