Why Calorimeter Heat Capacity Matters in HSC Chemistry

A reaction warms 100.0 g of solution by \(6.0^\circ\text{C}\). Using the usual specific heat capacity of an aqueous solution, the solution absorbs \(2.51\text{ kJ}\) of energy.

But there is something missing. The calorimeter itself also warms. The cup, lid, thermometer, and stirrer all absorb energy as their temperatures increase.

Suppose their combined heat capacity is \(42.0\ \text{J K}^{-1}\). They absorb another \(252\text{ J}\). So the reaction actually released approximately \(2.76\text{ kJ}\), not \(2.51\text{ kJ}\).

The calorimeter’s heat capacity matters because the solution is not the only part of the surroundings absorbing energy. If you ignore the calorimeter when it absorbs heat, you underestimate the energy released by an exothermic reaction and calculate an enthalpy change that is not negative enough.

All numerical data in the worked examples and questions below are illustrative.

01The solution is not the whole calorimeter

Imagine mixing hydrochloric acid and sodium hydroxide in a polystyrene cup.

The reaction releases chemical energy. That energy transfers to the surroundings, causing the temperature to rise.

It is tempting to assume that all the released energy goes into the solution. This is the simplest calorimetry model, but it leaves something out.

The cup also warms. So do the lid, thermometer probe, and any stirrer that is in thermal contact with the solution.

To calculate the energy released accurately, we need to account for all these components.

System and surroundings

Before using an equation, identify which part of the experiment you are studying.

• System: The chemical reaction taking place. This is where the chemical energy change occurs.
• Surroundings: Everything receiving or supplying energy to the reaction. In our school calorimeter, this includes the aqueous solution and the calorimeter apparatus.

For a well-insulated experiment, energy lost by the reaction is gained by its surroundings.

What does the sign of q mean?

The symbol \(q\) represents heat transferred into or out of a particular part of the system.

We use this sign convention:

• \(q>0\): The object or process gains energy.
• \(q<0\): The object or process loses energy.

Consider an exothermic reaction. The solution and calorimeter warm up, so both gain energy. The chemical reaction loses energy.

For an ideal insulated calorimeter:

\[ q_{\text{reaction}}+q_{\text{solution}}+q_{\text{cal}}=0 \]

Rearranging:

\[ \boxed{q_{\text{reaction}}=-(q_{\text{solution}}+q_{\text{cal}})} \]

This is the central relationship for including the calorimeter’s heat capacity.

Two different heat capacities

There is an important difference between the specific heat capacity of the solution and the heat capacity of the calorimeter.

The solution’s heat change is calculated using:

\[ q_{\text{solution}}=mc\Delta T \]

where:

• \(m\) is the mass of the solution, in grams (g)
• \(c\) is its specific heat capacity, in \(\mathrm{J\,g^{-1}\,K^{-1}}\)
• \(\Delta T\) is its temperature change, in K

For dilute aqueous solutions, we often approximate:

\[ c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}} \]

The calorimeter is treated differently:

\[ \boxed{q_{\text{cal}}=C_{\text{cal}}\Delta T_{\text{cal}}} \]

Here, \(C_{\text{cal}}\) is the heat capacity of the entire calorimeter assembly, measured in \(\mathrm{J\,K^{-1}}\).

Unlike specific heat capacity, it is not measured per gram. It describes the energy needed to change the temperature of the whole apparatus by one degree.

For example, a calorimeter with:

\[ C_{\text{cal}}=42.0\ \mathrm{J\,K^{-1}} \]

absorbs 42.0 J when its temperature increases by 1 K.

If its temperature increases by \(6.0\text{ K}\), it absorbs:

\[ q_{\text{cal}}=42.0\times6.0=252\text{ J} \]

Notice that we do not multiply by the mass of the solution again. The calorimeter’s heat capacity already describes the entire apparatus.

If you need to revise the underlying relationship between mass, specific heat capacity, and temperature change, see Specific Heat Capacity: Using Q = mcΔT in HSC Physics.

02How do we determine the calorimeter’s heat capacity?

You might be wondering where the value of \(C_{\text{cal}}\) comes from.

It is usually determined experimentally through a process called calibration.

Calibration means comparing an instrument’s response with a known quantity so that we can determine how it behaves during later measurements.

For a calorimeter, we can use water at known temperatures.

The calibration experiment

Picture a nested polystyrene cup containing a measured quantity of cold water.

The cup has a lid, a thermometer, and a stirrer. Before starting, the cold water and the calorimeter are allowed to reach the same initial temperature.

Next, a measured mass of hot water is added. The lid is replaced, the mixture is stirred, and the final temperature is recorded.

We know how much energy the hot water loses and how much energy the cold water gains.

If those two quantities are different, where did the remaining energy go?

Into the calorimeter itself.

Predict before measuring

Suppose we had no calorimeter at all. We simply mixed the two quantities of water in a perfectly insulated container that absorbed no energy.

The final temperature would be approximately \(37.1^\circ\text{C}\).

However, the actual experiment gives a final temperature of \(36.0^\circ\text{C}\).

Why is it lower?

The calorimeter started at \(20.0^\circ\text{C}\), so it also warmed up. Some of the hot water’s energy went into heating the apparatus rather than just the cold water.

That difference lets us calculate the calorimeter’s heat capacity.

The energy balance during calibration is:

\[ q_{\text{hot water}}+ q_{\text{cold water}}+ q_{\text{cal}}=0 \]

Once we know the heat gained by the calorimeter, we can divide it by its temperature change:

\[ C_{\text{cal}}= \frac{q_{\text{cal}}}{\Delta T_{\text{cal}}} \]

03Using calorimeter heat capacity to calculate enthalpy

Now we can use the calibrated calorimeter in an actual chemical reaction.

The key is to calculate the energy gained or lost by both the solution and the apparatus.

For an experiment where the solution and calorimeter start at the same temperature:

\[ q_{\text{solution}}=mc\Delta T \]

\[ q_{\text{cal}}=C_{\text{cal}}\Delta T \]

Adding these gives:

\[ q_{\text{surroundings}} =mc\Delta T+C_{\text{cal}}\Delta T \]

Then:

\[ \boxed{q_{\text{reaction}}=-q_{\text{surroundings}}} \]

However, calculating the reaction heat is not quite the final step. We still need to find the amount of substance that actually reacted.

Why the amount reacted matters

Enthalpy change is often reported per mole of reaction.

For a constant-pressure coffee-cup calorimeter:

\[ \Delta H=\frac{q_{\text{reaction}}}{n_{\text{reaction}}} \]

Here, \(n_{\text{reaction}}\) is the amount of reaction that occurred, in moles.

It is not necessarily the initial amount of every reactant. If one reactant is in excess, the limiting reactant determines how much reaction can occur.

Consider the neutralisation reaction:

\[ \ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)} \]

Hydrochloric acid and sodium hydroxide react in a 1:1 mole ratio.

If \(0.0500\text{ mol}\) of each reactant is present, \(0.0500\text{ mol}\) reacts, producing \(0.0500\text{ mol}\) of water.

That is the amount we use when calculating the molar enthalpy of neutralisation.

04Does the calorimeter always absorb heat?

No. This is another important distinction.

So far, we have considered exothermic reactions. The temperature increases, and the calorimeter gains energy.

But what happens during an endothermic reaction?

Imagine dissolving a salt in water. The process absorbs energy, causing the solution and calorimeter to cool.

Now:

\[ \Delta T<0 \]

Therefore:

\[ q_{\text{solution}}<0 \]

and:

\[ q_{\text{cal}}<0 \]

Both the solution and calorimeter lose thermal energy. The reaction gains that energy:

\[ q_{\text{reaction}} =-(q_{\text{solution}}+q_{\text{cal}}) \]

The reaction’s heat change is positive.

For example, suppose:

• \(100.0\text{ g}\) of solution cools by \(3.0\text{ K}\).
• \(C_{\text{cal}}=50.0\ \mathrm{J\,K^{-1}}\).
• \(c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\).

Then:

\[ q_{\text{solution}}=(100.0)(4.18)(-3.0) =-1254\text{ J} \]

And:

\[ q_{\text{cal}}=(50.0)(-3.0) =-150\text{ J} \]

Therefore:

\[ \begin{aligned} q_{\text{reaction}} &=-(-1254-150)\\ &=+1404\text{ J} \end{aligned} \]

The reaction absorbs \(1404\text{ J}\).

The important point is that the calorimeter can either gain or lose energy. Its heat change depends on its own temperature change, not on whether the reaction is exothermic or endothermic.

Common mistakes to avoid

05Experimental assumptions and limitations

Including the calorimeter’s heat capacity improves the energy calculation, but it does not make the experiment perfect.

There are several assumptions behind the method.

One particularly important limitation is heat loss to the external environment.

Including \(C_{\text{cal}}\) accounts for energy absorbed by the cup and its components. It does not account for energy that escapes into the surrounding air.

For an exothermic reaction, this external heat loss can make the measured temperature rise smaller than it should be. The resulting experimental enthalpy is then less negative than the actual value.

If temperature-time data are supplied, a cooling correction can help account for this additional energy transfer. See How to Use a Cooling Correction in HSC Chemistry.

A note on syllabus scope

The Chemistry Stage 6 Syllabus (2017) explicitly includes calorimetry and \(q=mc\Delta T\) in the study of energy changes in chemical reactions.

The detailed calibration of \(C_{\text{cal}}\) in this article is a practical extension of that measurement method. It should not be interpreted as a claim that a separate calorimeter-constant calculation is a named requirement in every syllabus edition.

The Chemistry 11-12 Syllabus (2025) begins implementation in 2028. Students should check the syllabus edition and course program that apply to their cohort.

• NESA: Chemistry Stage 6 Syllabus (2017)
• NESA: Chemistry 11-12 Syllabus (2025)

06Questions and solutions

The following questions move from straightforward calorimeter corrections to more challenging energy balances. All numerical data are illustrative.

Question 1

A student carries out an exothermic reaction in a coffee-cup calorimeter.

The following measurements are recorded:

• Mass of solution: \(90.0\text{ g}\)
• Temperature increase: \(4.50\text{ K}\)
• Calorimeter heat capacity: \(35.0\ \mathrm{J\,K^{-1}}\)
• Amount of limiting reactant consumed: \(0.0250\text{ mol}\)

Assume the solution and calorimeter initially have the same temperature.

Use \(c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\).

Calculate:

a. The heat gained by the solution.

b. The heat gained by the calorimeter.

c. The molar enthalpy change of the reaction.

Solution 1

The molar enthalpy change is:

\[ \boxed{\Delta H=-74.0\ \mathrm{kJ\,mol^{-1}}} \]

The reaction is exothermic because the surroundings gain energy.

Step 1: Calculate the heat gained by the solution.

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=(90.0)(4.18)(4.50)\\ &=1692.9\text{ J} \end{aligned} \]

Step 2: Calculate the heat gained by the calorimeter.

\[ \begin{aligned} q_{\text{cal}} &=C_{\text{cal}}\Delta T\\ &=(35.0)(4.50)\\ &=157.5\text{ J} \end{aligned} \]

Step 3: Calculate the total energy gained by the surroundings.

\[ \begin{aligned} q_{\text{surroundings}} &=1692.9+157.5\\ &=1850.4\text{ J} \end{aligned} \]

Therefore:

\[ q_{\text{reaction}}=-1.8504\text{ kJ} \]

Step 4: Calculate the molar enthalpy change.

\[ \begin{aligned} \Delta H &=\frac{-1.8504}{0.0250}\\ &=-74.016\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

To three significant figures:

\[ \boxed{\Delta H=-74.0\ \mathrm{kJ\,mol^{-1}}} \]

The important step is including both energy terms before dividing by the amount reacted.

Question 2

A student is calibrating a calorimeter by mixing hot and cold water.

The measurements are:

Assume no heat escapes to the external environment and use:

\[ c_{\text{water}}=4.18\ \mathrm{J\,g^{-1}\,K^{-1}} \]

Calculate the heat capacity of the calorimeter.

A student incorrectly assumes that the calorimeter initially had the same temperature as the cold water.

Calculate the value this incorrect assumption would produce, and explain the difference.

Solution 2

The correct calorimeter heat capacity is:

\[ \boxed{C_{\text{cal}}=69.7\ \mathrm{J\,K^{-1}}} \]

The incorrect assumption produces \(44.8\ \mathrm{J\,K^{-1}}\).

The difference occurs because the calorimeter’s initial temperature is not the same as the cold water’s initial temperature.

Step 1: Calculate the heat lost by the hot water.

\[ \begin{aligned} q_{\text{hot}} &=(50.0)(4.18)(34.0-65.0)\\ &=(50.0)(4.18)(-31.0)\\ &=-6479\text{ J} \end{aligned} \]

Step 2: Calculate the heat gained by the cold water.

\[ \begin{aligned} q_{\text{cold}} &=(100.0)(4.18)(34.0-20.0)\\ &=(100.0)(4.18)(14.0)\\ &=5852\text{ J} \end{aligned} \]

Step 3: Calculate the heat gained by the calorimeter.

Using conservation of energy:

\[ q_{\text{hot}}+q_{\text{cold}}+q_{\text{cal}}=0 \]

Therefore:

\[ \begin{aligned} q_{\text{cal}} &=-(-6479+5852)\\ &=627\text{ J} \end{aligned} \]

Step 4: Use the calorimeter’s actual temperature change.

The calorimeter starts at \(25.0^\circ\text{C}\), not \(20.0^\circ\text{C}\).

\[ \Delta T_{\text{cal}}=34.0-25.0=9.0\text{ K} \]

Therefore:

\[ \begin{aligned} C_{\text{cal}} &=\frac{627}{9.0}\\ &=69.7\ \mathrm{J\,K^{-1}} \end{aligned} \]

Step 5: Examine the incorrect assumption.

If the student incorrectly uses \(20.0^\circ\text{C}\) as the calorimeter’s initial temperature:

\[ \Delta T_{\text{cal,incorrect}}=34.0-20.0=14.0\text{ K} \]

They would calculate:

\[ \begin{aligned} C_{\text{cal,incorrect}} &=\frac{627}{14.0}\\ &=44.8\ \mathrm{J\,K^{-1}} \end{aligned} \]

This is incorrect because the calorimeter did not undergo a \(14.0\text{ K}\) temperature increase.

The tempting assumption is that everything inside the calorimeter starts at the cold water’s temperature. That is only valid if the apparatus has been allowed to reach thermal equilibrium with the cold water before the experiment.

Question 3

Two students investigate the same exothermic reaction using the same calibrated calorimeter.

Both experiments consume \(0.0300\text{ mol}\) of limiting reactant.

Their measurements are:

Assume:

\[ c_{\text{solution}}=4.18\ \mathrm{J\,g^{-1}\,K^{-1}} \]

Calculate the experimental molar enthalpy for each trial.

Explain why Trial B has a smaller temperature increase even though it releases approximately the same amount of energy.

Solution 3

The calculated molar enthalpies are:

\[ \boxed{\Delta H_A=-74.9\ \mathrm{kJ\,mol^{-1}}} \]

\[ \boxed{\Delta H_B=-75.6\ \mathrm{kJ\,mol^{-1}}} \]

The values are very similar, despite the different temperature increases.

Step 1: Calculate the energy gained by the surroundings in Trial A.

The solution absorbs:

\[ \begin{aligned} q_{\text{solution,A}} &=(80.0)(4.18)(6.00)\\ &=2006.4\text{ J} \end{aligned} \]

The calorimeter absorbs:

\[ \begin{aligned} q_{\text{cal,A}} &=(40.0)(6.00)\\ &=240.0\text{ J} \end{aligned} \]

Therefore:

\[ \begin{aligned} q_{\text{surroundings,A}} &=2006.4+240.0\\ &=2246.4\text{ J} \end{aligned} \]

Step 2: Calculate the molar enthalpy for Trial A.

\[ \begin{aligned} \Delta H_A &=\frac{-2.2464}{0.0300}\\ &=-74.88\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Step 3: Calculate the energy gained by the surroundings in Trial B.

The solution absorbs:

\[ \begin{aligned} q_{\text{solution,B}} &=(160.0)(4.18)(3.20)\\ &=2140.16\text{ J} \end{aligned} \]

The calorimeter absorbs:

\[ \begin{aligned} q_{\text{cal,B}} &=(40.0)(3.20)\\ &=128.0\text{ J} \end{aligned} \]

Therefore:

\[ \begin{aligned} q_{\text{surroundings,B}} &=2140.16+128.0\\ &=2268.16\text{ J} \end{aligned} \]

Step 4: Calculate the molar enthalpy for Trial B.

\[ \begin{aligned} \Delta H_B &=\frac{-2.26816}{0.0300}\\ &=-75.605\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

To three significant figures:

\[ \Delta H_B=-75.6\ \mathrm{kJ\,mol^{-1}} \]

Step 5: Explain the difference between the temperature increases.

Trial B contains twice as much solution.

A larger mass has a larger total heat capacity, so more energy is required to produce the same temperature increase.

Even though Trial B has a smaller temperature rise, it absorbs approximately the same total energy because it contains more solution.

The two calculated enthalpies are close. Their small difference could reflect experimental variation, although measurement uncertainties would be needed to assess their agreement formally.

The misconception is that temperature rise directly measures reaction energy. It does not. The temperature rise must be considered alongside the total heat capacity of the surroundings.

Question 4

A student dissolves a salt in water inside a calibrated calorimeter.

The following data are recorded:

• Mass of final solution: \(120.0\text{ g}\)
• Initial temperature: \(24.0^\circ\text{C}\)
• Final temperature: \(20.5^\circ\text{C}\)
• Calorimeter heat capacity: \(50.0\ \mathrm{J\,K^{-1}}\)
• Amount of salt dissolved: \(0.0250\text{ mol}\)

Assume:

• The calorimeter initially has the same temperature as the solution.
• The specific heat capacity of the solution is \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\).
• No heat escapes to the external environment.

Calculate the molar enthalpy of dissolution.

Explain why the calorimeter’s heat change is negative and why ignoring it gives an incorrect result.

Solution 4

The molar enthalpy of dissolution is:

\[ \boxed{\Delta H=+77.2\ \mathrm{kJ\,mol^{-1}}} \]

The positive value means the dissolution is endothermic.

The calorimeter loses energy because its temperature decreases.

Step 1: Calculate the temperature change.

\[ \Delta T=20.5-24.0=-3.5\text{ K} \]

The negative temperature change indicates cooling.

Step 2: Calculate the heat change of the solution.

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=(120.0)(4.18)(-3.5)\\ &=-1755.6\text{ J} \end{aligned} \]

The solution loses \(1755.6\text{ J}\).

Step 3: Calculate the heat change of the calorimeter.

\[ \begin{aligned} q_{\text{cal}} &=C_{\text{cal}}\Delta T\\ &=(50.0)(-3.5)\\ &=-175\text{ J} \end{aligned} \]

The calorimeter also loses energy.

Step 4: Calculate the energy absorbed by the reaction.

\[ \begin{aligned} q_{\text{reaction}} &=-(q_{\text{solution}}+q_{\text{cal}})\\ &=-(-1755.6-175)\\ &=+1930.6\text{ J} \end{aligned} \]

Converting to kilojoules:

\[ q_{\text{reaction}}=+1.9306\text{ kJ} \]

Step 5: Calculate the molar enthalpy.

\[ \begin{aligned} \Delta H &=\frac{1.9306}{0.0250}\\ &=77.224\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{\Delta H=+77.2\ \mathrm{kJ\,mol^{-1}}} \]

If the calorimeter were ignored, the student would calculate:

\[ \Delta H=\frac{1.7556}{0.0250} \]

\[ \Delta H=+70.2\ \mathrm{kJ\,mol^{-1}} \]

The result would underestimate the energy absorbed by the reaction.

The important distinction is that the calorimeter does not always absorb energy. In this experiment, it supplies some of the energy needed by the endothermic process.

Question 5

A student investigates the neutralisation of hydrochloric acid with sodium hydroxide.

The student uses an unusual setup in which the two reactant solutions and the calorimeter apparatus begin at three different temperatures.

The experimental conditions are:

Hydrochloric acid:

• Volume: \(40.0\text{ mL}\)
• Concentration: \(1.00\ \mathrm{mol\,L^{-1}}\)
• Initial temperature: \(20.0^\circ\text{C}\)

Sodium hydroxide:

• Volume: \(60.0\text{ mL}\)
• Concentration: \(1.00\ \mathrm{mol\,L^{-1}}\)
• Initial temperature: \(30.0^\circ\text{C}\)

Calorimeter apparatus:

• Initial temperature: \(34.0^\circ\text{C}\)
• Heat capacity: \(50.0\ \mathrm{J\,K^{-1}}\)

After mixing and stirring, the final temperature of the entire system is \(31.8^\circ\text{C}\).

Assume:

• Both solutions have a density of \(1.00\ \mathrm{g\,mL^{-1}}\).
• Both solutions have a specific heat capacity of \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\).
• The calorimeter heat capacity refers to the apparatus only.
• The reaction goes to completion.
• No heat escapes to the external environment.

Calculate the experimental molar enthalpy of neutralisation per mole of water formed.

Your explanation must address why:

1. The two solutions cannot simply be treated as having the same initial temperature.
2. The calorimeter’s heat change is negative, even though the reaction is exothermic.
3. The limiting reactant determines the amount used in the enthalpy calculation.

Solution 5

The experimental molar enthalpy of neutralisation is:

\[ \boxed{\Delta H=-57.9\ \mathrm{kJ\,mol^{-1}}} \]

The reaction is exothermic, but the calorimeter actually loses energy during the experiment.

This question is more challenging because the solutions and apparatus do not share one initial temperature. Each component must have its own temperature change.

Step 1: Calculate the amount of each reactant.

For hydrochloric acid:

\[ \begin{aligned} n(\ce{HCl}) &=CV\\ &=(1.00)(0.0400)\\ &=0.0400\text{ mol} \end{aligned} \]

For sodium hydroxide:

\[ \begin{aligned} n(\ce{NaOH}) &=CV\\ &=(1.00)(0.0600)\\ &=0.0600\text{ mol} \end{aligned} \]

The balanced equation is:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

The mole ratio is 1:1.

Hydrochloric acid is the limiting reactant because it has fewer moles.

Therefore:

\[ n(\ce{H2O})=0.0400\text{ mol} \]

This is the amount of reaction used to calculate the molar enthalpy.

Step 2: Calculate the heat change of the hydrochloric acid solution.

The hydrochloric acid solution warms from \(20.0^\circ\text{C}\) to \(31.8^\circ\text{C}\).

Its mass is:

\[ m_{\ce{HCl}}=40.0\text{ g} \]

Its temperature change is:

\[ \Delta T_{\ce{HCl}}=31.8-20.0=11.8\text{ K} \]

Therefore:

\[ \begin{aligned} q_{\ce{HCl}} &=mc\Delta T\\ &=(40.0)(4.18)(11.8)\\ &=1972.96\text{ J} \end{aligned} \]

The hydrochloric acid solution gains energy.

Step 3: Calculate the heat change of the sodium hydroxide solution.

Its mass is:

\[ m_{\ce{NaOH}}=60.0\text{ g} \]

Its temperature change is:

\[ \Delta T_{\ce{NaOH}}=31.8-30.0=1.8\text{ K} \]

Therefore:

\[ \begin{aligned} q_{\ce{NaOH}} &=(60.0)(4.18)(1.8)\\ &=451.44\text{ J} \end{aligned} \]

The sodium hydroxide solution also gains energy.

Notice that the two solutions have different temperature changes. Using one arbitrary initial temperature for both would give the wrong energy balance.

Step 4: Calculate the heat change of the calorimeter.

The calorimeter starts at \(34.0^\circ\text{C}\), which is higher than the final temperature.

Its temperature change is:

\[ \Delta T_{\text{cal}}=31.8-34.0=-2.2\text{ K} \]

Therefore:

\[ \begin{aligned} q_{\text{cal}} &=C_{\text{cal}}\Delta T_{\text{cal}}\\ &=(50.0)(-2.2)\\ &=-110\text{ J} \end{aligned} \]

The negative sign tells us that the calorimeter loses 110 J.

This is the conceptual trap in the question. A calorimeter does not automatically gain energy during an exothermic reaction. Its own initial and final temperatures determine the sign of its heat change.

Step 5: Calculate the total heat change of the surroundings.

Add the heat changes of all three components:

\[ \begin{aligned} q_{\text{surroundings}} &=q_{\ce{HCl}}+q_{\ce{NaOH}}+q_{\text{cal}}\\ &=1972.96+451.44-110\\ &=2314.4\text{ J} \end{aligned} \]

The surroundings gain \(2314.4\text{ J}\).

Therefore:

\[ q_{\text{reaction}}=-2314.4\text{ J} \]

Converting to kilojoules:

\[ q_{\text{reaction}}=-2.3144\text{ kJ} \]

Step 6: Calculate the molar enthalpy of neutralisation.

We established that \(0.0400\text{ mol}\) of water forms.

Therefore:

\[ \begin{aligned} \Delta H &=\frac{q_{\text{reaction}}}{n(\ce{H2O})}\\ &=\frac{-2.3144}{0.0400}\\ &=-57.86\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

To three significant figures:

\[ \boxed{\Delta H=-57.9\ \mathrm{kJ\,mol^{-1}}} \]

The tempting approach is to assume that every component has the same temperature change and that the calorimeter must absorb energy because the reaction is exothermic.

Neither assumption is valid.

The two solutions have different initial temperatures, and the calorimeter starts hotter than the final mixture. It therefore supplies some energy to the surroundings rather than absorbing it.

This is why the most reliable approach is to calculate the heat change of every component separately, add those changes, and then reverse the sign to obtain the reaction’s heat change.

Once you can account for the solution and calorimeter separately, the next experimental decision is whether any additional energy escaped into the room. That requires analysing heat loss and, where appropriate, applying a cooling correction rather than simply increasing the calorimeter’s heat capacity.