Full chapter

Thermal properties of matter

All 6 topics and the revision summary on one page.

Internal energy

Temperature describes an average property of the particles. Internal energy adds the motion and interaction energies of all the particles in the system.

A higher temperature corresponds to greater average random particle kinetic energy. Energy is measured in joules (J); a temperature reading in degrees Celsius is not a reading of the total energy stored.

Internal energy is an energy store made up of two particle contributions:

The total kinetic energy associated with the random motion of the particles.
The total potential energy between the particles, associated with their arrangement and interactions.

These are totals for the whole system, not the energy of one selected particle. The potential-energy contribution matters as well as the motion: changing the particles' arrangement can change internal energy without raising temperature.

Temperature is related to an average; internal energy is a total

Worked comparison

100 g and 200 g of the same liquid at 30°C

The two samples have the same average random kinetic energy per particle because they are the same substance at the same temperature.

The 200 g sample contains about twice as many particles. Adding their kinetic energies therefore gives about twice the total random kinetic energy, although the temperature has not doubled.

Internal energy also includes the total potential energy between those particles. Matching temperatures alone does not establish matching internal energies, and the temperature reading does not give a numerical value for either sample's internal energy.

For a fixed amount of one substance that stays in the same state, warming increases average random kinetic energy and increases internal energy. When comparing different samples, also consider the amount of material, the substance and its state. Temperature alone is not a complete energy comparison.

Keep motion of the whole body separate

A trolley moving across a room has a kinetic store associated with that overall motion. Its particles also move randomly and interact with one another; those contributions form its internal store.

Making the whole trolley move faster does not automatically mean its temperature has risen. Conversely, a stationary block can gain internal energy while its centre remains at rest.

Include both internal-energy contributions. "Total energy of moving particles" leaves out the potential energy between particles. "Average kinetic energy" describes the temperature relationship, not the full internal-energy store.

Optional check Two samples of the same liquid, 100 g and 200 g, are both at 30 degrees C. Which statement correctly compares their particle energies?

Next, heat capacity relates an energy transfer to a temperature change. Changes of state show why internal energy can also change at constant temperature.

Heat capacity and specific heat capacity

The energy needed for a temperature rise depends on what is being heated and how much material there is.

Heating transfers energy into a sample's internal store. This page considers a temperature change without a change of state, with specific heat capacity treated as constant over the interval.

A property of the body and a property per unit mass

Heat capacity, C: The energy needed to raise the temperature of the specified body by one degree Celsius or one kelvin. Its unit is J/°C or J/K. A larger amount of the same material has a larger heat capacity under the same conditions.
Specific heat capacity, c: The energy needed to raise the temperature of unit mass of a substance by one degree Celsius or one kelvin. A common unit is J/(kg °C), equivalent to J/(kg K). The word specific means that the energy is given per unit mass.

E = CΔT = mcΔTE is the energy transferred to the sample in J; m is its mass; ΔT is its temperature change. For a uniform sample under these conditions, C = mc.

Here ΔT denotes a temperature difference: final temperature minus initial temperature on one scale. It can be expressed in °C or K, matching the heat-capacity unit. A change of 1°C is the same size as a change of 1 K. This does not mean that a temperature of 1°C equals 1 K.

Use the temperature rise of the sample

A 0.20 kg metal sample warms from 20 to 50 °C without changing state. Its supplied specific heat capacity is 450 J/(kg °C).

ΔT = 50 - 20 = 30 °C.

E = mcΔT = 0.20 × 450 × 30 = 2700 J.

Heat capacity of this sample: C = mc = 90 J/°C.

A rise of 30 °C is also a rise of 30 K. This is a temperature interval, not a conversion of 50 °C to kelvin.

The sample warms from 20°C to 50°C. The marked interval is 30°C, or 30 K. Use this change in the energy calculation, not the final reading of 50°C.

Worked example

Warm a metal sample

A 0.20 kg sample has c = 450 J/(kg °C). It warms from 20°C to 50°C without changing state.

Find the interval: ΔT = 50 - 20 = 30°C.
Find the sample's heat capacity: C = mc = 0.20 x 450 = 90 J/°C.
Find the energy: E = mcΔT = 0.20 x 450 x 30 = 2700 J.

Each degree of temperature rise requires 90 J for this sample. The same material's specific heat capacity is 450 J per kilogram per degree; these are different quantities with different units.

Match the mass unit to the supplied value

A value of 4200 J/(kg °C) means that 1 kg requires 4200 J for a 1°C rise. The equivalent value is 4.2 J/(g °C), because 1 kg contains 1000 g. This is also 4.2 J/(g K), since a temperature interval of 1 K equals an interval of 1°C.

For 100 g of that material warming by 10°C, either use 0.100 x 4200 x 10 or use 100 x 4.2 x 10. Both give 4200 J. Using 100 with the kilogram value would mix incompatible units.

c = E / (mΔT)
ΔT = E / (mc)Rearrange according to the unknown. For equal energy input and equal material, a larger mass has a smaller temperature rise.

Separate energy reaching the sample from heater input

Power is energy transferred per unit time. If a 120 W heater transfers all its output to the metal sample above at a constant rate, the ideal time is t = E/P = 2700/120 = 22.5 s.

If some heater energy warms the apparatus or transfers to the surroundings, less reaches the sample in that time. Reaching the same 30°C rise then requires more input energy or a longer time. A heater's output is not automatically the sample's energy gain.

Determine a specific heat capacity

For a suitable metal sample, use an embedded heater of known constant power and a temperature sensor in good thermal contact with the sample. Insulation reduces transfer to the surroundings. Keep the sample below any change of state.

Measure the mass: use a balance for the sample itself, excluding separate supports and instruments.
Measure the temperature rise: record the initial and final sample temperatures over the chosen heating interval. Place the sensor in the sample's intended sensing position, rather than measuring the exposed heater. Choose suitable range and resolution, and allow for response lag.
Find the supplied energy: with constant heater power P and measured time t, E_input = Pt.
Use the energy account: the input warms the sample and apparatus and may escape to the surroundings. If these other transfers are negligible or have been estimated and subtracted, c = E_sample / (mΔT).

Supplied practical data

Correct the energy account

In a separate run, a 20 W heater operates for 180 s while a 0.20 kg sample warms by 30°C. Independent calibration supplies two estimates: 300 J warms the apparatus and 600 J transfers to the surroundings.

Heater input: E_input = 20 x 180 = 3600 J.
Energy reaching the sample: E_sample = 3600 - 300 - 600 = 2700 J.
Specific heat capacity: c = 2700 / (0.20 x 30) = 450 J/(kg °C).

Ignoring both other transfers gives 3600 / (0.20 x 30) = 600 J/(kg °C), which is too high. The correction values are supplied for this example. An unpowered setup that stays near room temperature would not automatically measure the heat loss from a sample following a different, hotter temperature history.

If all input is wrongly treated as energy reaching the sample, the numerator is too large and the calculated c tends to be too high. Better insulation and accounting for apparatus heating address that cause. Repeating the same imperfect setup may show variation but does not remove the systematic error.

For cooling, say which energy you are calculating

If the same 0.20 kg sample cools from 50°C to 20°C without changing state, the positive amount of energy transferred out is mc(50 - 20) = 2700 J. Under the stated model, the sample's internal-energy change is -2700 J. The minus sign describes a loss; it does not mean that a negative amount was delivered to the surroundings.

Choose the temperature interval and the system. E = mcΔT does not cover melting or boiling, and Pt is the heater input unless the question says all of it reaches the sample.

Optional check A 0.20 kg metal sample has a specific heat capacity of 450 J/(kg degree C). It warms from 20 to 50 degrees C without changing state. How much energy reaches the sample?

Changes of state

A substance can gain or lose internal energy while its temperature stays constant during a change of state.

Consider a pure substance undergoing the stated transition at constant pressure. Its temperature remains at the transition temperature while the amounts in the two states change. The particles continue moving; constant temperature does not mean zero motion.

State changes and the direction of energy transfer

For a pure substance at fixed pressure, temperature stays constant during each stated transition. Energy moves into or out of the substance.

Melting and boiling transfer energy into the substance. Solidification and condensation transfer energy out. Each reverse process has the opposite energy-transfer direction.

Four processes, with two transfer directions

Melting: solid to liquid: Energy enters the substance. Particles become able to move past one another instead of remaining near fixed positions. Their arrangement and interactions change, increasing the potential-energy contribution to internal energy.
Solidification: liquid to solid: Energy leaves the substance. Particles settle into an arrangement in which they vibrate about fixed positions. The potential-energy contribution decreases. The released energy can increase internal stores of the surroundings.
Boiling: liquid to gas: Energy enters the substance. Particles become widely separated as the liquid changes to gas. Energy is needed to separate particles against their attractions, so their potential energy increases.
Condensation: gas to liquid: Energy leaves the substance. Particles become close together in a liquid, and their potential energy decreases. The energy transferred out can increase the surroundings' internal stores.

During each transition at the stated conditions, the average random kinetic energy remains unchanged because the temperature remains unchanged. The energy change is associated with the changing particle arrangement and potential energy.

Melting does not break the atoms or molecules into smaller particles. It changes how they are arranged and how freely they move relative to their neighbours. A liquid's particles are still close together; becoming widely separated is characteristic of the gas state.

Read a temperature plateau as a state change

The following idealised values describe an unspecified pure sample at a constant pressure. Its melting point is 50°C, and energy continues to enter throughout the six minutes. These are supplied model values, not measurements of a named material.

Temperature while energy continues to enter the sample
Time / min	Temperature / °C
0	20
1	35
2	50
3	50
4	50
5	65
6	80

Energy continues entering during the melting plateau

Supplied idealised data for an unspecified pure substance at constant pressure. Its melting point is 50 °C.

0-2 min: the solid warms from 20 to 50 °C.

2-4 min: solid changes to liquid at 50 °C. Energy transfer continues.

4-6 min: the liquid warms from 50 to 80 °C.

The temperature rises to 50°C, stays there from 2 to 4 min while melting occurs, then rises again as the liquid warms. Energy continues entering during the horizontal part of the graph.

0 to 2 min: the solid warms from 20°C to 50°C. Its average particle kinetic energy increases.
2 to 4 min: melting takes place at 50°C. The amount of liquid increases and the amount of solid decreases. Internal energy increases while average random kinetic energy stays unchanged.
After 4 min: the sample is liquid and its temperature rises. Average particle kinetic energy increases again.

Worked explanation

Why does the reading stay at 50°C?

At 3 min, the sample is in the melting interval. Continued energy input changes the particle arrangement and potential energy as more solid becomes liquid. It does not increase average random kinetic energy during this transition, so the temperature remains at 50°C.

The flat section does not show that the heater has stopped or that the particles have stopped moving. Temperature and total internal energy are different quantities.

Explain the reverse process

If energy is removed from the same liquid, it first cools to its solidification temperature. While it solidifies at the same fixed pressure, energy continues to leave and the amount of solid increases, but its temperature stays constant until the transition is complete.

The same reasoning applies to boiling and condensation at the boiling/condensation temperature for that pressure. Boiling requires continuing energy input while liquid becomes gas; condensation transfers energy out while gas becomes liquid. A constant temperature does not mean the transfer has ended.

Measure temperature and time consistently

Keep the thermometer's sensing region in the sample, clear of the heated vessel, and use the same timing origin for all readings. Read at the chosen intervals while recording whether the sample is solid, liquid or a mixture. Do not infer state only from a single temperature reading.

A thermometer that responds slowly can lag behind a rising sample temperature and round off a change in slope. Poor placement can measure a locally hotter region rather than the sample represented by the model. Repetition does not by itself fix either cause.

State the plateau conditions. This explanation concerns a pure substance undergoing a transition at fixed pressure. It does not claim that every mixture melts at one sharp temperature or that all warming must include a flat interval.

Optional check A pure sample melts at 50 degrees C at constant pressure while energy continues to enter. Which particle explanation accounts for its constant temperature?

Latent heat gives the energy needed for a stated amount to change state. Before calculating, distinguish boiling from evaporation: both can make vapour, but they occur differently.

Boiling and evaporation

Boiling forms vapour throughout a liquid. Evaporation occurs at its surface and can happen below the boiling temperature.

Particles in a liquid move randomly with a range of kinetic energies. Attractions keep them close together, but some particles at the surface have enough energy to escape into the gas above.

Two ways for liquid to become vapour
Boiling	Evaporation
Vapour bubbles form throughout the liquid and rise.	Particles escape at the liquid surface; bubbles throughout the liquid are not required.
A pure liquid boils at its boiling temperature for the stated pressure.	It can occur below the boiling temperature, over a range of temperatures at which the liquid exists.
During boiling at fixed pressure, continued energy input changes the state without raising the temperature.	It can reduce the liquid's temperature if incoming energy does not replace the energy removed quickly enough.

Boiling bubbles contain the substance's vapour. They are not simply bubbles of liquid water or evidence that water has split into hydrogen and oxygen. The boiling temperature depends on pressure; 100°C is the familiar value for water at approximately normal atmospheric pressure, not a rule for every liquid under all conditions.

Why evaporation can cool a liquid

Some particles escape from the liquid surface

This particle model shows evaporation below the boiling temperature. All the liquid particles keep moving; only selected motion arrows are drawn.

Particles that escape take energy with them. The average kinetic energy of the particles remaining can fall, so the liquid cools.

Cooling depends on the energy account: incoming energy may replace the energy removed quickly enough to maintain the temperature. The arrows show particle motion, not a separate energy-transfer pathway.

Some higher-energy surface particles escape while the others continue moving and interacting. Evaporation takes place at the surface; this picture does not show vapour bubbles throughout the liquid.

A range of energies: particles do not all have the same kinetic energy, even at one temperature.
Escape: a surface particle needs sufficient energy to overcome the attractions holding it in the liquid.
What remains: preferential removal of higher-energy particles can reduce the average random kinetic energy of the remaining liquid.
Temperature: the lower average kinetic energy corresponds to a lower temperature, unless incoming energy offsets the loss.

The whole liquid does not need to reach its boiling temperature for some surface particles to escape. A wet surface can therefore dry in an ordinary room.

Worked explanation

Why evaporating sweat cools skin

Sweat on the skin changes from liquid to vapour. Energy needed for this change can transfer from the skin into the evaporating liquid, so the skin loses energy.

The cooling effect depends on evaporation, not merely on the presence of liquid. If surrounding air already contains much water vapour and little evaporates, this cooling route is reduced. Incoming energy can also partly or fully replace the energy lost, so an evaporating surface need not always fall in temperature.

Interpret a temperature comparison

Compare two similar temperature probes initially at the same room temperature. Cover one sensing region with a small wet covering and leave the other dry. Under the same airflow, evaporation from the wet covering can make that probe read a lower temperature.

The lower reading concerns the evaporating wet region. It does not show that the room air as a whole has reached that temperature. Keep probe type, initial conditions and airflow comparable, and allow for response time before interpreting the difference.

Cooling needs an energy explanation. Evaporation does not release a substance called cold. It removes energy, and the remaining liquid cools when other transfers do not replace that energy fast enough.

Optional check A liquid evaporates at its surface below its boiling temperature. Why can the liquid cool if energy from the surroundings does not replace the energy removed quickly enough?

Latent heat

An energy transfer can change the state of a sample while its temperature stays constant. Latent heat measures that transfer.

For a pure substance at the stated transition and fixed pressure, energy changes the particle arrangement and potential energy during the transition. Average random kinetic energy, and therefore temperature, remains unchanged.

Latent heat, L: The energy transferred when the specified sample changes state without a temperature change. It is a total energy, measured in joules.
Specific latent heat, l: The energy transferred per unit mass for that change of state without a temperature change. The lowercase letter l has units J/kg or J/g, depending on the stated mass unit.

E = mlUse the mass that changes state and the specific latent heat for that transition. For this sample's transition, its total latent heat L is ml.

Specific latent heat of fusion concerns solid becoming liquid; the same magnitude is released per unit mass in the reverse solidification process at the same conditions. Specific latent heat of vaporisation concerns liquid becoming gas; condensation transfers that energy out.

During melting, the arrangement changes so that particles can move past one another while remaining close together. During boiling, particles become widely separated against their attractions. Both processes increase the particle potential-energy contribution. During solidification or condensation, it decreases as energy is transferred out. The particles continue moving; latent heat is not energy used to stop them or split the molecules into smaller particles.

Worked example

Ice already at its melting point

A 0.050 kg sample of ice is already at 0°C under normal atmospheric pressure. Use the supplied specific latent heat of fusion l_f = 334000 J/kg.

Identify the stage: only melting is required; there is no warming interval in this question.
Use the mass that melts: E = ml_f = 0.050 x 334000 = 16700 J.
Describe the result: after this ideal transfer, the sample is liquid water at 0°C, not water at a higher temperature.

Equivalently, 334000 J/kg is 334 J/g, and 50 g x 334 J/g gives the same 16700 J. If 0.050 kg of water freezes at the same conditions, 16700 J is transferred out.

Calculate each stage separately

Now start with 0.10 kg of ice at -10°C and finish with water at 20°C, at normal atmospheric pressure. First warm the ice to 0°C, then melt it, then warm the water. The supplied heat capacities apply to their respective states.

Follow the same 0.10 kg sample through three stages

Ice at -10 °C becomes water at 20 °C. Use the supplied material values and a melting point of 0 °C at the stated pressure. The energy amounts below all reach the sample; warming the apparatus and transfers to the surroundings are excluded.

1. Warm the ice

c_ice = 2100 J/(kg °C); ΔT = 0 - (-10) = 10 °C.

E = mc_iceΔT = 0.10 × 2100 × 10 = 2100 J.

2. Melt the ice

Specific latent heat of fusion: l_f = 334000 J/kg.

E = ml_f = 0.10 × 334000 = 33400 J.

The lowercase l_f is energy per unit mass. The sample's temperature stays at 0 °C during melting.

3. Warm the water

c_water = 4200 J/(kg °C); ΔT = 20 - 0 = 20 °C.

E = mc_waterΔT = 0.10 × 4200 × 20 = 8400 J.

Total energy transferred in = 2100 + 33400 + 8400 = 43900 J.

The diagrams show the sequence. Box sizes and arrow lengths do not measure energy or time.

Choose the equation for each stage before substituting. The warming stages use mcΔT; melting uses ml. The cards are schematic and are not sized in proportion to their energy values. The total 43900 J is energy reaching the sample.

The melting contribution is 33400 J even though the temperature does not rise during that stage. The two warming contributions are 2100 J and 8400 J. Adding them gives 43900 J; neither one mcΔT calculation across -10°C to 20°C nor ml alone accounts for the complete change.

This total excludes energy warming the apparatus or escaping to the surroundings. An actual heater may need to supply more. The supplied material values are data for this example, not values that can be applied to every substance.

Determine specific latent heat of fusion

Use ice already at its melting point, with a heater transferring energy at a known constant rate. Collect and measure the mass of liquid produced over a measured time. Melting can also occur because the surroundings are warmer than the ice, so not all collected meltwater is necessarily due to the heater.

Establish the transition: let the ice reach its melting point and drain away water present before the timed collection. Keep some ice present throughout the measurement interval so the intended process remains melting.
Measure meltwater mass: collect the water in a dry weighed container; subtract its empty mass from its final mass. A balance reading that includes the container is not the mass that melted.
Estimate background melting: use a matched unpowered ice setup over the same time. Match the surroundings, exposed geometry, initial conditions and collection method as closely as possible.
Separate the heater contribution: estimate the additional mass melted by the heater as m_heated - m_control. Use its mass unit consistently with the intended l unit.
Relate energy and mass: if apparatus warming and other differences are negligible or corrected, l_f = Pt / (m_heated - m_control).

Supplied practical data

Separate heater melting from background melting

A 16.7 W heater operates for 200 s. The powered setup produces 14.0 g of collected meltwater; a matched unpowered control produces 4.0 g. Both contain ice at its melting point throughout. Assume equal background heating, negligible apparatus warming during collection and negligible energy loss from the heater or leads outside the melting region.

Heater energy: E = 16.7 x 200 = 3340 J.
Additional melted mass: 14.0 - 4.0 = 10.0 g = 0.0100 kg.
Specific latent heat: l_f = 3340 / 0.0100 = 334000 J/kg.

The 4.0 g correction estimates melting due to the surroundings. It does not, by itself, account for energy that leaves the heater through its leads or warms another part of the apparatus.

Ignoring background melting makes too much mass appear to have melted because of the heater. Dividing its energy by this excessive mass gives an estimate of l_f that is too low.

The control is an estimate, not an automatic correction for every transfer. It must experience comparable background heating. Meltwater left behind instead of collected, initial water included by mistake, apparatus warming and changing heater output can also affect the result. Repetition helps reveal variation but does not remove an unmatched control or a wrong energy account.

Choose the stage, then the equation. Use mcΔT for a temperature change within one state. Use ml for the mass that changes state at its transition temperature. Capital L is total energy; lowercase l is energy per mass.

Optional check A 0.10 kg sample of ice at -10 degrees C becomes water at 20 degrees C. Supplied values are c_ice = 2100 J/(kg degree C), l_f = 334000 J/kg and c_water = 4200 J/(kg degree C). Which total energy reaches the sample at the stated normal-pressure transition?

Cooling curves

A cooling curve records temperature against time. Read each interval by asking which state is present and how energy is leaving.

Net heating transfers energy from a hotter region to a cooler one. Temperature falls within a state, but can remain constant during a change of state at the stated conditions.

Interpret a liquid that cools and solidifies

The following supplied model readings describe an unspecified pure substance at fixed pressure. It starts as a liquid, and its solidification temperature is 50°C. Energy continues to leave the sample throughout. These are idealised values, not readings collected from an experiment.

Energy leaves during the solidification plateau

Supplied idealised readings for an unspecified pure substance at fixed pressure. The sample starts as a liquid. These readings illustrate interpretation; they were not collected from an experiment.

Supplied model readings
Time / min	Temperature / °C
0	80
1	65
2	50
3	50
4	50
5	35
6	20

0-2 min: the liquid cools from 80 to 50 °C.

2-4 min: liquid changes to solid at 50 °C. During solidification, both states are present.

4-6 min: the solid cools from 50 to 20 °C.

On the plateau, average particle kinetic energy stays unchanged while the interaction potential energy decreases. Internal energy falls as energy continues to leave the sample.

Temperature is plotted vertically and elapsed time horizontally. The sample cools as liquid, solidifies during the 50°C plateau from 2 to 4 min, then cools as solid. The straight segments are an idealised model, not a claim that real cooling rates are always constant.

During the sloping intervals, temperature and average random particle kinetic energy decrease. During solidification, the amount of solid increases while the amount of liquid decreases. Energy still leaves, but the particle potential-energy contribution decreases while average random kinetic energy stays unchanged.

Worked interpretation

What can be said at 3 min?

The point lies within the solidification interval. Both liquid and solid are present, the temperature is 50°C, and energy is still being transferred out.

The reading does not show that half the sample is solid simply because 3 min is halfway between 2 and 4 min. That fraction would need additional information about the energy-transfer rate and the energy required for the whole transition.

Sketch from the physical stages

Label both axes and units: temperature on the vertical axis, elapsed time on the horizontal axis.
Start in the given state: for this liquid, draw a falling temperature towards 50°C.
Mark the transition: draw the constant-temperature interval at 50°C and label it liquid plus solid. Continued energy loss makes more solid.
Continue after the transition: once the sample is all solid, draw a falling temperature again.

If plotting readings, place their actual values first. If sketching without numerical data, do not invent an exact duration or make both sloping regions equally steep unless the information supports it.

A cooling gas can similarly condense at its condensation temperature for the stated pressure before the liquid cools further. Identify the specified transition; not every horizontal part is melting, and a cooling curve need not contain every possible state change.

Use a temperature probe and data logger

A probe and data logger can record temperature automatically at selected time intervals. The probe measures the temperature of its sensing region, not heat, latent heat or the whole sample's internal energy directly.

Place the sensor: keep its sensing region in good thermal contact with the sample and clear of the vessel's base and walls. Maintain suitable contact as the sample changes state.
Choose the range and interval: the probe must cover the sample's temperatures. Select a sampling interval short enough to show the transition, and keep the same time origin.
Record and inspect: retain time units and temperature units, and observe the sample's state where possible. The graph must represent the recorded points, including any scatter.
Evaluate the response: a slow sensor can round off changes in slope. Recording more often does not make the sensor respond faster.

In the model above, readings only at 1 min and 5 min would miss the entire 2 to 4 min plateau. More frequent sampling can reveal the interval, but cannot fix poor contact or an unsuitable response time.

Real cooling rates may change as the temperature difference from the surroundings changes. A fixed reading from a disconnected or poorly placed sensor is not evidence of solidification. Check the sensor and the observed state before assigning a physical explanation to a flat trace.

A flat temperature is not a flat energy account. During the stated solidification interval, internal energy decreases even though temperature stays constant.

Optional check An idealised cooling curve for a pure substance at fixed pressure is horizontal at 50 degrees C from 2 to 4 min while it solidifies. What happens during that interval?

Revision summary

Internal energy: Total kinetic energy of random particle motion plus total potential energy between particles. Temperature concerns an average; it does not give a sample's total internal energy.
Temperature change: E = CΔT = mcΔT, with C = mc. Use a temperature interval, no change of state, and c treated as constant over the interval. Match kg with J/(kg °C), or g with J/(g °C).
Change of state: E = ml. Use the mass that changes state and the correct fusion or vaporisation value. Lowercase l is energy per mass; the sample's total latent heat L is in J.
Heater input: With constant power, E_input = Pt. It equals energy reaching the sample only when apparatus heating and other transfers are negligible or accounted for.

Follow the particles and the energy

Melting and boiling take in energy; solidification and condensation transfer it out. For a pure substance at its transition temperature and fixed pressure, average random kinetic energy stays unchanged while particle arrangement and potential energy change.
Evaporation occurs at the surface and can happen below boiling temperature. Preferential escape of higher-energy particles can cool the liquid if incoming energy does not replace the loss.
Cooling-curve slopes show falling temperature within a state. A transition plateau contains two states and can involve continuing energy loss.
For a change involving several states, separate the warming, cooling and state-change stages before choosing equations.

Keep the distinctions clear

Quantities and conditions that are easy to confuse
Distinction	Remember
C and c	Heat capacity C belongs to the specified body, in J/°C. Specific heat capacity c is per unit mass, for example J/(kg °C).
Temperature and interval	Here ΔT is final minus initial temperature on one scale. From 20°C to 50°C, ΔT = 30°C = 30 K. Use the difference with a matching heat-capacity unit, not the final reading of 50°C.
Energy out and energy change	An amount transferred out is positive when stated as an amount. The sample's corresponding internal-energy change is negative.
Boiling and evaporation	Boiling forms vapour bubbles throughout the liquid at its boiling temperature for the pressure. Evaporation occurs at the surface over a range of temperatures.
A plateau and no transfer	During the stated phase change, energy changes particle potential energy. Constant temperature does not mean particles stop or energy transfer stops.

Measurement reminders

Identify the sample mass, measure a temperature difference and distinguish heater input from the sample's gain. For melting, measure the mass that changes state and account for background melting. A control needs comparable conditions. For a cooling curve, keep the probe in good contact, choose a useful sampling interval and retain the actual readings. Repetition does not correct poor placement, response lag or a wrong energy account.

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