Energy in and out: heat, temperature, and exothermic and endothermic change
By the end of this lesson you will be able to
- Distinguish between heat and temperature, and convert between joules, kilojoules, megajoules, kilocalories and kilowatt-hours.
- Define system and surroundings, and classify a process as exothermic or endothermic from a temperature change.
- Describe the energy changes in physical changes (melting, boiling, dissolving) and compare their size with chemical changes.
- Explain, at the level of ions and water molecules, why some salts cool water when they dissolve and others warm it.
- Distinguish qualitative from quantitative measurements of energy change, and calculate the energy a sample of water gains or loses.
You met this before
- Energy is measured in joules (J); it is never made or destroyed, only transferred from one store to another — the law of conservation of energy (Year 1).
- Temperature measures how hot something is — the average kinetic energy of its particles; heat is energy transferred because of a temperature difference, always from hotter to colder (Year 1 and Year 3 Heat).
- An exothermic change transfers energy to the surroundings, which get hotter; an endothermic change takes energy from the surroundings, which get colder (Year 3 Energy and chemical change).
- Breaking bonds takes in energy; making bonds gives energy out. Whether a change is exo- or endothermic depends on which is bigger (Year 3, “bonds that cost and bonds that pay”).
- Melting and boiling take in energy without a change of temperature; freezing and condensing give it out (Year 3 Heat).
A student mixes two colourless solutions in a test tube. The thermometer goes from 21.0 °C to 17.5 °C. State whether the change is exothermic or endothermic, which way energy moved, and one other observation that would tell you a chemical reaction happened.
Reveal
Endothermic: the temperature of the mixture fell by 3.5 °C, so energy moved from the water and the test tube (the surroundings) into the reacting particles (the system).
Signs of a chemical change: a gas given off, a precipitate, a colour change, or a new smell. A temperature change on its own does not prove a reaction — dissolving and melting change temperature too, as this lesson shows.
Revise it in full: Year 3 · Energy and chemical change · Lesson 4 — Energy in and out: exothermic and endothermic change · Year 1 · Energy · Lesson 4 — Thermal energy: temperature, heat and how it moves · Year 3 · Energy and chemical change · Lesson 1 — Physical or chemical?
The puzzle
A netball player goes down with a twisted ankle. The first-aider takes a white plastic pouch from the kit, squeezes it hard until something inside pops, shakes it for ten seconds — and it is suddenly as cold as ice. Nothing was plugged in, nothing was frozen beforehand, and the pouch has sat in a hot kit bag all afternoon.
A different pouch, in a hiker’s pack, does the opposite: squeeze, shake, and it becomes too hot to hold. Both pouches contain only water and a white salt. Where did the “cold” come from, and where did the heat come from?
The cold pack contains 30 g of ammonium nitrate and 100 g of water in separate compartments. Predict what happens to the temperature if the pack is made with 60 g of salt instead, and say whether “cold” is a thing that flows into the ankle.
Compare your answer
Twice the salt takes about twice as much energy from the water, so the temperature falls about twice as far — until the water nears 0 °C or the salt stops dissolving.
“Cold” does not flow. Energy flows from the warmer ankle into the pack, because the dissolving salt has already taken energy from the water in the pack and left it colder than the skin. By the end of this lesson you will be able to calculate that energy.
The science — structured notes
1.1 Heat is energy; temperature is not
Two words that everyday speech mixes up must be kept apart in chemistry. Temperature is a measure of how hot something is: it is proportional to the average kinetic energy of its particles, and it is measured in degrees Celsius (°C) or kelvin (K) with a thermometer.
Heat (or thermal energy transferred) is energy that moves from a hotter object to a colder one because of their temperature difference. It is measured in joules (J).
A swimming pool at 25 °C holds far more thermal energy than a cup of tea at 80 °C, because it has far more particles — yet energy flows from the tea into the pool water if you pour it in, because the tea is at the higher temperature. Temperature decides the direction of flow; the amount of energy depends also on the mass and the material (Lesson 2).
| Unit | Symbol | Value in joules | Where you meet it |
|---|---|---|---|
| joule | J | 1 J | SI unit; energy to lift a 100 g apple by 1 m |
| kilojoule | kJ | 1 000 J (10³ J) | enthalpy changes per mole; food labels |
| megajoule | MJ | 1 000 000 J (10⁶ J) | energy per kilogram of fuel |
| calorie | cal | 4.184 J | older chemistry books |
| kilocalorie (food “Calorie”) | kcal | 4 184 J | food labels beside kJ |
| kilowatt-hour | kWh | 3 600 000 J (3.6 MJ) | electricity bills; batteries |
A litre of petrol releases about 34 MJ when it burns. Express this in kJ, in kcal and in kWh.
kJ. 1 MJ is 1 000 kJ.
34 MJ × 1 000 = 34 000 kJ (3.4 × 10⁴ kJ)
kcal. 1 kcal is 4.184 kJ, so divide.
34 000 kJ ÷ 4.184 kJ kcal⁻¹
= 8 130 kcal (8.1 × 10³ kcal, about four days of food for an adult)
kWh. 1 kWh is 3.6 MJ, so divide.
34 MJ ÷ 3.6 MJ kWh⁻¹
= 9.4 kWh
Check. Bigger unit, smaller number, every time.
“A spark from a grinder at 1 000 °C carries more heat than a bath of water at 40 °C, because it is hotter.”
Reveal answer
False. The spark is hotter, but it is a speck of metal of perhaps a milligram; the energy it can give up as it cools is tiny, which is why a spark landing on your hand barely stings.
The bath holds 100 kg or more of water, and cooling it by just 1 °C releases about 400 000 J. Temperature is intensive (it does not depend on how much there is); thermal energy is not.
1.2 System and surroundings: exothermic and endothermic
To talk about energy changes without confusion, chemists split the world in two. The system is the particles that are changing — the reactants turning into products, or the salt dissolving. The surroundings are everything else: the water they are dissolved in, the cup, the thermometer, the air.
A thermometer dipped in the solution measures the temperature of the surroundings.
- Exothermic: energy is transferred from the system to the surroundings. The surroundings get hotter; the temperature reading rises. Combustion, respiration, neutralisation, rusting and dissolving calcium chloride are exothermic.
- Endothermic: energy is transferred from the surroundings into the system. The surroundings get colder; the reading falls. Photosynthesis, thermal decomposition, dissolving ammonium nitrate and the reaction of citric acid with sodium hydrogencarbonate are endothermic.
The energy change of the system at constant pressure is called the enthalpy change, symbol ΔH. Energy lost by the system gives a negative ΔH (exothermic); energy gained gives a positive ΔH (endothermic). Lesson 3 turns ΔH into a number per mole; for now, learn the sign.
“When the magnesium ribbon dissolved in the acid the tube got hot, so the magnesium and acid gained energy, and the reaction is endothermic.”
Reveal answer
The tube getting hot means the surroundings (the water in the acid, the glass) gained energy. That energy came out of the system — the magnesium atoms and hydrogen ions becoming magnesium ions and hydrogen gas.
The reaction is exothermic, with a negative ΔH. The error is treating the thermometer as if it measured the reacting particles.
1.3 Physical changes have energy changes too
An energy change does not prove a chemical reaction. Every change of state involves energy, because attractions between particles are overcome or formed:
- Melting and boiling (and evaporation) are endothermic: energy is needed to overcome the attractions holding the particles together. Ice at 0 °C takes in 6.0 kJ per mole to become water at 0 °C.
- Freezing and condensing are exothermic: the same attractions reform and give out the same amount of energy. Steam condensing on your skin releases 40.7 kJ per mole — which is why a steam burn is far worse than a splash of boiling water.
- Dissolving can go either way (Section 1.4).
What distinguishes a chemical change is that new substances form, with new bonds. The energy change is usually much larger per mole, because covalent and ionic bonds are much stronger than the attractions between molecules.
Sweat cools you on a hot afternoon even when the air is at 38 °C, hotter than your skin. Explain, in terms of energy and particles, how evaporation can cool the skin, and say whether evaporation is a physical or a chemical change.
Reveal answer
Evaporation is a physical change: water molecules escape from the liquid surface as water vapour, and no new substance forms.
To escape, a molecule must overcome the attractions to its neighbours, so evaporation is endothermic — the energy is taken from the remaining liquid and the skin beneath it, which cool. The fastest molecules leave first, lowering the average kinetic energy of those left behind. The skin keeps losing energy to the sweat as long as sweat evaporates, whatever the air temperature.
1.4 Dissolving salts: two energy changes in one
When an ionic salt dissolves, two things happen, and they pull in opposite directions:
- Breaking up the lattice — separating the positive and negative ions against their strong electrostatic attraction. This takes in energy (endothermic).
- Hydrating the ions — water molecules are polar, so their slightly negative oxygen ends cluster round positive ions and their slightly positive hydrogen ends round negative ions. Forming these ion–water attractions gives out energy (exothermic).
The overall enthalpy change of solution is the difference. If the lattice costs more than hydration pays back, dissolving is endothermic and the water cools. If hydration pays back more, dissolving is exothermic and the water warms.
Why is calcium chloride so different from ammonium nitrate? The Ca²⁺ ion is small and carries a double charge, so it attracts water molecules very strongly — and one formula unit releases three ions (Ca²⁺ and two Cl⁻), each of which becomes hydrated. The hydration pay-back is huge.
The NH₄⁺ and NO₃⁻ ions are larger and singly charged, so they attract water only weakly; the lattice costs more than hydration returns.
Anhydrous copper(II) sulfate is white; when water is added it turns blue and the tube becomes warm. Predict whether dissolving the blue hydrated copper(II) sulfate, CuSO₄·5H₂O, will warm the water as much, and give the reason.
Reveal answer
No — dissolving the hydrate warms the water far less (it is in fact slightly endothermic, about +12 kJ mol⁻¹, while the anhydrous salt gives roughly −70 kJ mol⁻¹).
In the hydrate the Cu²⁺ ion is already surrounded by water molecules, so most of the hydration energy has already been released when the crystals formed. Dissolving it still costs the lattice energy, but little extra hydration is left to pay it back.
1.5 Qualitative and quantitative: from “it feels cold” to a number
A qualitative observation describes without numbers: “the tube felt cold”, “the mixture fizzed and warmed”. It tells you the direction of the energy change, but not its size, and your hand is a poor instrument — it senses the rate of energy flow, not temperature.
A quantitative measurement gives a number with a unit: “the temperature fell from 20.0 °C to 14.2 °C”. The temperature change, ΔT, is final minus initial temperature:
ΔT = T(final) − T(initial)
ΔT = 14.2 °C − 20.0 °C
ΔT = −5.8 °C (the minus sign tells you it fell)
Knowing the mass of water and the fact that every gram of water needs 4.18 J to warm by 1 °C, you can turn ΔT into the energy transferred with Q = m × c × ΔT. Lesson 2 builds this equation properly; here is its first use.
2.00 g of ammonium nitrate is stirred into 50.0 g of water. The temperature falls from 20.0 °C to 17.1 °C. Calculate the energy taken from the water, and state which way it moved.
Temperature change. ΔT = 17.1 − 20.0 = −2.9 °C; the size of the change is 2.9 °C.
Equation. Q = m × c × ΔT
Q = 50.0 g × 4.18 J g⁻¹ °C⁻¹ × 2.9 °C
Q = 606 J, which is 0.61 kJ
Direction. 606 J left the water (the surroundings) and went into the dissolving salt (the system): endothermic, ΔH positive.
A sports cold pack mixes 30.0 g of ammonium nitrate (M = 80.04 g mol⁻¹) with 100 g of water at 25.0 °C. Dissolving takes in 25.7 kJ per mole. Estimate the lowest temperature the pack reaches, assuming all the energy comes from the water.
Moles. n = m ÷ M
n = 30.0 g ÷ 80.04 g mol⁻¹
n = 0.375 mol
Energy taken in. Q = 0.375 mol × 25 700 J mol⁻¹
Q = 9 630 J
Temperature fall. ΔT = Q ÷ (m × c)
ΔT = 9 630 J ÷ (100 g × 4.18 J g⁻¹ °C⁻¹)
ΔT = 23.0 °C
Answer. 25.0 − 23.0 = about 2 °C. Real packs reach about 2–5 °C: some energy flows in from the plastic, the air and your hand while the salt dissolves, so the real fall is a little smaller.
5.00 g of anhydrous calcium chloride is dissolved in 50.0 g of water and the temperature rises from 20.0 °C to 36.6 °C. Calculate the energy released into the water and state the sign of ΔH.
Reveal answer
ΔT = 36.6 − 20.0 = 16.6 °C
Q = 50.0 × 4.18 × 16.6
Q = 3 470 J (3.47 kJ)
The water warmed, so energy left the system: exothermic, ΔH negative. (Lesson 3 shows that this is about −77 kJ per mole of CaCl₂ — a little less than the data-book −81.3 kJ mol⁻¹, because some energy escaped from the cup.)
What if the same 2.00 g of ammonium nitrate were dissolved in 100.0 g of water instead of 50.0 g? Predict the temperature change and explain.
Reveal answer
The salt takes in the same 606 J, because the same number of moles dissolve. That energy is now shared by twice the mass of water, so the fall is half as big: about 1.45 °C instead of 2.9 °C.
This is why the mass of water is a control variable in every calorimetry experiment, and why a small temperature change is harder to measure precisely.
1.6 Chemical changes: which reactions release energy, and which absorb it
| Process | Type | What you observe | Where it matters |
|---|---|---|---|
| Combustion: CH₄ + 2O₂ → CO₂ + 2H₂O | Exothermic | Flame, heat, light | Cooking, power stations, engines (Lessons 4–5) |
| Respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O | Exothermic | Body warmth | Every living cell |
| Neutralisation: H⁺ + OH⁻ → H₂O | Exothermic | Solution warms by a few °C | Antacids, soil treatment (Lesson 2) |
| Displacement: Zn + Cu²⁺ → Zn²⁺ + Cu | Exothermic | Blue fades, copper forms, solution warms | Lesson 3 |
| Iron rusting (hand warmers) | Exothermic | Slow, steady warmth | Hand warmers use iron powder, salt and air |
| Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ | Endothermic | Needs light energy | All food chains |
| Thermal decomposition: CaCO₃ → CaO + CO₂ | Endothermic | Needs continuous strong heating | Lime and cement kilns |
| Citric acid + sodium hydrogencarbonate | Endothermic | Fizzing, solution cools | Sherbet feels cool on the tongue |
Classify each as exothermic or endothermic and as physical or chemical: (i) water vapour forming dew on grass at dawn; (ii) limestone being heated to make quicklime; (iii) a car battery charging; (iv) quicklime (CaO) reacting with water to form slaked lime.
Reveal answer
(i) Condensation: exothermic, physical — attractions between water molecules re-form.
(ii) Thermal decomposition: endothermic, chemical — CaCO₃ → CaO + CO₂ needs heat all the time.
(iii) Charging stores energy in the chemicals of the cell: endothermic, chemical (electrical energy in).
(iv) CaO + H₂O → Ca(OH)₂: exothermic, chemical — enough to boil some of the water; it is the reaction in some self-heating food cans.
Sodium acetate hand warmers contain a supersaturated solution. Clicking a metal disc starts crystallisation and the pack warms to about 54 °C; boiling the pack for ten minutes resets it. Explain which process in the cycle is exothermic, which is endothermic, whether each is physical or chemical, and why the pack never gets hotter than about 58 °C.
Reveal answer
Crystallisation is exothermic: ions (and water molecules) lock into the crystal lattice and the attractions formed give out energy. Boiling to reset is endothermic: energy is taken in to break the lattice apart and dissolve the crystals again. Both are physical changes — sodium acetate trihydrate is the same substance before and after.
The limit is about 58 °C, the melting point of sodium acetate trihydrate; in practice energy lost to the air keeps a real pack a few degrees lower, around 54 °C. While solid and liquid coexist, the energy released by crystallising holds the temperature near the melting point, just as ice and water stay at 0 °C — a phase change at constant temperature.
Try it — dissolving salts: how much colder, how much hotter?
The lab stirs a weighed mass of salt into 50.0 cm³ of water at 20.0 °C and records the lowest (or highest) temperature on a digital thermometer reading to 0.1 °C.
You choose the salt — ammonium nitrate, anhydrous calcium chloride or sodium chloride — and the container: a polystyrene cup with a lid, or an open glass beaker. The readout gives the size of the temperature change; the lab tells you whether it was a rise or a fall.
Suppose your mean temperature falls for ammonium nitrate in the cup were: 2.0 g → 2.9 °C; 4.0 g → 5.8 °C; 6.0 g → 8.8 °C; 8.0 g → 11.7 °C; 10.0 g → 14.6 °C.
(a) Show with a numerical test that the fall is proportional to the mass.
(b) Use the 10.0 g result to calculate ΔH of solution in kJ mol⁻¹ with its sign.
(c) The data-book value is +25.7 kJ mol⁻¹. Calculate the percentage difference and explain which way the heat exchange with the room pushed your value.
Reveal answer
(a) Fall ÷ mass = 1.45, 1.45, 1.47, 1.46, 1.46 °C g⁻¹ — constant; doubling 2.0 → 4.0 g doubles 2.9 → 5.8 °C. Proportional, through the origin.
(b) Energy taken from the water:
Q = 50.0 × 4.18 × 14.6
Q = 3 051 J
n = 10.0 ÷ 80.04 = 0.1249 mol
ΔH = +3 051 J ÷ 0.1249 mol
ΔH = +24 400 J mol⁻¹ = +24.4 kJ mol⁻¹
(c) Difference = (25.7 − 24.4) ÷ 25.7 × 100 = 5.1 %.
The solution was colder than the room, so energy flowed in from the air and the cup while the salt dissolved; the fall was smaller than it should be and ΔH comes out too small. A systematic error — all five points are low by the same fraction, so the line stays straight but is too shallow.
Common mistakes to avoid
- “Gave out cold” — cold does not flow. Say energy was taken from the surroundings.
- Reading the thermometer as the system — the thermometer is in the surroundings. It rises for exothermic, falls for endothermic.
- Heat and temperature as synonyms — temperature in °C, heat (energy) in J.
- “Needs heating to start, so endothermic” — a reaction that needs only a spark to start and then keeps going is exothermic.
- “An energy change means a chemical reaction” — melting, boiling and dissolving change temperature too; a new substance is the test.
- Wrong sign of ΔH — exothermic is negative (the system lost energy); endothermic is positive.
- Unit slips — kJ versus J (factor 1 000); kWh is 3.6 MJ, not 1 000 J.
Study smart: how this is examined
Study routine. Today: write the two definitions (with “system”, “surroundings”, the thermometer direction and the sign of ΔH) on one card, and redo both EXAMPLES from memory. In two days: explain dissolving of NH₄NO₃ and CaCl₂ to someone using the words lattice, hydration, energy in and energy out.
In a week: run the dissolution lab as a full investigation, including the ΔH calculation from the gradient, then do the Think task and the quiz.
How we found out — from “caloric” to the joule
In the 1760s the Glasgow physician and chemist Joseph Black noticed that a block of ice sitting in a warm room stayed at the freezing point for hours while it melted, even though heat was clearly flowing into it. He concluded that the heat had been absorbed without raising the temperature — “latent” (hidden) heat — and so drew the first clear line between the quantity of heat and the temperature.
Black also found that equal masses of different substances needed different amounts of heat to warm by the same amount: the idea behind specific heat capacity (Lesson 2).
In 1782–1783 Antoine Lavoisier and Pierre-Simon Laplace built an ice calorimeter: a chamber surrounded by packed ice. Whatever heat an experiment released melted some ice, and they weighed the meltwater. They compared the heat from a guinea pig breathing inside with the heat from burning charcoal, and argued that respiration is a slow combustion.
Like most scientists of the time, Lavoisier believed heat was a weightless fluid, “caloric”.
Caloric theory fell in the 1840s. The English brewer and physicist James Prescott Joule showed with falling weights driving a paddle wheel in water that a fixed amount of mechanical work always produced the same amount of heating — heat was energy, not a substance.
The unit of energy now carries his name. The words exothermic and endothermic were coined later in the nineteenth century by the French chemist Marcellin Berthelot, who spent years measuring the heat of reactions.
Where it is used now
- First aid. Instant cold packs (ammonium nitrate or urea with water) for sprains; they need no freezer, which matters at a school sports day or on a rural clinic’s shelf.
- Keeping food cool without electricity. The clay “pot-in-pot” cooler, popularised in northern Nigeria in the 1990s by the teacher Mohammed Bah Abba, keeps vegetables fresh for days: water in wet sand between two pots evaporates, an endothermic change that takes energy from the inner pot.
- Roads and mines. Calcium chloride de-ices roads and suppresses dust on gravel roads; its exothermic dissolving helps it work in cold weather.
- Self-heating food. Some self-heating cans and military ration heaters use exothermic reactions — quicklime with water, or magnesium–iron alloy with salty water — to warm a meal with no flame.
- Hand warmers. Iron-powder packs (slow exothermic oxidation) and reusable sodium acetate packs (exothermic crystallisation) are sold worldwide for cold mornings and mountain hikes.
Science now — and the scientist you need to be
What is happening in 2026
Storing heat is now as important as making it. Solar and wind power produce electricity when the weather allows; industries and homes need heat at other times. One answer is thermal energy storage: heating a large mass of material when electricity is cheap and releasing the energy later.
In Finland a company switched on what it described as the first commercial “sand battery” in 2022, heating sand with surplus electricity to warm a town’s district-heating water.
A second answer uses exactly the physical changes of this lesson. Phase-change materials — waxes and salt hydrates — absorb large amounts of energy as they melt and release it as they freeze, at a fixed temperature. They are used in cool boxes for vaccines, in building panels and in battery packs.
Chemists screen thousands of candidate salts and mixtures for the right melting point and a large enthalpy change, and machine-learning tools are increasingly used to suggest candidates worth testing in the laboratory.
Every one of these projects rests on the measurement you learn in Lesson 2: a careful calorimeter reading turned into joules per gram.
The mindshift
What you need to be able to do
Classify any change from a temperature reading and say which way energy moved; convert between energy units; explain dissolving with lattice and hydration; and turn a temperature change into joules with Q = mcΔT. Those four skills carry the rest of the unit.
See an example
Typical results: salt, a small fall of around 1 °C (slightly endothermic, as the data book says); sugar, a very small fall; baking soda, a fall of about 1–2 °C (sodium hydrogencarbonate dissolves endothermically).
A kitchen thermometer reading to 1 °C can barely see the salt effect — a quantitative measurement is only as good as its instrument, which is Lesson 2’s topic. An assistant that says “dissolving salt is exothermic” is wrong; one that says “slightly endothermic, about +4 kJ mol⁻¹” matches your data.
Key terms
| Temperature | How hot something is; proportional to the average kinetic energy of its particles; measured in °C or K. |
| Heat (thermal energy transferred) | Energy that moves from a hotter to a colder object because of a temperature difference; measured in J. |
| System | The particles that are reacting or changing. |
| Surroundings | Everything else: the water, the container, the thermometer, the air. |
| Exothermic | Transfers energy from the system to the surroundings; the temperature of the surroundings rises; ΔH negative. |
| Endothermic | Takes energy from the surroundings into the system; the temperature of the surroundings falls; ΔH positive. |
| Enthalpy change, ΔH | The energy change of the system at constant pressure, usually per mole, in kJ mol⁻¹. |
| Enthalpy of solution | The energy change when one mole of a substance dissolves in water. |
| Lattice breaking / hydration | Separating the ions of a crystal (energy in); water molecules surrounding the freed ions (energy out). |
| Qualitative / quantitative | A description without numbers / a measured value with a unit. |
Think like a scientist
Criterion C A company wants a reusable hand warmer that uses anhydrous calcium chloride and water. A student measures the temperature rise when different masses of CaCl₂ are dissolved in 50.0 cm³ of water in a polystyrene cup with a lid, starting at 20.0 °C, reading a digital thermometer to 0.1 °C. Her hypothesis was: “If the mass of CaCl₂ is doubled, the temperature rise will double, because each mole dissolved releases the same energy.”
| Mass of CaCl₂ / g | Rise, trial 1 / °C | Rise, trial 2 / °C | Rise, trial 3 / °C |
|---|---|---|---|
| 1.00 | 3.3 | 3.4 | 3.2 |
| 2.00 | 6.7 | 6.6 | 6.8 |
| 3.00 | 9.9 | 10.1 | 7.2 |
| 4.00 | 13.4 | 13.2 | 13.3 |
| 5.00 | 16.5 | 16.8 | 16.6 |
(a) Identify the anomalous result and suggest a specific cause related to the properties of calcium chloride. [2]
(b) Calculate the mean rise at each mass, to an appropriate precision. [2]
(c) Describe how you would present the processed data as a graph, and describe the relationship shown, with a numerical test. [3]
(d) Calculate ΔH of solution of CaCl₂ from the 5.00 g result (M = 110.98 g mol⁻¹), with its sign. [3]
(e) Evaluate the hypothesis using the data. [2]
(f) The data-book value is −81.3 kJ mol⁻¹. Explain one systematic error that accounts for the difference and suggest one improvement. [3]
Model answer
(a) Trial 3 at 3.00 g, 7.2 °C — about 3 °C below the other two trials (9.9 and 10.1 °C) and off the steady rise of about 3.3 °C per gram.
Cause: anhydrous calcium chloride absorbs water from the air (it is deliquescent). If the sample for that trial came from a jar left open, part of it was already hydrated, so part of its hydration energy had been released before it reached the cup, and 3.00 g contained less CaCl₂ — a smaller rise. (Also acceptable: a lump that had not fully dissolved when the reading was taken.)
(b) Means: 3.3, 6.7, 10.0 (9.9 and 10.1 only, the anomaly excluded), 13.3 and 16.6 °C — one decimal place, matching the thermometer.
(c) Mass of CaCl₂ / g on the horizontal axis from 0 to 5.00, mean temperature rise / °C on the vertical axis from 0 to 18, even scales, points plotted as crosses, a straight best-fit line drawn with a ruler.
The rise increases steadily with mass, from 3.3 °C at 1.00 g to 16.6 °C at 5.00 g. Rise ÷ mass = 3.3, 3.35, 3.33, 3.33, 3.32 °C g⁻¹ — constant — so the rise is directly proportional to the mass and the line passes through the origin.
(d) Energy released to the water:
Q = m × c × ΔT
Q = 50.0 g × 4.18 J g⁻¹ °C⁻¹ × 16.6 °C
Q = 3 470 J
n = 5.00 g ÷ 110.98 g mol⁻¹ = 0.04505 mol
ΔH = −3 470 J ÷ 0.04505 mol
ΔH = −77 000 J mol⁻¹ = −77.0 kJ mol⁻¹ (negative because the water warmed: exothermic).
(e) Supported: doubling the mass from 1.00 to 2.00 g raised the rise from 3.3 to 6.7 °C, and from 2.00 to 4.00 g from 6.7 to 13.3 °C — doubling within 0.1 °C — and the ratio rise ÷ mass is constant at 3.3 °C g⁻¹ across the range.
(f) The solution became warmer than the room, so energy flowed out through the cup wall and lid into the air while the salt dissolved and the student waited for the highest reading. The measured rise is therefore too small and ΔH too small in size (−77.0 against −81.3 kJ mol⁻¹, about 5 % low). Systematic: every trial loses a similar fraction.
Improvement: record the temperature every 30 s for five minutes after mixing, extrapolate the cooling line back to the moment of mixing, and use that extrapolated maximum; or nest two polystyrene cups with a tight lid. “Repeat more times” does not remove a systematic error.
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth. Take c(water) = 4.18 J g⁻¹ °C⁻¹ and 1 cm³ of water = 1.00 g.
Select the endothermic processes from this list: condensing steam; dissolving ammonium nitrate; burning ethanol; melting ice; neutralising an acid.
Reveal answer
Dissolving ammonium nitrate [1] and melting ice [1]. Condensing, burning and neutralising are exothermic.
A phone battery stores 15 Wh. Convert this to kJ and to kcal.
Reveal answer
15 Wh × 3 600 J Wh⁻¹ = 54 000 J = 54 kJ [1]
54 kJ ÷ 4.184 kJ kcal⁻¹ = 12.9 kcal (about 13 kcal — less than one biscuit) [1]
When ammonium nitrate dissolves, the beaker feels cold. Explain this observation in terms of the energy transferred between system and surroundings.
Reveal answer
Dissolving ammonium nitrate is endothermic: breaking up the ionic lattice takes in more energy than hydrating the ions gives out [1]. The energy needed is taken from the surroundings — the water, the beaker [1] — so their particles lose kinetic energy and the temperature falls; the hand then loses energy to the colder beaker, which is felt as cold [1].
A 5 cm³ splash of boiling water and a 200 L bath at 45 °C are both at a higher temperature than your skin. Explain which one can transfer more energy to you, using the terms heat and temperature correctly.
Reveal answer
The splash has the higher temperature, so energy flows from it faster at first, but it has very little mass: cooling 5 g from 100 to 37 °C releases only about 1.3 kJ [1].
The bath has a lower temperature but a mass of about 200 kg; cooling it by just 1 °C releases about 840 kJ. The bath can transfer far more heat: the amount of heat depends on mass and temperature change, not on temperature alone [1].
75.0 cm³ of water cools from 22.5 °C to 16.0 °C when a salt dissolves in it. Calculate the energy transferred and state its direction.
Reveal answer
ΔT = 16.0 − 22.5 = −6.5 °C [1]
Q = 75.0 × 4.18 × 6.5
Q = 2 040 J (2.04 kJ) [1]
2.04 kJ moved from the water into the dissolving salt: endothermic [1].
Explain why dissolving anhydrous calcium chloride is exothermic while dissolving ammonium nitrate is endothermic.
Reveal answer
In both, separating the ions of the lattice takes in energy and hydrating the ions gives out energy; the overall change is the difference [1].
Ca²⁺ is small and doubly charged, and each formula unit gives three ions to hydrate, so strong ion–water attractions release more energy than the lattice costs: exothermic [1].
NH₄⁺ and NO₃⁻ are large and singly charged; their weaker hydration releases less energy than breaking the lattice needs: endothermic [1].
7. A cold pack must cool 120 g of water from 25.0 °C to 5.0 °C. Ammonium nitrate (M = 80.04 g mol⁻¹) takes in 25.7 kJ mol⁻¹ when it dissolves, but tests show only 85 % of that energy comes from the water; the rest leaks in from outside. Calculate the minimum mass of ammonium nitrate the pack must contain.
Reveal answer
Energy to take from the water: Q = 120 × 4.18 × 20.0
Q = 10 030 J [1]
The salt must take in more, because only 85 % comes from the water:
Energy taken in by the salt = 10 030 ÷ 0.85 = 11 800 J [1]
n = 11 800 J ÷ 25 700 J mol⁻¹ = 0.459 mol [1]
m = 0.459 mol × 80.04 g mol⁻¹ = 36.8 g, so about 37 g [1]
8. A kitchen worker’s hand is hit by 2.0 g of steam at 100 °C, which condenses and then cools to skin temperature, 37 °C. Another worker is splashed by 2.0 g of water at 100 °C. Condensing steam releases 2 260 J g⁻¹. Calculate the energy delivered to the skin in each case and explain why steam burns are more serious.
Reveal answer
Water only cools: Q = 2.0 × 4.18 × 63
Q = 527 J [1]
Steam first condenses: 2.0 × 2 260 = 4 520 J [1]
then the water formed cools by 63 °C: another 527 J
Total = 5 047 J ≈ 5.0 kJ, nearly ten times as much [1]
Condensation is exothermic: as the steam turns to liquid on the skin, the attractions between water molecules re-form and release the energy that was absorbed to boil it — at 100 °C, right at the skin surface [1].
Summary
- Temperature (°C, K) measures how hot; heat (J) is energy transferred because of a temperature difference. Conversions: 1 kJ is 1 000 J, 1 kcal is 4.184 kJ and 1 kWh is 3.6 MJ.
- The system is what changes; the surroundings — including the thermometer — are everything else. Exothermic: energy out, surroundings warm, ΔH negative. Endothermic: energy in, surroundings cool, ΔH positive.
- Physical changes have energy changes too: melting, boiling and evaporation take energy in; freezing and condensing give it out. They are usually much smaller per mole than reactions.
- Dissolving a salt = lattice breaking (energy in) + hydration (energy out): NH₄NO₃ is endothermic (+25.7 kJ mol⁻¹), CaCl₂ exothermic (−81.3 kJ mol⁻¹), NaCl nearly neutral (+3.9 kJ mol⁻¹).
- Quantitative measurement turns “it feels cold” into ΔT, and Q = mcΔT turns ΔT into joules.
Next time → Measuring heat: specific heat capacity, Q = mcΔT in full, and the calorimeter — how to measure an energy change accurately, what limits the accuracy of your equipment, and a neutralisation lab that finds the energy released when an acid meets an alkali.