Thermal processes and efficiency

Ice receives energy, yet its temperature stays almost constant while melting. A gas being compressed can warm without receiving heat. To understand these cases, distinguish the change of state, the way energy is transferred, and the result of a complete cycle.

Place a spoon in hot water: energy passes to the cooler spoon because of the temperature difference. This is heat transfer, which can occur by conduction, convection, or radiation. Heat Q measures transferred energy, not a substance inside an object. Stored internal energy is U. To calculate a temperature change, we need the energy required by the particular material under the chosen conditions.

Heating twice the mass through the same temperature rise usually takes twice the energy. With constant specific heat capacity c, this gives Q=mc\Delta t. Here m is in kg, \Delta t is the temperature change in °C, and c is in J/(kg·K); temperature changes in °C and K have equal numerical values. The value of c belongs to the stated process conditions, especially for gases. The formula applies without phase changes or other unaccounted energy transfers; during cooling, Δt and Q are negative.

Consider ice below its melting temperature. At first, incoming energy raises its temperature. Once melting begins, the structure changes: particles can change neighbours. A pure crystalline substance melts at constant temperature under constant pressure while both phases coexist. Energy is still entering, and the molecules themselves are not being broken apart. A single mc\Delta t calculation therefore cannot describe the whole journey from cold ice to water.

For material already at its melting temperature, the specific latent heat of fusion \lambda is the required energy per kilogram, in J/kg. Melting mass m completely requires Q=m\lambda. Twice the mass requires twice the energy under the same conditions; zero ΔT does not prevent this. During the reverse process, freezing, the same energy is transferred to the surroundings: for the material itself, Q=-m\lambda. The sign indicates transfer direction, not the absence of a phase change.

Turning liquid into vapour changes the arrangement and interactions of particles; expansion may also involve work. The specific latent heat of vaporisation r combines the required energy per kilogram under specified conditions: Q=mr, with r in J/kg. Vaporising twice the mass requires twice the energy at the same r. A pure substance boils at constant temperature under constant pressure; surface evaporation can also occur at other temperatures that change during the process.

During condensation, vapour gives energy back to its surroundings: for the condensing mass, Q=-mr under the same conditions. For a more complex path, separate physically different stages. Warm cold ice using its heat capacity, melt it, then warm the water using the heat capacity of water. Add the energies of the stages, not their temperatures. Check that each transition is complete before treating the whole mass as liquid and applying the next stage.

Where can the energy come from? Complete combustion of fuel mass m releases Q=mq, where q is the specific energy of combustion in J/kg. Doubling the fully burned mass doubles the energy released. Here Q is the positive magnitude of released energy; for the system giving it out, the transfer would have a negative sign. Only some reaches the heated object: energy can warm the air, walls, and escaping combustion products. Released energy need not equal useful energy received.

Now thermally insulate the gas and move the piston. No heat crosses the boundary: Q=0. This is an adiabatic process. The first law becomes \Delta U=-A=A\prime. It does not say U is constant: work still transfers energy. Good insulation or a sufficiently rapid process can make an experiment approximate this model; speed alone is not the definition. Heat transfer must be negligible.

During adiabatic expansion, the gas pushes the piston and does positive work. No heat replaces this energy, so U decreases and an ideal gas of fixed composition cools. Compression does work on the gas, increasing U and temperature. The work condition matters: in free expansion into a vacuum, the gas may do no work, so cooling of an ideal gas no longer follows. Increasing volume alone is insufficient.

After a single expansion, the gas may leave the piston at the end of the cylinder. An engine needs a repeatable sequence of processes: a cycle that returns the working substance to its initial state. Then \Delta U_{\text{цикл}}=0: its internal energy cannot decrease cycle after cycle. Define positive magnitudes Q_1 for heat received from the hot reservoir and Q_2 for heat released to the cold reservoir. This differs from using a single signed Q at each stage.

For a complete cycle, the first law gives A=Q_1-Q_2. Some energy from the hot reservoir leaves as work; the rest goes to the cold reservoir. No energy remains accumulated inside after the cycle. Doubling both heats while keeping their ratio doubles the work. Q₂ is subtracted because it is defined as a positive magnitude of outgoing heat. Energy conservation sets the balance; ruling out complete conversion of heat from one reservoir into cyclic work also requires the second law of thermodynamics.

To compare engines, find the fraction of energy from the hot reservoir that becomes work: \eta=A/Q_1=(Q_1-Q_2)/Q_1. This is dimensionless; multiplying by 100 gives a percentage. At the same Q₁, less outgoing Q₂ means higher efficiency, but physics limits how far Q₂ can be reduced. Dividing by Q₂ would compare work with rejected energy rather than the energy input. With A = 0, positive heat input yields no useful cyclic work.

Why must even an ideal engine reject some energy? The direction of thermal processes imposes another constraint beyond energy conservation. Let the reservoirs have constant temperatures T_1>T_2, both in K. A reversible process is a limiting process that can be reversed without leaving any net change in the system and surroundings together. Such an engine sets the upper limit between two reservoirs. Removing mechanical friction alone does not make every thermal stage reversible.

For a reversible Carnot engine, the limit is \eta_{\text{Карно}}=(T_1-T_2)/T_1=1-T_2/T_1. As the temperatures approach each other, it approaches zero: equal temperatures provide no such thermal resource for an engine. At fixed T₂, raising T₁ raises the limit, but the formula does not promise this efficiency for a real device. Its efficiency is found from work and Q₁; between the same reservoirs, it cannot exceed the Carnot limit. The ratio requires absolute temperatures.

In one cycle, an engine receives 800 J of heat and rejects 500 J to the cold reservoir. Find its work and efficiency.

The working substance returns to its initial state: ΔUcycle=0.

Work A=Q₁−Q₂=800−500=300 J.

Efficiency η=A/Q₁=300/800=0.375=37.5%.

Check: 37.5% of the incoming energy becomes work; the other 62.5% goes to the cold reservoir.

Work: 300 J; efficiency: 37.5%.

Heating and phase changes use energy differently, so calculate them as separate stages.

No heat transfer does not imply constant temperature: work can change internal energy.

Over a cycle, U returns to its initial value and A = Q₁ − Q₂. The Carnot limit depends on two temperatures and is not automatically the efficiency of a real engine.

Next, we describe electrical interaction using charge and field.