Thermodynamic Systems

01

Internal energy and thermal equilibrium

Choose which matter belongs to the system. Its internal energy describes its microscopic energy, while heating and work describe energy crossing its boundary.

Define the system and its state

A thermodynamic system is the matter selected for an energy account. Its boundary separates it from the surroundings. For example, a gas inside a cylinder can be the system, with the piston and the room outside it. In a calorimeter, the selected system might instead include the sample, heater and vessel together.

Internal energy U is associated with the distribution of microscopic kinetic and potential energies of the system's particles. It excludes the kinetic energy of the whole system moving together and its gravitational potential energy relative to the surroundings.

At equilibrium, the macroscopic state determines U. Specify the amount, composition and phase as well as the relevant pressure, volume and temperature. A temperature reading alone does not determine the total internal energy of every possible sample.

Internal energy is a state quantity: its change between the same initial and final states does not depend on how the change occurred. Heating Q and work W are transfers during a process. Their separate values can depend on the route, even when the initial and final states are the same. A hot object has internal energy; it does not contain a stored quantity of heat Q.

Temperature describes a particle average

In the classical particle model, thermodynamic temperature is proportional to mean microscopic kinetic energy. For an ideal gas, the precise translational result is:

T is in kelvin. The particle-energy derivation concerns an average per particle, not equal energy for every particle. Increasing the number of particles at the same temperature does not change this average, but can increase the total energy.

For a monatomic ideal gas in this model, internal energy is the total translational kinetic energy:

N is the particle number and n the amount in mol. At the same T, doubling N doubles this U. Two such samples can therefore have the same temperature and different internal energies.

Keep the monatomic ideal-gas condition beside this total-energy equation. Molecular gases and other phases can have additional microscopic energy contributions, so U = 3NkT/2 is not a universal formula for every material.

Thermal contact and equilibrium

When systems at different temperatures are put in thermal contact without a competing imposed transfer, net energy moves by heating from higher to lower temperature. They approach a common temperature. At thermal equilibrium, there is no net heating between them.

The direction of net heating follows temperature, not which object has more total internal energy. At equal temperature, microscopic exchanges may continue in both directions while their net effect is zero.

A system can also receive energy without a temperature rise. During a phase change under the stated conditions, its microscopic potential-energy contribution can change while temperature remains constant.

Use the zeroth law

If 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. This is the zeroth law of thermodynamics.

This makes a calibrated thermometer useful for comparing temperatures. Allow it to settle in good thermal contact. An early reading may describe the sensor while the sample is still at a different temperature.

A thermometer with substantial heat capacity can itself change a small sample's temperature. It then measures their changed equilibrium. A sensor with small heat capacity, suitable contact and a stable reading reduces this disturbance; equal readings do not imply that the thermometer exchanged equal amounts of energy with different samples.

02

Heating, work and the first law

Internal energy can change through heating, work or both. Use one work convention throughout the calculation and identify which transfers enter the chosen system.

Set the signs before substituting

For the systems considered here, with no change in the whole system's bulk kinetic or gravitational potential energy, the first law is:

Q is positive for net energy transferred into the system by heating. W_on is positive for net work done on it. A positive ΔU means its internal energy increases.

If work done by the gas is given instead, W_by = -W_on, so the same law becomes ΔU = Q - W_by. Do not add positive work by the gas to Q in the work-on convention.

Derive constant-pressure displacement work

Choose the gas as the system. It expands against a constant absolute external pressure p_ext. For piston area A, the external resisting force has magnitude p_extA. Moving the boundary outward through Δx increases the gas volume by ΔV = AΔx.

For expansion, ΔV is positive: the gas does positive work and W_on is negative. For compression, ΔV is negative and work on the gas is positive. Use the total stated external pressure, including the surroundings and load, rather than only a gauge-pressure difference.

In a slow, frictionless piston process, the gas pressure is effectively equal to the external pressure. The area under that constant-pressure p-V path then gives the work magnitude. A general gas-pressure curve cannot automatically give work for an irreversible process in which gas and external pressures differ.

Estimate the work scale

Pressure of order 10⁵ Pa acting through a volume increase of order 10^-3 m³ gives work of order 10² J. If heating supplies a few hundred joules, an internal-energy increase of a few hundred joules is plausible, with expansion work taking some energy out.

Apply the same convention to different processes

In the reverse path, compression work enters the gas but more energy leaves by heating. The latter two rows are independent examples; they are not stated to reach the same endpoints as the expansion.

An insulated system can warm when work is done on it. Insulation makes Q zero under the stated model; it does not make ΔU zero or ensure constant temperature. A rigid boundary removes volume-displacement work, but electrical or other work could still enter unless excluded.

For expansion into a vacuum, p_ext = 0, so the displacement work is zero even though the volume increases. A volume increase alone therefore does not establish a positive work output.

03

Temperature change and calorimetry

Specific heat capacity relates a sample's heating to its temperature change. An apparatus input may also warm the vessel or supply losses, so isolate the sample's share before calculating c.

Define the property and its conditions

Specific heat capacity c is the energy required by heating per unit mass per unit temperature rise, without a change of state and under the stated process conditions.

Use c in J/(kg K), m in kg and the temperature change in K. A Celsius temperature interval has the same numerical size, so J/(kg °C) gives the same numerical c. An absolute temperature is not interchangeable with a temperature change.

Use a supplied approximately constant c over the interval. Its value can depend on temperature and process. Heating a gas at fixed pressure and at fixed volume are different conditions; do not treat one gas value as process-independent.

Choose the calorimeter boundary

For a system containing the sample, heater, vessel and temperature sensor, electrical input through the heater is work on that assembly. Energy lost to the surroundings by heating is a separate outward transfer. If the sample alone is selected instead, energy passing from the heater into the sample can be heating across its boundary.

Both descriptions can be consistent, but do not count the same input as both electrical work and heating in one account. A calibrated input-energy reading lets the thermal calculation proceed without assuming how the electrical instrument obtained it.

Decide whether corrections matter

Roughly 0.3 kg warming by about 10 K with a few kilojoules input suggests c of order 10³ J/(kg K). Apparatus warming of a few hundred joules plus a loss of several hundred joules is of order 1 kJ, which is not automatically negligible beside a 6 kJ input.

Make the measurement represent the sample

• Mass: measure the sample separately or subtract the empty container mass. Record the balance range, resolution and zero or tare.
• Temperature: establish sensor contact, record its range and resolution, and use stable readings. Stir a suitable liquid so one sensor reading represents it; avoid treating a local heater hot spot as the whole sample's temperature.
• Input: measure the energy delivered over a defined interval. State whether the heater and vessel belong to the chosen system.
• Apparatus response: use independently determined heat capacity and justify the component temperature changes. Equal endpoint changes do not require every component to share one temperature at every instant.
• Loss: reduce avoidable transfer and estimate the remaining loss under comparable conditions. Record what evidence supports that correction.

Keep the vessel stable and the heater and sensor securely supported so a pulled lead cannot tip or spill the sample. Switch off before adjusting the heater, and let hot parts cool before handling them.

A cooling record or matched reference can help estimate background transfer only if relevant temperatures, exposed areas and surroundings are comparable. A no-heater control at a different temperature does not automatically reproduce the loss during heating. Correction values must come from calibration or other evidence, not be chosen to produce an expected c.

Apply energy conservation to thermal contact

04

Phase change and latent heat

Decide whether the energy changes temperature or changes phase. At a phase change, energy can enter while temperature stays constant under the stated pressure and conditions.

Define specific latent heat

Specific latent heat l is the energy required by heating per unit mass for a stated change of phase at constant temperature under stated conditions. Its unit is J/kg.

Specific latent heat of fusion concerns melting; specific latent heat of vaporisation concerns liquid becoming vapour. Their values need not be equal. Freezing and condensation transfer energy out under the corresponding reversed conditions.

Here lowercase l denotes the specific quantity. Some references use L for it; check the definition and unit. A value in J/kg is not the total energy for an arbitrary sample.

During a phase change, energy can change microscopic potential energy and supply expansion work without raising temperature. Q = mcΔT describes warming within a phase, so setting ΔT = 0 in that equation does not prove that no energy entered during a phase change.

Split a multi-stage change

Before calculating precisely, roughly 0.1 kg multiplied by a supplied fusion latent heat of order 3 × 10⁵ J/kg needs energy of order 3 × 10⁴ J. Melting should therefore dominate the following account compared with its short warming stages.

Check whether all the material changes phase

Suppose a separate 0.0200 kg ice sample is already at its melting point and receives 5000 J, with negligible other transfers. Complete melting would require:

About 5.03 g of ice remains. Ice and water coexist at the melting temperature under the stated conditions; there is insufficient energy to melt everything and then warm the liquid.

Account for work during vaporisation

Latent heating need not all become an internal-energy increase when the sample expands significantly. Choose a fixed sample as the system and use ΔU = Q + W_on.

Measure fusion latent heat with a matched control

A melting investigation can compare water collected with a measured heater input against a comparable no-heater control. Begin with ice and apparatus at the melting temperature, drain pre-existing water before timed collection, and use the same collection geometry and duration. Measure collected mass with a tared collector and record the input and timing resolutions.

In the supplied comparison, measured heater power is 16.7 W = 16.7 J/s for 300 s. The heated run collects 20.00 g of water; the comparable control collects 5.00 g.

This calculation assumes complete collection, equal background transfer in the two runs, and that the measured heater input accounts for the additional melting rather than appreciable apparatus warming or an unmatched loss. The control is a conditional correction, not a guarantee that every error has disappeared.

• Ice initially below its melting point: part of the input first warms it. Treating all that input as melting energy can make the inferred l too high.
• Water retained in the apparatus: the collected extra mass can be smaller than the mass actually melted. Dividing input by that underestimated mass can also give too large an l.
• Different background conditions: a heater can change temperature gradients or losses. A control that no longer matches these conditions cannot simply be subtracted as an equal background.

Water leaving the vessel crosses its boundary. Do not apply a closed, fixed-mass first-law account to that vessel while ignoring the outflow. The matched input-and-mass calculation above states the required assumptions without treating the draining vessel as such a system.

05

Revision summary

Choose the system, identify whether energy crosses its boundary as heating or work, then decide whether its temperature, phase or both change.

State quantities and process transfers

Internal energy U is associated with microscopic kinetic and potential energies. It excludes the whole system's bulk kinetic and external gravitational potential energies. At equilibrium the macroscopic state determines U, so ΔU between the same endpoints is independent of the route. Q and W describe transfers during a process and can depend on that route.

Thermodynamic temperature relates to mean microscopic kinetic energy in the classical model. For an ideal gas, mean translational energy is 3kT/2. Only for the stated monatomic ideal-gas model may its full internal energy here be written:

Equal temperature does not imply equal total U. Net heating is from hotter to colder; thermal equilibrium has no net heating. By the zeroth law, if A and B each equilibrate with C, then A and B are in equilibrium with each other.

Keep the work convention visible

Heating into the system makes Q positive. Expansion makes W_on negative; compression makes it positive. In the supplied expansion, 450 J enters by heating, 180 J leaves as work and U increases by 270 J.

Use absolute external pressure. A slow frictionless process permits the gas-pressure p-V area interpretation. Insulation means Q = 0, not necessarily ΔU = 0. A rigid boundary removes volume work but does not exclude all possible work transfers. Expansion into zero external pressure has no displacement work.

Select the thermal equation by stage

Specific heat capacity is heating required per unit mass per unit temperature rise under stated conditions. Specific latent heat is heating required per unit mass for the specified phase change at constant temperature under stated pressure and conditions. Use compatible mass units and the appropriate property value.

For 0.0800 kg ice at -5.0°C becoming water at 25.0°C, the supplied stages require 840 + 26720 + 8400 = 35960 J. A horizontal temperature segment can have continuing energy input. An energy-axis graph does not directly give a time duration.

If energy is insufficient for a full phase change, calculate the mass changed using Q/l before attempting a temperature rise. If volume work is appreciable, use the first law too: the separate vaporisation example has Q = 18000 J but ΔU = 16800 J after 1200 J work output.

Check a measurement's energy destinations

A heater-and-sample assembly receives electrical work; the sample alone may receive heating from that heater. Do not count the same input twice. In the corrected calorimetry example:

Apparatus warming uses its actual endpoint temperature change; heat-loss corrections need independent evidence under comparable conditions. Assigning all input to the sample gives too high a c in this model.

The fusion comparison uses 5010 J for an extra 15.00 g melted, giving l_f = 3.34 × 10⁵ J/kg. It requires matched background transfer, correct initial thermal conditions and complete collection. The draining vessel alone is not a closed fixed-mass system.

Quantities and units

An unqualified W in the plus-sign first law means work on the system. The unit W in a power value means watt. Likewise, specific heat capacity c here is not the microscopic speed c used in a kinetic-gas derivation.