Thermodynamic systems and the first law

A thermodynamic statement is meaningless until you say what it is about. The system is the matter or region you have chosen; everything else is surroundings; the boundary is the surface between them, and it may be real (a cylinder wall) or imagined (a fixed volume of air in a room). A closed system exchanges energy but not matter across its boundary; an open system exchanges both; an isolated system exchanges neither. The boundary also decides the label on every transfer. Run 2.0 A through a heater coil at 12 V for 60 s and 1440 J leave the battery. Draw the boundary around the water and the coil together and those 1440 J enter as electrical work: charge is pushed through a potential difference across the boundary, and no temperature difference is involved. Draw the boundary around the water alone and the same 1440 J enter as heat, driven by the temperature difference between the hot coil and the water. Nothing physical changed; the bookkeeping category did, because heat and work are names for how energy crosses a surface, not for what it becomes inside.

A state function is a quantity fixed by the equilibrium state alone, like the altitude of a point on a map. For a simple compressible system the state is set by two variables, say T and V, and U = U(T, V) has one value there however the system arrived. Heat and work are like the distance walked: they belong to the journey, not the place. That is why the law is written with δQ and δW rather than dQ and dW; there is no function Q(T, V) of which δQ is the change, and writing a change in Q, as if heat were stored, is the error the notation guards against. The operational test is the cyclic integral. Take a system around any closed loop of states and ∮ dU = 0, while ∮ δQ and ∮ δW are in general non-zero and equal to each other. A gas can take in 1250 J of heat and later give back 1370 J on a different route home, its internal energy returning to exactly where it began: the 120 J difference was work put in along the way. After the fact the gas carries no record of either number, only its state.

Joule wrapped a paddle wheel in an insulated tank of water, drove it with falling weights and measured the temperature rise. The essential result was not the mechanical equivalent of heat but this: for a given pair of end states, the adiabatic work required is the same whether it is delivered by paddles, by a rubbed plate or by a resistor carrying current. Work that does not depend on the path is the change of a function of state, so U_B − U_A is defined as the adiabatic work done on the system from A to B. Take 0.500 kg of water (c = 4184 J kg⁻¹ K⁻¹) warmed by 1.20 K in an insulated, rigid tank: ΔU = 0.500 × 4184 × 1.20 = 2.51 × 10³ J, and every adiabatic route between those two states costs exactly that much work. Now remove the insulation and reach the same final state with only 1800 J of paddle work. The deficit is the heat: Q = ΔU + W = 2510 + (−1800) = 710 J. No instrument measured 710 J of heat flowing; heat was inferred from a state change and a work total. That is what heat means in thermodynamics.

Two conventions are in use and both are correct. Most introductory physics texts, many engineering texts, and this site write ΔU = Q − W with W the work done by the system, so an expanding gas has W > 0 and spends its internal energy. Chemistry texts, and many advanced physics texts, write ΔU = Q + W with W the work done on the system, so the same expansion has W < 0. The physics is identical; only the sign of W flips. Test it on one process. A gas absorbs 500 J and expands, doing 300 J of work on its piston. Physics convention: ΔU = 500 − 300 = +200 J. Chemistry convention: the work done on the gas is −300 J, so ΔU = 500 + (−300) = +200 J. The dangerous move is to import a formula from one convention into a calculation running on the other, which silently doubles or cancels the work term. Two habits prevent it: write the subscript, W_by or Wₒₙ, on the first line of every problem, and check the sign of ΔU against the physics before you trust it. A gas that is compressed and insulated must end with more internal energy; if your arithmetic says less, the convention slipped, not the gas.

The first law makes two demands. The system must be closed: mass crossing the boundary carries energy with it and the balance needs extra terms. And both end states must be equilibrium states, because U(T, V) is defined only where the system has one T and one V. It makes no demand on the path. Take 1.00 mol of argon at 300 K in one half of a rigid, insulated vessel, the other half evacuated, and break the partition. The gas rushes across with no uniform pressure or temperature during the process; it pushes on nothing, so W = 0, and the insulation gives Q = 0. Once the gas is quiet the law reads ΔU = 0, and since an ideal gas has U = (3/2) n R T the temperature is 300 K again, at twice the volume and half the pressure. Nothing here was quasi-static and the first law did not care. That restriction belongs to a different statement, W = ∫p dV with p the gas’s own pressure, which needs a defined p at every instant. The tidy result ends with the ideal-gas model: a real gas expanding freely cools slightly, because its U depends weakly on V through intermolecular attraction.

Divide the differential form by dt and the first law becomes a power balance: dU/dt = Q̇ − Ẇ, every term in watts, where Q̇ is the rate of heating into the system and Ẇ the power the system delivers outward. Take the water and element of a kettle as the system, with 0.80 kg of water and a 1.2 kW element. Electrical work enters at 1200 W, so before losses matter dU/dt = 1200 W and the water warms at 1200/(0.80 × 4184) = 0.36 K s⁻¹. Near the boil the hot body loses about 150 W to the room as heat, so dU/dt = 1200 − 150 = 1050 W and the rate falls to 0.31 K s⁻¹. A steady state is the case dU/dt = 0: a soldering iron at constant tip temperature sheds heat at exactly the electrical power it draws, and a resistor at constant temperature is a device that converts work entering to heat leaving with nothing stored. The rate form also keeps a steady state apart from equilibrium. A body at steady temperature is not one where nothing is happening: its boundary transfers have come into balance while energy still flows through it, so the soldering iron is in a steady state, not in equilibrium with the room, which would mean no net transfer at all.