Free Pearson Edexcel International A Level Physics lesson
Thermal energy and temperature
Follow energy through warming and changes of state.
What you’ll learn
- Separate a thermal state, a microscopic energy store and a transfer process.
- Use the energy reaching the stated sample and distinguish absolute temperature from temperature change.
- Treat warming and phase change as separate stages in the same energy balance.
Temperature is not total internal energy
Internal energy belongs to the microscopic state of a system. It includes the disordered kinetic energy of particles and the potential energy associated with their interactions. The kinetic energy of a whole moving beaker is not part of its internal energy: moving the beaker without changing its temperature does not necessarily change the molecular disorder inside it.
Temperature describes the thermal state and determines the direction of spontaneous energy transfer when systems can exchange energy. Two samples at the same temperature can have very different internal energies because their particle numbers, substances and phases differ. Heating is an energy transfer caused by a temperature difference; heat is not a substance stored in an object.
During melting, supplied energy can increase molecular potential energy as particles become less strongly constrained, while the temperature remains approximately constant. During warming within one phase, average random kinetic energy generally increases. Define the system before comparing energies, because energy used to warm the vessel is outside a budget that includes only its contents.
Internal energy = microscopic random kinetic energy + interaction potential energy
Symbols and units
- U is total internal energy in joules (J): microscopic random kinetic energy plus interaction potential energy, excluding bulk motion of the whole sample.
- N is the number of gas particles, a dimensionless count; T is absolute temperature in kelvin (K).
- For the ideal monatomic-gas example, k is Boltzmann’s constant, 1.38 × 10⁻²³ J K⁻¹, and the mean translational energy per particle is 3kT/2 in J.
How the equation explains the idea
- Use the stated classical ideal monatomic model: interactions and internal molecular modes do not contribute, so U = N × (3kT/2). This is a special case of the internal-energy definition.
- At the same T, the mean energy 3kT/2 is unchanged. Replacing N by 2N gives U₂ = 2N × (3kT/2) = 2U₁.
- Therefore U₂/U₁ = 2 even though the temperatures are equal. For a real phase change, interaction potential energy can also change, so temperature alone cannot determine U.
Worked example
Two equal-temperature samples
- Two samples of the same ideal monatomic gas contain N and 2N particles at the same T.
- Each particle has the same mean random kinetic energy.
- The 2N sample therefore has twice the internal energy in this ideal model.
Equal temperature does not imply equal total internal energy.
Keep this idea. Separate a thermal state, a microscopic energy store and a transfer process.
Warming uses a temperature difference
Specific heat capacity c relates an energy transfer to a temperature rise within a phase: ΔE = mcΔθ. Its unit is J kg⁻¹ K⁻¹. A temperature difference of one degree Celsius is the same size as one kelvin, so a difference calculated from Celsius readings is valid here. The absolute temperature itself is a different quantity.
In an electrical heating experiment, VIt is the electrical input if voltage and current are constant. Some input warms the heater and container and some escapes to the surroundings. Equating all input to the sample’s mcΔθ usually overestimates c. Insulation, a close-fitting lid and a well-coupled temperature sensor reduce losses, but an explicit loss correction is better evidence than simply calling the apparatus insulated.
If c varies significantly with temperature, a single value gives only an approximate result over a range. A heating curve can separate a slope during warming from a plateau during a phase change. Read both temperature endpoints and account for their uncertainties; repeated measurements reduce random scatter but do not repair a persistent sensor calibration error.
ΔE = mcΔθ constant electrical input: E = VIt
Symbols and units
- ΔE is the energy gained by the sample in J; m is its mass in kg; c is specific heat capacity in J kg⁻¹ K⁻¹.
- Δθ = θfinal − θinitial is a temperature change in K or °C; these difference units have equal size. It is not an absolute temperature.
- In E = VIt, E is electrical input in J, V is heater voltage in V, I is current in A and t is time in s; voltage and current must remain constant.
- Puseful is the power reaching the sample in watts (W = J s⁻¹), after container heating and other losses are allowed for.
How the equation explains the idea
- The block stays in one phase with approximately constant c. Its temperature change is 43 − 18 = 25 °C = 25 K, so the relevant change is 25, not the final reading 43.
- Multiply energy per kilogram per kelvin by mass and rise: ΔE = mcΔθ = 0.40 × 900 × 25 = 9000 J.
- From ΔE = Puseful t, divide by useful power: t = 9000/30 = 300 s. Using all electrical power here would underestimate the time if some input misses the block.
- Conversely, c = ΔE/(mΔθ). Substituting uncorrected VIt for a smaller sample energy gain makes the inferred c too large.
Conditions. Computed from ΔT = Pt/(mc). Both samples have c = 1000 J kg⁻¹ K⁻¹ and receive useful power P = 20 W, with no phase change or heat loss. Mass is the only changed input.
Time / s; Temperature rise ΔT / K.
The gradient is P/(mc): doubling mass halves the warming rate. At 100 s the 0.20 kg sample has risen by 10 K, while the 0.40 kg sample has risen by 5 K. These are temperature differences, not absolute temperatures.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Warm an instrument block
- A 0.40 kg block has c = 900 J kg⁻¹ K⁻¹ and warms from 18 °C to 43 °C.
- Δθ = 25 K, so ΔE = 0.40 × 900 × 25 = 9000 J.
- At useful heating power 30 W, the minimum heating time is 9000/30 = 300 s.
The block gains 9.00 kJ.
Keep this idea. Use the energy reaching the stated sample and distinguish absolute temperature from temperature change.
Phase changes have an energy cost
Specific latent heat L is the energy transferred per unit mass for a stated phase change, under stated conditions. Use ΔE = LΔm with the mass that actually changes phase. Fusion means melting in this thermal context; nuclear fusion is a different process. Vaporisation normally requires more energy per kilogram than melting because particles must separate much further.
A heating curve may show warming, a phase-change interval, then further warming in the new phase. Calculate these stages separately and add the energies. A horizontal interval does not mean that the heater has stopped doing work. At the phase-change temperature, increasing the phase fraction changes internal potential energy without necessarily increasing mean molecular kinetic energy.
Evaporation can occur below the boiling point, while boiling occurs throughout a liquid when its vapour pressure matches the external pressure. A classroom latent-heat measurement must specify what mass is measured, subtract any background melting or evaporation and handle hot surfaces under supervision. The numerical model is not a practical safety procedure.
ΔE = LΔm staged heating: Etotal = Σ(mcΔθ) + Σ(LΔm)
Symbols and units
- ΔE is energy transferred during the specified phase change in J; L is specific latent heat in J kg⁻¹.
- Δm is the mass that actually changes phase in kg, not necessarily the entire sample mass.
- In the staged sum, m is the mass warmed in that stage in kg, c is its specific heat capacity in J kg⁻¹ K⁻¹ and Δθ is that stage’s temperature change in K.
- Etotal is the sum of stage energies in J; Σ means add the separate contributions. 1 kJ = 1000 J.
How the equation explains the idea
- The example starts at the melting temperature, so the useful input goes into a change of phase. There is no initial mcΔθ warming term in this stated budget.
- Rearrange ΔE = LΔm by dividing by L: Δm = ΔE/L. Convert both values consistently: 12.0 kJ = 12000 J and 300 kJ kg⁻¹ = 300000 J kg⁻¹.
- Calculate Δm = 12000/300000 = 0.0400 kg = 40.0 g. The energy changes the microscopic arrangement while the coexistence temperature stays approximately fixed.
- If warming occurs before or after the change of phase, calculate each mcΔθ separately and add it to LΔm; a temperature plateau does not mean zero energy transfer.
Conditions. Hypothetical pure material: m = 0.10 kg, initial T = 260 K, melting point = 280 K, csolid = 1000 J kg⁻¹ K⁻¹, cliquid = 2000 J kg⁻¹ K⁻¹, and L = 200000 J kg⁻¹. Constant external pressure, equilibrium phases and no losses; this is a model, not a measured substance.
Useful energy supplied / J; Temperature / K.
The first 2000 J warms the solid by 20 K. The next 20000 J melts it at 280 K. A further 4000 J warms the liquid by 20 K. The plateau width is mL, while each sloping segment has gradient 1/(mc). At constant useful power, energy is proportional to elapsed time.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Finish melting a sample
- A heater supplies 12.0 kJ usefully to a substance at its melting temperature.
- Its specific latent heat of fusion is 300 kJ kg⁻¹.
- Δm = 12000/300000 = 0.0400 kg.
Forty grams melts with no required temperature rise.
Keep this idea. Treat warming and phase change as separate stages in the same energy balance.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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