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Calorimetry Mc Delta T

Learn and apply Calorimetry Mc Delta T in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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Calorimetry and Molar Enthalpy: Orientation

Calorimetry questions are method marks: if you write the same chain every time (q = mcΔ T → q_reaction = -qₛₒₗᵤₜᵢₒₙ → divide by moles), you stop losing marks to sign and unit slips.

Keep Enthalpy Changes and Energy Profiles and the Energetics and Thermodynamics hub in view, because most questions mix definitions with cycle reasoning.

What this page is really testing

  • Can you move cleanly from temperature data to Δ H in kJ mol⁻¹?
  • Can you justify the sign of Δ H from the observed temperature change?
  • Can you state assumptions (mass, density, heat loss) without overclaiming precision?

Definitions (Must Know)

A. Heat energy change, q

q is the heat energy transferred (units: J).

B. Specific heat capacity, c

c is the energy needed to raise the temperature of 1 g of a substance by 1 K (units: J g⁻¹ K⁻¹).

For dilute aqueous solutions, you often use c ≈ 4.18 J g⁻¹K⁻¹ (water).

C. Temperature change, Δ T

Δ T = T_final - Tᵢₙᵢₜᵢₐₗ

D. Enthalpy change per mole, Δ H

For a reaction at constant pressure, the enthalpy change per mole is: Δ H = q_reaction/n

In calorimetry, you usually calculate q for the solution, then use q_reaction = -qₛₒₗᵤₜᵢₒₙ (assuming no heat loss).

Detailed Explanations

A. Why the sign flips (because → therefore)

If the solution warms up, the solution gained heat from the reaction. Therefore the reaction released heat:

  • qₛₒₗᵤₜᵢₒₙ > 0
  • q_reaction = -qₛₒₗᵤₜᵢₒₙ < 0
  • so Δ H < 0 (exothermic)

B. Calorimetry workflow (method marks)

  1. Identify the mass of solution, m (in g).
  2. Calculate Δ T = T_final - Tᵢₙᵢₜᵢₐₗ.
  3. Calculate qₛₒₗᵤₜᵢₒₙ = mcΔ T (in J).
  4. Find moles reacted, n (usually the limiting reagent).
  5. Calculate Δ H = -qₛₒₗᵤₜᵢₒₙ/n and convert to kJ mol⁻¹.

Mini example:

  • m = 50.0 g, Δ T = +6.50 K, c = 4.18
  • qₛₒₗᵤₜᵢₒₙ = (50.0)(4.18)(6.50) = 1.36 × 10³ J

C. Turning volume into mass (common assumption)

If you are given volumes of dilute aqueous solutions, exam questions often let you assume:

  • density ≈ 1.00 g cm⁻³
  • so 50.0 cm³ ≈ 50.0 g

State the assumption if it is not explicitly given.

Worked Examples

Modelled example 1

Calculate heat gained by a solution

Core

Problem

25.0 cm³ of 1.00 mol dm⁻³ acid reacts and raises the temperature of 50.0 g of solution by 6.50°C. Calculate qₛₒₗᵤₜᵢₒₙ using c = 4.18 J g⁻¹K⁻¹.
Study the worked solution

  1. Identify the signed temperature change

    Method

    Use the temperature rise as a positive Δ T.

    Reason

    The solution gains heat, so its temperature-change term is positive.

    Working

    Δ T = +6.50 K.

  2. Substitute into q equals mc delta T

    Method

    Multiply the mass, specific heat capacity and temperature change.

    Reason

    qₛₒₗᵤₜᵢₒₙ = mcΔ T measures the heat gained by the solution.

    Working

    qₛₒₗᵤₜᵢₒₙ = (50.0)(4.18)(6.50) = 1.36 × 10³ J.

Guided practice 2

Convert solution heat to molar enthalpy

About 7 min

Problem

In Example 1, 0.0250 mol of limiting reagent reacted. Calculate Δ H in kJ mol⁻¹.

Try this before viewing the solution

Hints

Hint 1: reaction boundary
The heat released by the reaction is the negative of the heat gained by the solution.
Hint 2: per mole
Divide -1.36 × 10³ J by 0.0250 mol, then convert J to kJ.
View solution step by step

  1. Change system boundary

    Method

    Reverse the sign when moving from solution heat to reaction heat.

    Reason

    Energy lost by the reaction is gained by the surroundings measured here.

    Working

    q_reaction = -1.36 × 10³ J.

  2. Calculate the molar value

    Method

    Divide by the amount reacted and convert to kilojoules.

    Reason

    Enthalpy change is reported per mole of reaction as written.

    Working

    Δ H = -(1.36 × 10³)/0.0250 = -5.44 × 10⁴ J mol⁻¹ = -54.4 kJ mol⁻¹.

Common misconception 3

Correct a calorimetry sign error

Find and correct the mistake

Learner claim

A solution warms and has qₛₒₗᵤₜᵢₒₙ = +2.10 kJ. A learner reports q_reaction = +2.10 kJ because the measured temperature change is positive. Diagnose the claim.

Choose the reaction sign

The reaction heat is

View solution step by step

  1. Name the measured system

    Method

    Recognise that mcΔ T gives the solution’s heat change.

    Reason

    The thermometer follows the surroundings of the reacting chemicals.

    Working

    qₛₒₗᵤₜᵢₒₙ = +2.10 kJ.

  2. Apply energy conservation

    Method

    Reverse the sign for the reaction.

    Reason

    q_reaction = -qₛₒₗᵤₜᵢₒₙ.

    Working

    q_reaction = -2.10 kJ, so the reaction is exothermic.

Examiner practice 4

Interpret a temperature fall

3 marks

Problem

The temperature of a 100 g solution falls by 4.20°C during a reaction. Calculate qₛₒₗᵤₜᵢₒₙ using c = 4.18 J g⁻¹K⁻¹ and state the sign of Δ H. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Assign delta T

    1 mark

    Method

    Use final minus initial temperature.

    Reason

    A fall gives a negative change.

    Working

    Δ T = -4.20 K.

  2. Calculate solution heat

    1 mark

    Method

    Substitute the signed change.

    Reason

    The negative value records heat lost by the solution.

    Working

    qₛₒₗᵤₜᵢₒₙ = (100)(4.18)(-4.20) = -1.76 × 10³ J.

  3. State the reaction sign

    1 mark

    Method

    Reverse the solution’s energy direction.

    Reason

    The reaction absorbs the heat lost by the solution.

    Working

    Δ H > 0; the reaction is endothermic.

Challenge 5

Predict the effect of heat loss

Minimal support

Problem

An exothermic reaction loses some heat to the room before the maximum temperature is recorded. Predict the effect on the measured Δ T, the calculated magnitude of q, and the calculated Δ H.

Try this before viewing the solution

Hints

Hint 1: temperature record
Heat escaping to the room means the solution reaches a lower recorded maximum.
Hint 2: propagate the error
Follow the smaller Δ T through q = mcΔ T and then Δ H = -q/n.
View solution step by step

  1. Follow the measured change

    Method

    Reduce the recorded temperature rise.

    Reason

    Not all released energy remains in the measured solution.

    Working

    Measured Δ T is too small.

  2. Propagate to q

    Method

    Use the proportional relationship in q = mcΔ T.

    Reason

    With m and c unchanged, a smaller temperature rise gives a smaller heat magnitude.

    Working

    |q| is underestimated.

  3. Propagate to molar enthalpy

    Method

    Keep the exothermic sign but reduce its magnitude.

    Reason

    Dividing the underestimated heat by the same amount reacted cannot restore the missing energy.

    Working

    Δ H is calculated as less negative than the true value.

Mind Stretchers

Connect This To

Practice Route

  1. Re-do Examples 1-3 and include a one-line sign explanation each time.
  2. Complete the Energetics Thermodynamics quiz under timed conditions.
  3. Use Exam Skills to tighten command-word responses for practical-evaluation questions.

Mind stretcher 1Extension

An acid reacts completely with excess base. The enthalpy change of neutralisation is given as Δ H = -57.1 kJ mol⁻¹. When 25.0 cm³ of the acid is mixed, the temperature of 50.0 g solution rises by 7.20°C. Calculate the concentration of the acid (assume density = 1.00 g cm⁻³ and c = 4.18).

Show Hint

Separate heat gained by the solution from heat released by the reaction, then divide by the amount that actually reacts.

Show Answer

Mark scheme:

  • qₛₒₗᵤₜᵢₒₙ = mcΔ T = (50.0)(4.18)(7.20) = 1.50 × 10³ J
  • q_reaction = -qₛₒₗᵤₜᵢₒₙ = -1.50 × 10³ J
  • Δ H = -57.1 kJ mol⁻¹ = -5.71 × 10⁴ J mol⁻¹
  • n = q_reaction/(Δ H) = (-1.50 × 10³)/(-5.71 × 10⁴) = 2.64 × 10⁻² mol
  • V = 25.0 cm³ = 0.0250 dm³
  • c = n/V = (2.64 × 10⁻²)/0.0250 = 1.06 mol dm⁻³

Mind stretcher 2: Comparing two experimental enthalpiesExtension

Question. Burning 0.460 g ethanol heats 200 g water by 8.50 °C. Calculate the experimental molar enthalpy of combustion using c = 4.18 J g⁻¹K⁻¹ and M(C₂H₅OH) = 46.0 g mol⁻¹, then suggest why its magnitude is below the data-book value.

Show Hint

The ethanol amount is exactly 0.0100 mol; the reaction sign is opposite to the water’s heat change.

Show Answer

q = 200(4.18)(8.50) = 7106 J = 7.106 kJ. Therefore Δ H_c = -7.106/0.0100 = -711 kJ mol⁻¹. Heat loss, incomplete combustion and heating the apparatus make the measured temperature rise too small.