Thermodynamics · Lesson 2 of 7
Heat, Work and the First Law
“Track energy transfers, apply a consistent sign convention and distinguish path-dependent heat and work from internal energy.”
• Distinguish temperature, internal energy, heat and work. • Apply Q = ΔU + W with W defined as work done by the system. • Derive piston work and interpret its sign and units. • Explain how different paths can have the same ΔU but different Q and W. • Solve energy balances for vaporisation, adiabatic changes and energy-transfer rates.
Energy stored inside matter
A gas may be stationary as a whole while its molecules move, rotate and vibrate. Internal energy U includes the microscopic kinetic and interaction potential energies associated with the internal state, evaluated relative to the motion of the centre of mass. It excludes kinetic energy simply due to the entire container moving through the room, and excludes the external gravitational potential energy of the whole system. Translational, rotational and vibrational molecular contributions can all matter; their detailed distribution is studied later.
In the ideal-gas model, intermolecular interaction energy is neglected and internal energy for a fixed amount and composition depends only on temperature: U = U(T). This does not imply U is always directly proportional to T with the same coefficient for every gas. Nor does it make temperature identical to total internal energy. Different sample sizes or substances at one temperature can have different U. For liquids and solids, internal interaction energies are important, and U need not depend on T alone.
Heat and work describe transfers, not contents
Heat Q is energy transferred because of a temperature difference. When a hotter object contacts a colder gas, energy enters the gas as heat. Work W describes energy transfer by other mechanisms, such as moving a piston or stirring a liquid. Once transferred, energy contributes to the system’s energy; it is not stored with a label saying “heat” or “work”. A state has internal energy. A process transfers heat and performs work.
The old caloric idea treated heat as a conserved invisible fluid. Rumford’s cannon-boring observations in 1798 challenged that view: continuing mechanical work generated continuing warming, enough to boil surrounding water, and the effect depended on work input rather than the sharpness of the drill. Friction converts organised mechanical energy into internal energy; it does not reveal a pre-existing supply of caloric. The physically correct spontaneous direction of heat transfer is from higher to lower temperature.
| Quantity | Physical meaning | State or process? |
|---|---|---|
| T | Temperature; indicates thermal-equilibrium condition and direction of heat flow | State variable, unit K |
| U | Internal energy associated with the internal state | State variable, unit J |
| Q | Energy transferred as heat due to a temperature difference | Path quantity, unit J |
| W | Energy transferred as work; positive here for work done by the system | Path quantity, unit J |
Conservation of energy with a sign convention
For a closed system whose bulk kinetic and external potential energy changes can be neglected, define Q as positive when heat enters and W as positive when the system does work on its surroundings. Heat leaving has Q < 0. Compression work supplied by the surroundings has W < 0. The energy retained internally is what enters as heat minus what leaves as work. This gives the first law.
Problem
A gas absorbs 250 J while doing 90 J of work. In another process, it loses 80 J of heat while 120 J of compression work is done on it. Find ΔU in each case.
- 1.First process: Q = +250 J and W = +90 J, so ΔU = 250 − 90 = +160 J.
- 2.Second process: Q = −80 J and W = −120 J, so ΔU = −80 − (−120) = +40 J.
- 3.Losing heat need not make U decrease if a larger amount of energy enters through work. Always attach signs to transfers before substituting.
Deriving pressure–volume work
Let a piston have cross-sectional area A and move outward by a small distance dx. The boundary displaces volume dV = A dx. Against external pressure Pext, the opposing force is Pext A, and the gas does work dW = Pext A dx = Pext dV. For a quasi-static frictionless piston, Pext differs only infinitesimally from the gas pressure P, so the limiting expression is dW = P dV. In a rapid irreversible process, one must use the external resisting pressure; a uniform gas pressure may not exist.
Expansion gives dV > 0 and positive work; compression gives dV < 0 and negative work. The product Pa·m³ equals (N/m²)·m³ = N·m = J. A process at fixed volume has zero pressure–volume work, but need not have zero electrical or stirring work. In the gas-process examples below, pressure–volume work is the only work unless stated otherwise.
The same endpoints can require different heat and work
Internal energy is a state function. If a gas starts in the same equilibrium state A and finishes in the same equilibrium state B, ΔU is fixed even if the route differs. But work depends on the pressures against which each volume change occurs. Expanding at a high pressure transfers more work than expanding by the same volume change at a low pressure. Q must then differ so that Q − W still equals the same ΔU.
Problem
An ideal gas changes from A: (1.0×10⁵ Pa, 1.0×10⁻³ m³) to B: (2.0×10⁵ Pa, 2.0×10⁻³ m³). Route X first expands at the initial pressure and then heats at fixed volume. Route Y first heats at the initial volume and then expands at the final pressure. Compare their work and heat.
- 1.In route X the expansion work is (1.0×10⁵)(1.0×10⁻³) = 100 J. The fixed-volume leg does no boundary work.
- 2.In route Y the fixed-volume leg does no boundary work, while the expansion work is (2.0×10⁵)(1.0×10⁻³) = 200 J.
- 3.Both routes have identical endpoint states, hence the same ΔU. Therefore QY − QX = WY − WX = 100 J.
- 4.The equality concerns the difference in heat input; finding either absolute heat value would also require the gas’s internal-energy relation.
Problem
A gas goes from A to B adiabatically while 22.3 J of work is done on it. On another path from A to B, it absorbs 9.35 cal. Find the net work done by the gas on that second path, using 1 cal = 4.19 J.
- 1.On the adiabatic path Q = 0 and W = −22.3 J. Therefore ΔU = 0 − (−22.3) = +22.3 J.
- 2.This ΔU belongs to the endpoints and is the same on the other path.
- 3.The other heat input is Q = 9.35×4.19 = 39.1765 J.
- 4.W = Q − ΔU = 39.1765 − 22.3 = 16.8765 J ≈ 16.9 J. The positive sign means net work is done by the gas.
Vaporisation: heat input need not equal ΔU
Problem
At atmospheric pressure 1.013×10⁵ Pa, 1 g of liquid water occupies 1 cm³ and its vapour occupies 1671 cm³. The latent heat input is 2256 J. Find expansion work and ΔU, neglecting bulk kinetic and potential energy changes.
- 1.The volume increase is 1671 − 1 = 1670 cm³ = 1.670×10⁻³ m³. Convert the change, rather than just the final volume, into SI units.
- 2.The pressure is constant, so W = PΔV = (1.013×10⁵)(1.670×10⁻³) = 169.171 J ≈ 169.2 J.
- 3.ΔU = Q − W = 2256 − 169.171 = 2086.829 J ≈ 2086.8 J.
- 4.Most input energy increases internal energy associated with changing molecular arrangement and separation. About 169 J is transferred to the surroundings as expansion work.
- 5.The liquid and vapour are at the same boiling temperature. The large ΔU illustrates why the ideal-gas statement U = U(T) must not be applied to a liquid-to-vapour phase transition.
The first law as a rate equation
Problem
A heater supplies energy at 100 J s⁻¹ under the stated heat-input convention. The system simultaneously performs work at 75 J s⁻¹. Find dU/dt.
- 1.The inward heating rate is +100 W; the outward work rate is +75 W.
- 2.dU/dt = 100 − 75 = 25 J s⁻¹ = 25 W.
- 3.The system’s internal energy grows by 25 J each second if these rates remain constant. If the heater is included within the system boundary, its electrical input would instead be classified as work entering; a consistent energy balance gives the same energy increase.
Quick Quiz
Quiz
A gas loses 40 J as heat while 70 J of work is done on it. What is ΔU?
Q = −40 J, W = −70 J, and ΔU = Q − W = +30 J.
Which statement describes a state property?
U is a state variable. Heat and work describe transfers along a process.
For a quasi-static piston expansion at 2.0×10⁵ Pa through 3.0×10⁻³ m³, the boundary work done by the gas is
W = PΔV = 600 J; expansion makes it positive.
Two paths connect identical initial and final equilibrium states. Which quantity must be identical?
Q − W equals ΔU, which depends only on endpoints.
What can raise a box’s kinetic energy without necessarily changing its internal energy?
Centre-of-mass kinetic energy is excluded from U. Internal dissipation or other state changes would have to be specified separately.
Practice Problems
- For each process compute ΔU with the chapter convention: (a) 180 J heat enters and 60 J work leaves; (b) 50 J heat leaves and 90 J compression work enters. Explain the signs.
- Derive dW = Pext dV for a piston of area A moving a distance dx. State the condition that permits replacing Pext by the gas pressure.
- A gas expands quasi-statically at 1.50×10⁵ Pa from 2.0 L to 5.0 L while absorbing 800 J. Find its work and internal-energy change; use 1 L = 10⁻³ m³.
- An adiabatic route between two states requires 35 J of work on a gas. A second route between the same states absorbs 80 J of heat. Find work done by the gas on the second route.
- Vaporising a sample requires 3000 J and increases its volume by 2.0×10⁻³ m³ against constant pressure 1.0×10⁵ Pa. Find W and ΔU. Explain why latent heat input is not all retained internally.
- A system receives heating at 240 J s⁻¹ while doing work at 150 J s⁻¹ for 20 s. Find the energy transfers and ΔU. State the assumptions that permit this rate balance.
Key Takeaways
• U is a state function; Q and W are energy transfers dependent on a process. • With heat into the system and work by it positive, ΔU = Q − W. • Piston work is ∫Pext dV, or ∫P dV for the quasi-static frictionless limit. • Endpoint equality fixes ΔU, while paths can have different heat and work. • Bulk motion is distinct from internal energy; phase change can change U at constant temperature.