Heat, work, internal energy, and the first law

A resistor, a gas and a piston

A sealed cylinder contains gas and an immersed electrical resistor. During one complete interval, the power source supplies 1 200 J to the resistor, 150 J leaves for the room, and the gas pushes a piston outward, doing 300 J of work. The gas and resistor have no change in bulk motion or height, and the resistor has no net stored-energy change. How much does the internal energy of the gas increase? Is the electrical supply a heat input?

Before choosing a formula, draw a boundary. If the system includes the resistor, electricity crosses its boundary as work. If the system is the gas alone and the excluded resistor is hotter than the nearby gas, energy crosses that boundary as heat. The final gas energy increase is the same. The classification follows what crosses the chosen boundary.

On a first reading, follow the definitions, signs, two boundaries, pressure–volume comparison and three adiabatic cases. “Further” sections return to differential notation, a finite pressure imbalance and flowing matter. The needed thermodynamic terms are introduced here. Published preparation is “Physical systems, states, boundaries, and environments”, “Work, kinetic energy, potential energy, and power”, and “Conservation of mass, momentum, and energy”. Calculus is useful for the return path, but the tables also show the main comparisons.

Choose the system and its stored energy

A system is the matter or region selected for study; its boundary separates it from the surroundings. A closed material system has no matter transfer across that boundary, although heat and work may cross it. An open control volume admits matter flow. An isolated system exchanges neither matter nor energy. An adiabatic boundary prevents heat transfer during the interval; it may still transmit work.

An equilibrium state has no macroscopic tendency to change under its constraints. A uniform gas at rest can be described by properties such as pressure p, volume V and absolute temperature T in kelvin. A process connects states. Intermediate states need not be equilibrium states: a rapidly released gas can have pressure gradients and organized motion even when both endpoints are well defined.

Internal energy U is energy stored in the material after separately accounting for the bulk kinetic and external gravitational energy specified in the model. It includes microscopic kinetic energy and interactions, and can include molecular rotation, vibration, phase and chemical contributions. It is not simply “the amount of heat” or always proportional to temperature. Even at the same temperature, different phases or compositions may have different U.

For a translating, nonrotating closed device in a fixed laboratory frame and a prescribed uniform gravitational field, write E = U + K + Φ, with K = mc²/2 and Φ = mgz. Here c is the device speed, m its mass, z the height of its center of mass and g the gravitational acceleration. Include additional changing macroscopic energy stores if the device requires them. Use the same frame and gravitational bookkeeping for the entire balance.

Heat and work describe transfers

Heat Q is energy transferred across the boundary because of a temperature difference. Conduction, convection at a boundary and thermal radiation can provide that transfer. Work is energy transfer through other organized interactions, such as a moving piston, a rotating shaft or an electrical connection. Work and heat have the same energy unit, the joule, J. They describe a process over an interval, rather than a stock kept inside the system.

We take Q positive into the system and W_by positive for work done by the system on the surroundings. Thus an electrical or shaft work input is a negative contribution to W_by. Expansion work is positive when the system pushes outward; compression work is negative. A different common convention takes W_on positive into the system. The conversion is W_on = −W_by, not a change in the physics.

The closed-system first law is ΔE = ΔU + ΔK + ΔΦ = Q − W_by. The symbol Δ means final minus initial. If bulk kinetic and gravitational changes vanish and no other store changes, this reduces to ΔU = Q − W_by. First list every transfer and store, then justify any zero. “Closed” alone does not set Q or W_by to zero.

A watt is one joule per second: 1 W = 1 J/s. The symbol W_by denotes an amount of work here; the unit W denotes watt. A dot, as in Q̇, denotes a rate. Keep amounts, rates and per-mass quantities distinct.

The same device, two boundaries

Boundary A encloses the gas and resistor. The piston, room and power source are outside. The electrical transfer is work input, so W_by,electric = −1 200 J; the piston contribution is +300 J. Their sum is W_by = −900 J. Heat loss gives Q = −150 J. Therefore ΔU_gas+resistor = −150 − (−900) = 750 J.

Boundary B encloses only the gas. Over the full interval, the resistor transfers all 1 200 J to the gas across their hotter-to-cooler interface. Together with the 150 J loss to the room, net Q = 1 050 J. The gas does 300 J of piston work, giving ΔU_gas = 1 050 − 300 = 750 J. The resistor has zero net endpoint energy change, so the two stored-energy answers agree.

These balances assume no additional lead conduction, radiation loss or changing field energy. For the electrically connected boundary, use the usual closed-system circuit idealization: the bulk contents are sealed, equal charge enters and leaves, there is no charge accumulation, and matter-carried energy other than electrical work is negligible. A substantial material or charge accumulation would need a more complete model.

Moving a boundary changes which interactions are internal and which cross it. Energy passed from resistor to gas is internal to boundary A, so it is not also entered as a heat input there. Adding both the 1 200 J electrical work and the same 1 200 J internal heat transfer to that ledger would count one transfer twice.

Keep motion and height in the ledger

A 2 kg sealed device accelerates from 1 m/s to 5 m/s while rising 3 m in a uniform field with g = 9.8 m/s². It loses 20 J by heat and receives a net 200 J of non-gravitational work. With our signs, Q = −20 J and W_by = −200 J, so its total stored energy rises by 180 J.

Only part of that rise is internal: ΔK = 2(5² − 1²)/2 = 24 J and ΔΦ = 2 × 9.8 × 3 = 58.8 J. Thus ΔU = 180 − 24 − 58.8 = 97.2 J. Using ΔU = Q − W_by without those terms would give a wrong internal-energy increase.

Here gravitational potential energy is included on the stored-energy side. Do not also include gravity as an external work input in W_by. One can use another consistent convention with gravity represented as a transfer, but combining both conventions counts the same energy twice.

Pressure work at a moving boundary

Imagine gas pushing a piston of cross-sectional area A through a small outward displacement dx. If p_b is the normal pressure at the gas–piston boundary, the gas force is p_b A and the volume change is dV = A dx. Hence the pressure-work amount done by the gas is δW_by = p_b dV. The total is W_by,PV = ∫ p_b dV along the actual boundary motion.

When the gas passes through nearly uniform equilibrium states and a frictionless piston is controlled with a negligibly small pressure imbalance, p_b can be replaced by the bulk equilibrium pressure p. Then W_by,PV = ∫ p dV, the signed area under a p–V path. Pressure has units Pa and volume m³, so Pa m³ = J; conveniently, 1 kPa L = 1 J.

The surrounding load pressure and gas-boundary pressure need not be identical while a piston accelerates. Their work difference can change piston kinetic energy or be dissipated by friction. Use the force at the chosen boundary, or include the piston and its energy transfers explicitly. The “Further” finite-imbalance example shows a case where the integrated load work can be used after checking these conditions.

A slowly varying process is called quasistatic. Slowness permits equilibrium properties to be used, but does not by itself remove piston friction or other dissipation. A reversible process can be undone with both system and surroundings restored; it requires more than a sequence of almost-equilibrium gas states. No reversibility assumption is needed for the general endpoint first law.

Same endpoints, different work and heat

Take a fixed amount of an ideal monatomic gas with pV = nRT and nR = 1 J/K. Add a caloric model, U = (3/2)nRT = (3/2)pV. This extra relation specifies internal energy; the pressure–volume–temperature equation alone would not do so. The model assumes the classical monatomic regime with no changing molecular internal excitations or interactions.

Compare two deliberately controlled quasistatic expansion paths from state A, V = 1 L and p = 300 kPa, to state C, V = 3 L and p = 100 kPa. Let v = V/(1 L), a dimensionless volume coordinate. Path D has p_D/kPa = 400 − 100v. Path B has p_B/kPa = 400 − 100v + 60(v − 1)(3 − v). A frictionless piston tracks each path with negligible pressure imbalance, and pressure–volume work is the only work.

Both endpoints have pV = 300 J, T = 300 K and U = 450 J. Therefore ΔU = 0 for either path, with equal zero endpoint changes in bulk K and Φ. Yet the area under D is 400 J, while the area under B is 480 J. The first law then gives Q_D = 400 J and Q_B = 480 J. The larger outward work requires a larger net heat input even though the gas ends in the same state.

These paths are not isothermal. At V = 2 L, path D has T = 400 K while path B has T = 520 K. Along B the pressure first rises slightly, reaching about 301.67 kPa near 1.167 L, then falls. The formula, rather than a monotonic sketch, defines the curve. Control of both mechanical loading and thermal contact is implicit; no time history is assigned.

A state change does not record its history

U is a state function: once the complete material state is specified, its value relative to a chosen reference does not depend on the route taken. Its change is U_final − U_initial. Heat and work are path quantities: their totals depend on the process, and the system does not possess values “Q in the final state” or “W in the final state”. Write Q and W_by for the interval, rather than ΔQ and ΔW_by.

The two gas paths show why endpoints alone usually cannot determine Q or W_by separately. Here they determine only Q − W_by = 0. To find either transfer, provide a path, a measured transfer, a constraint such as adiabaticity, or another physically sufficient relation.

A constant shift U′ = U + C changes no energy balance for a fixed material system: ΔU′ = ΔU. In this teaching model the selected reference is formally U = 0 at T = 0. That reference choice is bookkeeping; it does not assert that a classical ideal gas model remains valid down to zero temperature.

An ideal monatomic gas has U depending only on T in this model, so equal endpoint temperatures imply equal U for fixed amount. The inference cannot be applied to a general substance with changing volume, phase, composition or other internal variables without the appropriate caloric relation.

Returning around a cycle

Expand along B from A to C, then compress back along D. Reversing D reverses dV and the work sign, so the return work is −400 J. The complete cycle has W_by,net = 480 − 400 = 80 J. It traces a clockwise loop in this p–V diagram; its enclosed signed area gives that net pressure work.

The full state returns, including the mechanical and gravitational stores, so ΔE_cycle = 0. Consequently Q_net = W_by,net = 80 J. The net heat is not the sum of positive heat inputs: it includes every signed transfer, including the −400 J net heat on the return leg. Individual legs can contain both positive and negative heat-transfer portions.

If only U returned while a device gained speed or height, ΔU_cycle = 0 would not establish ΔE = 0. Likewise, a p–V loop gives only pressure–volume work; electrical or shaft work must be added if present. This example assigns no hot/cold reservoirs or efficiency. Energy balance alone neither proves that any proposed cycle is feasible nor determines its efficiency.

Adiabatic does not mean unchanged

Adiabatic means no energy crosses the boundary as heat during the interval, so Q = 0. It does not mean constant temperature, constant internal energy, reversibility or isolation. For a closed system with unchanged bulk stores, the first law becomes ΔU = −W_by. Work output lowers U; work input raises it.

In a rigid insulated vessel, a shaft does 240 J of stirring work on the sealed contents. Include the paddle hardware or require its endpoint stored energy to be unchanged; the contents begin and end at rest. Volume work is zero, but shaft work is W_by = −240 J. Thus ΔU = +240 J despite Q = 0. A rigid boundary eliminates moving-volume work, not every work mode.

Now let the same ideal monatomic gas model initially occupy 1 L at 300 kPa inside a rigid insulated two-compartment enclosure. The other compartment is evacuated. Remove a partition with negligible stored energy, and let the gas equilibrate throughout 2 L. There is no external work, no heat transfer and no changing endpoint bulk store. Therefore ΔU = 0, T_final = 300 K and p_final = 150 kPa.

During this free expansion the gas is not a sequence of uniform equilibrium states. A line joining the endpoints on a p–V plot would invent an equilibrium path; integrating that line would invent work. The gas pushes into vacuum without doing external work, while the rigid outer wall does not move. A real gas need not keep the same temperature at fixed U; the equal temperature here follows the specific ideal-gas caloric model.

Further: an irreversible finite-load expansion

Return to 1 L of the ideal monatomic gas at 300 kPa, T = 300 K and U = 450 J. This time a horizontal frictionless piston faces a constant external resisting pressure p_ext = 100 kPa. Release the restraint. The apparatus is adiabatic, the piston starts and finishes at rest, and internal gas dissipation allows settling. There is no other work or retained piston energy change.

For the gas alone, the work is ∫p_b dV. The piston energy balance gives ∫(p_b − p_ext)dV = ΔK_piston = 0 because it is frictionless with rest endpoints. Therefore the total gas work equals p_ext(V_final − V_initial). This is an equality of the complete integrals; it does not say p_b = p_ext at every instant of the accelerating motion.

At final mechanical equilibrium p_final = p_ext. With V_final expressed in litres, U_final = (3/2)(100 kPa)V_final. The first law gives (3/2)(100)V_final − 450 = −100(V_final − 1), using 1 kPa L = 1 J. Thus 250V_final = 550 and V_final = 2.2 L. It follows that T_final = 220 K, U_final = 330 J and W_by = 120 J.

The gas cools because it exports work while receiving no heat. There is no uniform intermediate gas p(V) supplied here and no reversible adiabatic relation may be inserted. Piston friction, nonzero endpoint piston speed, external damping or energy leakage would require additional terms or a different boundary. The free-expansion result cannot be reused just because both cases are adiabatic.

Further: exact and inexact changes

For neighboring equilibrium states of the same material system, write dU = δQ − δW_by when bulk stores do not change. The d in dU denotes an exact state differential: its integral depends only on endpoints. The δ notation marks an infinitesimal transfer whose integral generally depends on the path. It does not mean heat or work is numerically smaller than another small quantity.

For only quasistatic pressure work, δW_by = p dV. Around the complete loop, ∮dU = 0 while ∮δW_by = ∮p dV = 80 J and ∮δQ = 80 J. A candidate state function W = pV cannot represent the work, because d(pV) = p dV + V dp rather than p dV alone.

The two analytic work integrals can be checked directly using dimensionless v: W_D/J = ∫₁³(400 − 100v)dv = [400v − 50v²]₁³ = 400. Their difference is (W_B − W_D)/J = 60∫₁³(v − 1)(3 − v)dv = 80. This yields W_B = 480 J without estimating areas from a drawing.

For two adiabatic processes with the same complete initial and final states and unchanged endpoint bulk stores, Q = 0 fixes the same net W_by = −ΔU even if the work mechanisms differ. This special restriction is compatible with work being path dependent in general. In nonequilibrium motion, use the complete endpoint balance even when an equilibrium differential description of the interior is unavailable.

Further: follow a stream through a device

A fixed control volume describes an open device without following the same particles. Its stored energy E_CV can change while mass carries energy across inlets and outlets. In the simple model used here, transported energy per mass includes specific internal energy u, bulk kinetic energy c²/2 and gravitational potential gz. “Specific” means per unit mass; c is stream speed in the fixed laboratory frame, and z is elevation.

At an inlet, upstream pressure also does work to push fluid into the region; at an outlet the fluid does pressure work to push downstream fluid away. Combining this pressure flow work with u defines specific enthalpy h = u + p/ρ, where ρ is mass density. Total enthalpy is H = U + pV for the corresponding uniform simple system. Enthalpy is a state property and not a new form of heat. Inserting h at the ports already counts pressure flow work once.

For a stationary fixed control volume with one uniform inlet and one uniform outlet, the rate balance is dE_CV/dt = Q̇ − Ẇ_other,by + ṁ_in(h_in + c_in²/2 + gz_in) − ṁ_out(h_out + c_out²/2 + gz_out). Ẇ_other,by is work output excluding the pressure flow work included in h. The boundary and stream model here requires no additional field, species-conversion or other transported-energy term.

At steady state the stored mass and energy do not accumulate: ṁ_in = ṁ_out = ṁ and dE_CV/dt = 0. Rearranging gives Ẇ_other,by = Q̇ + ṁ[h_in − h_out + (c_in² − c_out²)/2 + g(z_in − z_out)]. Steady describes no time change at each location; inlet and outlet properties can still differ. A startup or filling tank requires the accumulation term.

Use h in J/kg, c²/2 and gz in J/kg, and ṁ in kg/s to get watts. If h is in kJ/kg, divide c²/2 and gz by 1 000 before combining them. Dividing the complete power equation by ṁ gives energy per mass, in J/kg or kJ/kg, not energy per time per mass. The stored-energy term remains E_CV; replacing it with stored enthalpy would silently alter the balance.

A flow ledger with a kinetic-energy change

A steady device carries 0.05 kg/s through one inlet and one outlet at the same elevation. The supplied common-reference enthalpies are h_in = 280 kJ/kg and h_out = 190 kJ/kg. Stream speeds are 40 m/s and 100 m/s. Heat loss is 0.2 kW, so Q̇ = −0.2 kW. Find the non-flow-work output under the stream assumptions above.

The specific kinetic energies are 40²/2 = 800 J/kg = 0.8 kJ/kg and 100²/2 = 5 000 J/kg = 5 kJ/kg. Thus 4.2 kJ/kg of the 90 kJ/kg enthalpy drop becomes additional bulk kinetic energy. The available stream-energy drop is 90 − 4.2 = 85.8 kJ/kg.

Therefore Ẇ_other,by = −0.2 + 0.05 × 85.8 = 4.09 kW. An equivalent rate ledger is 14.04 kW entering with the stream = 9.75 kW leaving with the stream + 0.20 kW heat loss + 4.09 kW work output. Dropping the kinetic term would predict 4.30 kW, an overestimate of 0.21 kW.

No equation of state or efficiency has been inferred from the assigned h values. Enthalpy differences must use a consistent material reference. Changing a common reference by the same additive amount does not affect this steady same-mass-flow result; inconsistent inlet and outlet reference choices do.

Choose the next question

Use this sequence on a new problem: name the material system or control volume and time interval; list changing stores; choose a sign convention; label every crossing transfer; then add the material and process relations needed to solve. Finally check dimensions, signs and whether the inferred result fits the stated boundary.

The first law is energy conservation. It does not supply an equation of state, a caloric relation, a heat-transfer rate or a direction-of-process criterion. It does not make an energy-balanced proposal automatically feasible. Entropy and the second law impose further restrictions; energy balance remains necessary.

The neighboring subject “Isothermal, adiabatic, isobaric, and isochoric processes” compares named constraints in more detail. “Heat capacities and calorimetry” develops relations between energy transfer and temperature change. “Entropy and the second law” and “Engines, refrigerators, and heat pumps” treat direction and performance limits. “Enthalpy, Helmholtz energy, and Gibbs energy” develops thermodynamic potentials beyond the flow-work use of h here. These are follow-on map topics, not assumed completed preparation.

Worked examples

Try it yourself

1. A closed system has ΔU = ΔK = ΔΦ = 0 and does 75 J of net work on its surroundings. What is Q?

   Show the answer and reasoning

   Q = +75 J.

   Rearrange the first law: Q = ΔE + W_by = 0 + 75 J. A zero internal-energy change does not require each transfer to vanish.

2. A complete closed-device cycle returns every energy store to its initial value and has net Q = −30 J. What is its net W_by, and does it give net work output?

   Show the answer and reasoning

   W_by,net = −30 J; it requires 30 J net work input.

   With ΔE = 0, Q_net = W_by,net. A negative W_by means surroundings do net work on the device.

3. A rigid insulated vessel is stirred through a shaft. Is the energy entering its sealed contents heat because the contents become warmer?

   Show the answer and reasoning

   It enters as shaft work. Adiabatic means Q = 0.

   Classify the boundary interaction, not the eventual temperature change. After stirring decays, transferred work can remain as internal energy. A caloric relation is needed to calculate a temperature change.

4. Someone joins the free-expansion endpoints (1 L, 300 kPa) and (2 L, 150 kPa) by a straight line and integrates the area. Does this give the work?

   Show the answer and reasoning

   No. For the stated rigid insulated enclosure, W_by = 0.

   The straight line invents intermediate uniform equilibrium pressures. The endpoint values do not specify a path, and the actual expansion into vacuum has no external moving-boundary work.

5. Replace U by U′ = U + 700 J for the fixed amount in the two controlled gas paths. What are the endpoint U′ values and ΔU′? Do Q and W_by change?

   Show the answer and reasoning

   U′_A = U′_C = 1150 J; ΔU′ = 0. Q and W_by do not change.

   Both endpoint energies increase by the same constant, which cancels in the difference. A reference change does not physically transfer energy.

6. For the moving closed device, a calculation gives ΔU = 180 J. Identify the missing terms and correct it.

   Show the answer and reasoning

   Subtract ΔK = 24 J and ΔΦ = 58.8 J: ΔU = 97.2 J.

   The 180 J is the total stored-energy gain. Gravity has already been assigned to the potential-energy store, so no additional gravity-work term is added.

7. An engineer uses h at an inlet and outlet, then adds the same p/ρ flow-work terms again. What is wrong?

   Show the answer and reasoning

   Pressure flow work has been counted twice.

   The definition h = u + p/ρ already combines internal energy and flow work. Use u with explicit flow work, or h with that part excluded from Ẇ_other,by.

8. Can a steady device have c_in = 40 m/s and c_out = 100 m/s? Does dE_CV/dt = 0 imply Q̇ = Ẇ_other,by?

   Show the answer and reasoning

   Yes, different port speeds are compatible with steady state. Zero accumulation alone does not imply equal heat and work rates.

   Steady means no time change at a fixed location. Streams can have different enthalpy, speed or height, and the net mass-carried energy appears in the control-volume balance.

9. A piston moves very slowly while sliding friction dissipates energy. Does quasistatic gas behavior prove the whole process reversible?

   Show the answer and reasoning

   No.

   Nearly uniform equilibrium gas states permit a quasistatic description, but friction prevents restoring both system and surroundings by an arbitrarily small reversal. Reversibility needs the relevant dissipative effects removed.

10. A proposed cyclic device satisfies Q_net = W_by,net. Is that sufficient to establish that the proposed heat exchanges are physically possible?

    Show the answer and reasoning

    No. It establishes the first-law energy condition only.

    The direction of heat exchanges and achievable performance also face second-law restrictions. This article supplies no reservoir temperatures or entropy analysis from which to decide those questions.

What to carry forward

• Choose the boundary before naming heat or work; one physical transfer can receive a different classification when the boundary changes.
• Internal energy is stored energy of the material state. Heat and work are transfers over a process, and endpoints usually determine only their difference.
• Retain bulk kinetic, potential and any other changing energy stores; justify every omitted term.
• A p–V area gives pressure work only for the appropriate actual boundary path. Nonequilibrium endpoints do not define a quasistatic curve.
• Adiabatic, rigid, quasistatic, reversible and isolated impose different conditions.
• Flowing matter transports energy. Enthalpy includes pressure flow work, and steady state removes accumulation rather than the stream terms.