Calorimetry Calculations

What this lesson covers in the specification

Incandio is aligned to this specification. It is not published by, endorsed by or affiliated with Pearson, and it reproduces none of Pearson's wording — the statement numbers are given so you can check every lesson against your own copy.

• 3.3 — Calculating heat energy change using Q = mcΔT
• 3.4 — Calculating molar enthalpy change ΔH from the heat energy change

1 · Understand it

No exam language yet. The only question this section answers is: do I actually understand what is happening?

The last page produced a temperature change. This page turns it into an energy, and then into a number that can be compared between reactions — which is what makes the measurement worth taking.

The first step uses Q = mcΔT. Q is the heat energy transferred in JOULES, m is the mass of the solution being heated in GRAMS, c is the SPECIFIC HEAT CAPACITY, and ΔT is the temperature change in degrees. For dilute aqueous solutions c is taken as 4.18 J/g/°C, the value for water, because the solutions are mostly water.

Think of it like working out a bill from how far a crowd moved

If you knew that pushing one person a metre costs a fixed effort, then knowing how many people there were and how far they moved would give you the total effort spent. Q = mcΔT is that sum. The mass says how much stuff there is, the temperature change says how far it was pushed, and the specific heat capacity is the fixed cost per gram per degree. Nothing about the chemistry appears in this equation at all — it is pure bookkeeping about the WATER, which is why the same equation serves every reaction on the previous page.

The mass in that equation is the one people get wrong, and the rule is worth stating flatly: m IS THE MASS OF THE SOLUTION BEING HEATED, not the mass of the chemical reacting. If 25 cm³ of acid is mixed with 25 cm³ of alkali, the solution being heated is 50 cm³, and since the density of a dilute aqueous solution is close to 1 g/cm³, that is 50 g.

From a temperature rise to an energy

25 cm³ of hydrochloric acid is mixed with 25 cm³ of sodium hydroxide solution. The temperature rises by 6.5 °C. Calculate the heat energy released. (c = 4.18 J/g/°C.)

1. Total volume of solution = 25 + 25 = 50 cm³.
2. Density of a dilute solution ≈ 1 g/cm³, so m = 50 g.
3. Q = mcΔT = 50 × 4.18 × 6.5.
4. = 1358.5 J, which is about 1360 J or 1.36 kJ.
5. Note that the mass used was the whole 50 g of solution, not the mass of acid or of alkali.

Answer: About 1360 J released.

That figure is still not comparable between reactions, because it depends on how much you happened to use. Double the quantities and you get double the energy, which tells you nothing new about the chemistry. What is needed is the energy PER MOLE, and that is the MOLAR ENTHALPY CHANGE, written ΔH.

Statement 3.4 — from Q to ΔH

1. Calculate Q in joules using Q = mcΔT.
2. Convert Q to KILOJOULES by dividing by 1000, because enthalpy changes are quoted in kJ/mol.
3. Work out the NUMBER OF MOLES of the reactant the question specifies — usually the one that is not in excess.
4. Divide: ΔH = Q in kJ ÷ moles.
5. Attach the SIGN. Exothermic reactions have a NEGATIVE ΔH, because the chemicals have LOST energy. Endothermic reactions have a POSITIVE ΔH.
6. The sign is the step most often dropped, and an exothermic ΔH quoted as positive is simply wrong.

That sign convention repays a moment's thought, because it is the mirror image of the thermometer reading and that is exactly why it confuses people. The temperature went UP, and ΔH is NEGATIVE. Both are correct, because they describe different things: the thermometer describes the SURROUNDINGS gaining energy, and ΔH describes the CHEMICALS losing it.

Completing the calculation to a molar enthalpy change

In the experiment above, 1360 J was released and the acid was 25 cm³ of 1.0 mol/dm³ hydrochloric acid, which was fully neutralised. Calculate ΔH in kJ/mol.

1. Convert the energy: 1360 J ÷ 1000 = 1.36 kJ.
2. Moles of acid = concentration × volume in dm³ = 1.0 × (25 ÷ 1000) = 0.025 mol.
3. ΔH = 1.36 ÷ 0.025 = 54.4 kJ/mol.
4. The reaction was exothermic, so the sign is negative: ΔH = −54.4 kJ/mol.
5. The accepted value for neutralisation is about −57 kJ/mol, so the measured figure is a little low — which is heat loss doing exactly what the last page predicted.

Answer: ΔH = −54.4 kJ/mol, slightly below the true value because of heat loss.

That last observation is worth keeping. A measured enthalpy change that comes out less negative than the accepted value is the expected result, because heat loss always reduces the temperature rise. If a measured value came out MORE negative than the true one, something would be wrong with the experiment rather than merely imperfect about it.

2 · Grade 9 Notes

A different job from the section above. You have already understood it; this is the precise set of things to LEARN — definitions to reproduce word for word, processes in order, equations with units, and the answers that score full marks.

Q = mcΔT

Q is the heat energy in joules (J), m is the mass of the SOLUTION in grams (g), c is the specific heat capacity in J/g/°C, ΔT is the temperature change in °C

Units: J, g, J/g/°C, °C. Take c as 4.18 J/g/°C for dilute aqueous solutions, and the density as 1 g/cm³ so that volume in cm³ gives mass in g.

ΔH = Q ÷ moles

ΔH is the molar enthalpy change in kilojoules per mole (kJ/mol), Q is the heat energy in kilojoules (kJ), moles is the amount of the specified reactant

Units: kJ/mol, kJ, mol. Convert Q from J to kJ by dividing by 1000. EXOTHERMIC is NEGATIVE; ENDOTHERMIC is POSITIVE.

The quantities in these calculations

Q

The heat energy transferred, in joules. Calculated from the temperature change of the solution.

Specific heat capacity, c

The energy needed to raise 1 g of a substance by 1 °C. Take it as 4.18 J/g/°C for dilute aqueous solutions.

ΔT

The temperature change — the final temperature minus the starting temperature.

Molar enthalpy change, ΔH

The energy change per mole of reactant, in kJ/mol. Negative for exothermic, positive for endothermic.

The full calculation, in order

1. Find the mass of SOLUTION being heated — the total volume in cm³, taken as grams.
2. Q = m × 4.18 × ΔT, giving joules.
3. Divide by 1000 to get kilojoules.
4. Find the moles of the reactant specified, usually with moles = concentration × volume in dm³.
5. ΔH = kJ ÷ moles.
6. Attach the sign: NEGATIVE for exothermic, POSITIVE for endothermic.

Why the temperature rises while ΔH is negative

A. The THERMOMETER measures the SURROUNDINGS, which have GAINED energy — so the temperature goes UP.

B. ΔH describes the CHEMICALS, which have LOST energy — so ΔH is NEGATIVE. Both describe the same exothermic reaction.

Where the marks are lost

• Using the mass of the REACTANT instead of the mass of the SOLUTION
• Forgetting to add the volumes when two solutions are mixed — 25 + 25 is 50 g, not 25 g
• Leaving Q in joules and dividing by moles, giving an answer 1000 times too large
• OMITTING THE SIGN, or giving an exothermic reaction a positive ΔH
• Using the moles of the reactant in excess rather than the one fully used up

Model answer [3 marks]

50 cm³ of solution rises in temperature by 8.0 °C. Calculate the heat energy released, taking c as 4.18 J/g/°C. [3]

The mass of solution is 50 g, since the density of a dilute aqueous solution is taken as 1 g/cm³. Using Q = mcΔT, the heat energy is 50 × 4.18 × 8.0 = 1672 J, which is about 1670 J or 1.67 kJ.

Model answer [3 marks]

A reaction releases 2.20 kJ and 0.050 mol of the limiting reactant was used. Calculate ΔH, including its sign. [3]

The molar enthalpy change is the energy divided by the number of moles: 2.20 ÷ 0.050 = 44 kJ/mol. The reaction released energy and is therefore exothermic, so the enthalpy change is negative: ΔH = −44 kJ/mol.

Not this: The mass in Q = mcΔT is the mass of the chemical that reacted.

This: It is the mass of the SOLUTION BEING HEATED. When two solutions are mixed, add their volumes — 25 cm³ plus 25 cm³ gives 50 g of solution, and that is the mass whose temperature rose.

Mark-losing trap. m is the mass of the SOLUTION, not of the reactant. Add both volumes when two are mixed.

Mark-losing trap. Convert Q to kJ before dividing by moles, or the answer is 1000 times too large.

Mark-losing trap. EXOTHERMIC means ΔH is NEGATIVE, even though the temperature went up.

3 · Prove it — the five questions

The five questions climb Grade 6 → Grade 7 → Grade 8 → Grade 9 → Grade 9 challenge, and are marked inside Incandio on your own device, by rule, with an authored diagnosis of the mistake you actually made. The mark schemes stay in the app so that the practice is worth doing; the questions themselves are here.

1. Grade 6 · Calculate [2 marks] — Calculate the heat energy transferred when 100 g of water rises in temperature by 5.0 °C. Take c as 4.18 J/g/°C.
2. Grade 7 · State [2 marks] — 25 cm³ of acid is mixed with 25 cm³ of alkali. What mass should be used as m in Q = mcΔT?
3. Grade 8 · Calculate [3 marks] — A reaction releases 1.50 kJ of energy and 0.030 mol of the limiting reactant was used. Calculate ΔH in kJ/mol, including its sign.
4. Grade 9 · Calculate [6 marks] — Select every statement that belongs in a full-mark calculation of ΔH for a neutralisation in which 50 cm³ of 1.0 mol/dm³ acid is mixed with 50 cm³ of 1.0 mol/dm³ alkali, giving a temperature rise of 6.8 °C. (c = 4.18 J/g/°C.)
5. 9+ · Analyse [6 marks] — Two students measure ΔH for the same neutralisation. One uses 25 cm³ of each solution and gets −53 kJ/mol; the other uses 100 cm³ of each and gets −56 kJ/mol. The accepted value is −57 kJ/mol. Select every statement that belongs in a full-mark analysis.

The people behind this science

Two ways into the same idea — the one whose name is the unit in every one of these answers, and the one who insisted the accounting must balance. Inside Incandio each of them answers knowing exactly which lesson you have just finished.

James Prescott Joule — the one whose name is the unit in every one of these answers

Every answer on this page is in joules or kilojoules, and the unit is named for the man who established that heat and mechanical work are the same currency. His famous experiment let falling weights turn a paddle in an insulated container of water and measured the resulting temperature rise, showing that a fixed amount of work always produces the same amount of heat. That equivalence is what allows a chemical reaction's output to be quoted in the same units as the energy of a moving object, and it took him years of increasingly precise measurement to establish.

• “What is a joule, and why is heat measured in the same units as work?”
• “How did paddles and falling weights measure heat?”
• “Why did it take so many attempts to get the number right?”
• “What was the caloric theory, and why was it wrong?”
• “How did you keep heat from leaking out of your apparatus?”

Antoine Lavoisier — the one who insisted the accounting must balance

Calorimetry is bookkeeping, and Lavoisier is the person who made chemistry into a subject where the books must balance. He established conservation of mass by weighing everything before and after in sealed vessels, and he applied the same accounting discipline to heat with his ice calorimeter. He is the right figure to ask why a discrepancy in a calculation should be treated as information rather than noise — a missing quantity means something has been overlooked, which is exactly how heat loss is diagnosed on this page.

• “Why must everything be accounted for in a chemical experiment?”
• “How did you measure heat as well as mass?”
• “What do you do when the numbers do not balance?”
• “How precisely could you measure in your laboratory?”
• “What did careful accounting let you discover that others missed?”

Carry on through the course

• Previous lesson: Exothermic and Endothermic Reactions
• Next lesson: Energy Diagrams and Bond Energies
• Exothermic and Endothermic Reactions — Reactions that warm their surroundings and reactions that chill them — and why the thermometer tells you which.
• Energy Diagrams and Bond Energies — Why breaking bonds costs energy, why making them repays it, and why the balance decides everything.
• Rates of Reaction — Four things that make a reaction faster, and two ways of measuring whether they did.
• Collision Theory — One explanation that accounts for every factor on the previous page.
• Catalysts — A substance that speeds a reaction up, takes part in it, and is there unchanged at the end.
• All of Chemistry · Chemistry key terms · required practicals · equations · Incandio Science · The 4CH1 specification