Calorimetry Discussion Examples | Effects of Heat of Neutralization, Heat of Dissolution, and Heat Loss

What Is Calorimetry?

Calorimetry is an experiment in which the amount of heat generated or absorbed is determined from the temperature change associated with a reaction or dissolution.
For example, if the temperature rises when an acid and a base are mixed, heat is considered to have been generated by neutralization.
Likewise, if the temperature rises when a substance dissolves in water, the dissolution is considered exothermic, while if the temperature decreases, the dissolution is considered endothermic.

In calorimetry, heat generated in the reaction system is considered to warm the solution and container, or heat may be taken from the solution.
Therefore, the temperature change is measured accurately, and the heat quantity is calculated from the mass of the solution, its specific heat, and the temperature change.

Main Items to Include in the Results

In the results of calorimetry, organize the temperature before mixing, the maximum or minimum temperature, the temperature change, the mass of the solution, the specific heat, the calculated heat quantity, and the heat per amount of substance.
If temperature changes were measured over time, preparing a temperature-time graph makes the discussion easier to write.

Main Items to Include in the Results

• Type of sample or solution used
• Volume and concentration of each solution
• Temperature before mixing
• Maximum or minimum temperature after mixing
• Temperature change ΔT
• Mass of the solution
• Specific heat
• Heat capacity of the container
• Calculated heat q
• Amount of substance reacted
• Heat of reaction per 1 mol
• Comparison with literature or theoretical values
• Error rate
• Temperature-time graph

Reference Experimental Values for Calorimetry and Examples of Analysis of Heat of Neutralization and Heat of Dissolution

Here, reference experimental values are organized for discussing heat of neutralization, heat of dissolution, and heat loss from temperature changes measured using a calorimeter.
Temperature rise and fall, heat calculations, heat of reaction per amount of substance, calorimeter constant, differences from theoretical values, and sources of error are summarized in a form that is easy to use in reports.

In calorimetry, heat generated or absorbed by reactions or dissolution is observed as a temperature change of the solution and calorimeter.
A temperature rise is considered to indicate an exothermic reaction, while a temperature decrease is considered to indicate an endothermic reaction.
However, in actual experiments, heat escapes to the surroundings and the calorimeter itself absorbs heat, so it is important to consider corrections.

Reference Experimental Conditions

Basic Equation for Heat Calculation

The heat received or lost by the solution is determined using the following equation.

q = mcΔT

Here, q is heat, m is the mass of the solution, c is the specific heat, and ΔT is the temperature change.
For aqueous solutions, the specific heat is often approximated as 4.18 J/(g·K) and the density as 1.00 g/mL.

The numerical value of a temperature difference is the same whether expressed in K or °C.
For example, if the temperature rises from 25.0°C to 31.8°C, ΔT can be treated as 6.8 K.

Example Measurement of Heat of Neutralization

The following is a reference example of the temperature change when 50.0 mL of 1.00 mol/L HCl is mixed with 50.0 mL of 1.00 mol/L NaOH.

Example Calculation of Heat of Neutralization

The heat received by the solution is determined as follows.

q = mcΔT = 100 × 4.18 × 6.8 = 2842 J

The amounts of HCl and NaOH that reacted in the neutralization reaction are both 0.0500 mol.
Therefore, the heat of neutralization per 1 mol is as follows.

Heat of neutralization = 2842 J ÷ 0.0500 mol = 56840 J/mol = 56.8 kJ/mol

Because neutralization is an exothermic reaction, when the sign of the enthalpy change is included, it can be written as ΔH = −56.8 kJ/mol.

Comparison of Heat of Neutralization Between Strong Acid-Strong Base and Weak Acid-Strong Base

In the neutralization of a strong acid and strong base, the main reaction is H⁺ + OH⁻ → H₂O.
In the case of a weak acid, however, the process in which the weak acid ionizes before neutralization is involved, so the heat of neutralization may become slightly smaller.

The heat of neutralization of a strong acid and strong base is almost constant because in every case the reaction is essentially the formation of water from hydrogen ions and hydroxide ions.
For weak acids and weak bases, energy changes associated with ionization are also involved, so the heat of neutralization may appear smaller.

Example Measurement of Heat of Dissolution: Ammonium Chloride

The following is a reference example in which the solution temperature decreases when ammonium chloride is dissolved in water.

Because the temperature decreased, the dissolution of NH₄Cl is considered to have proceeded endothermically.

Example Calculation of Heat of Dissolution

The magnitude of heat lost by the solution is calculated.
The total mass of the solution is approximated as 100.0 g of water + 5.35 g of NH₄Cl = 105.35 g.

q = mcΔT = 105.35 × 4.18 × (−3.8) = −1674 J

Because the solution temperature decreased, the solution lost 1674 J of heat.
Conversely, the dissolution process is considered to have absorbed this heat, so

Heat of dissolution = +1674 J ÷ 0.100 mol = +16.7 kJ/mol

Including the sign, the dissolution of NH₄Cl can be expressed as an endothermic process with ΔH_sol = +16.7 kJ/mol.

Comparison of Exothermic and Endothermic Dissolution

Dissolution involves both an endothermic process required to break the crystal lattice and an exothermic process in which ions are hydrated.
If the heat released by hydration is larger, dissolution becomes exothermic, while if the heat absorbed to break the lattice is larger, dissolution becomes endothermic.

Example Considering the Calorimeter Constant

In reality, not only the solution but also the calorimeter absorbs heat.
When the heat absorbed by the calorimeter is taken into account, the calorimeter constant C_cal is used.

q = mcΔT + C_calΔT

For example, in a neutralization experiment where the mass of the solution is 100 g, the specific heat is 4.18 J/(g·K), the temperature change is 6.8 K, and the calorimeter constant is 30 J/K,

q_solution = 100 × 4.18 × 6.8 = 2842 J

q_cal = 30 × 6.8 = 204 J

q_total = 2842 + 204 = 3046 J

When this correction is applied, the heat of neutralization becomes as follows.

Heat of neutralization = 3046 J ÷ 0.0500 mol = 60.9 kJ/mol

When the calorimeter constant is taken into account, the heat generated by the reaction is estimated to be larger than the uncorrected value.

Example of the Effect of Heat Loss

In an exothermic reaction, if heat escapes to the surroundings during the reaction, the observed maximum temperature becomes lower and the heat of reaction may be underestimated.

When heat loss occurs in an exothermic reaction, the observed temperature rise becomes smaller and the absolute value of the calculated heat of reaction becomes smaller.

Example of Temperature Correction Using a Cooling Curve

In an exothermic reaction, after the maximum temperature is reached, the temperature decreases because heat escapes to the surroundings.
If the temperature change is recorded over time, a corrected temperature that accounts for heat loss can be estimated.

By extrapolating the cooling curve back to the reaction time, the maximum temperature under ideal adiabatic conditions can be estimated.
Because the corrected temperature change becomes larger, the absolute value of the heat of reaction also becomes larger.

Example of Difference Between Experimental and Theoretical Heat of Neutralization

Assuming a reference value of −57.3 kJ/mol for the heat of neutralization of a strong acid and strong base, the difference from the experimental value is considered.

The difference from the theoretical value changes depending on whether corrections are applied and on the operating conditions.
If the experimental value is smaller than the theoretical value, heat loss or insufficient measurement of the temperature rise may be possible causes.

Comparison of Heat of Neutralization at Different Concentrations

As the concentration increases, the total amount of heat generated increases, but the heat of neutralization per 1 mol should not change greatly.
However, at high concentrations, the approximation that the specific heat and density are the same as those of water becomes less accurate.

Example of How to Write the Results

When 50.0 mL of 1.00 mol/L HCl and 50.0 mL of 1.00 mol/L NaOH were mixed, the temperature increased from 25.0°C to 31.8°C.
The temperature change was 6.8°C.
When the solution mass was taken as 100 g and the specific heat as 4.18 J/(g·K), the heat received by the solution was calculated as 100 × 4.18 × 6.8 = 2842 J.

The amounts of HCl and NaOH that reacted were both 0.0500 mol, and the amount of water produced by neutralization was also 0.0500 mol.
Therefore, the heat of neutralization per 1 mol was 2842 J ÷ 0.0500 mol = 56.8 kJ/mol.
Because the neutralization reaction releases heat, the reaction enthalpy including the sign is ΔH = −56.8 kJ/mol.

When NH₄Cl was dissolved in water, the temperature decreased from 25.0°C to 21.2°C.
This indicates that heat was absorbed from the surrounding aqueous solution during the dissolution process, so the dissolution of NH₄Cl is considered endothermic.
The calculated heat of dissolution was +16.7 kJ/mol.

Points for Connecting the Results to the Discussion

In a discussion of calorimetry, it is important not only to determine heat from temperature changes but also to explain the signs of exothermic and endothermic processes, heat per amount of substance, heat loss, calorimeter constant, and differences from theoretical values.

• Can you calculate the heat received by the solution using q = mcΔT?
• Can you determine the heat of reaction per 1 mol from the amount of reactant?
• Can you explain that ΔH is negative for an exothermic reaction and positive for an endothermic reaction?
• Can you explain why the heat of neutralization of strong acids and strong bases is almost constant?
• Can you discuss the energy change associated with ionization in weak acids and weak bases?
• For heat of dissolution, can you explain the relationship between lattice energy and hydration energy?
• Can you take into account heat absorbed by the calorimeter and heat lost to the surroundings?
• Can you explain the concept of a cooling curve and corrected temperature?
• Can you discuss the difference from the theoretical value in terms of heat loss, insufficient stirring, temperature reading, and approximations of specific heat and density?
• Can you explain that the approximation of using the same specific heat and density as water may become inaccurate for high-concentration solutions?

Example Discussion

In this experiment, the temperature change caused by neutralization of an acid and base was measured, and the heat of neutralization was determined.
When HCl and NaOH were mixed, the temperature increased from 25.0°C to 31.8°C, indicating that the neutralization reaction was exothermic.
The heat received by the solution was calculated as 2842 J from q = mcΔT, and dividing by the amount of substance that reacted, 0.0500 mol, gave a heat of neutralization of 56.8 kJ/mol.
Because the reaction is exothermic, the value including the sign is ΔH = −56.8 kJ/mol.

The heat of neutralization of a strong acid and strong base is almost constant because the essential reaction is the formation of water from H⁺ and OH⁻.
The heats of neutralization of HCl with NaOH and HNO₃ with NaOH were similar because both are almost completely ionized in aqueous solution.
In contrast, the heat of neutralization of acetic acid with NaOH was slightly smaller.
This is considered to be because acetic acid is a weak acid and energy is required for its ionization during neutralization.

In the measurement of heat of dissolution, the temperature decreased when NH₄Cl was dissolved in water.
This indicates that heat was absorbed from the surroundings during the dissolution process.
Dissolution involves both the heat absorbed to break the crystal lattice and the heat released when ions are hydrated.
For NH₄Cl, the heat absorbed to break the lattice is considered to have exceeded the heat released by hydration, resulting in an overall endothermic dissolution.

Possible reasons for the difference between the experimental and theoretical values include heat loss to the surroundings, heat absorbed by the calorimeter, insufficient stirring, delayed temperature readings, and approximation of the specific heat and density as the same as water.
In an exothermic reaction, if heat escapes to the surroundings, the observed maximum temperature becomes lower and the absolute value of the heat of reaction is underestimated.
In addition, if the temperature is not uniform immediately after the reaction, the maximum temperature may not be measured correctly.
Therefore, it is important to improve insulation, stir sufficiently, and record the temperature change over time.

Taking the calorimeter constant into account allows correction for the fact that some of the heat generated by the reaction is used to warm the calorimeter itself.
The corrected heat of reaction becomes larger than the uncorrected value and may become closer to the actual heat of reaction.
However, because there is also error in measuring the calorimeter constant, the validity of the correction must also be considered.

Summary

In calorimetry, heat is determined from temperature change using q = mcΔT, and the heat of reaction per 1 mol is calculated by dividing by the amount of reactant.
In an exothermic reaction, the temperature rises and ΔH is negative.
In an endothermic reaction, the temperature decreases and ΔH is positive.

This reference example covered heat of neutralization, heat of dissolution, exothermic and endothermic dissolution, calorimeter constant, heat loss, cooling curves, differences from theoretical values, and the effects of different concentrations.
In a report, it is useful to discuss not only the numerical value of the temperature change but also the sign of the reaction, heat transfer, heat loss, and the effects of the apparatus in relation to one another.

Basic Equation for Heat Calculation

In calorimetry, the heat received or lost by a solution is often determined using the following equation.
If the mass is m, the specific heat is c, and the temperature change is ΔT, the heat q is expressed as follows.

q = mcΔT

Here, m is the mass of the solution, c is the specific heat, and ΔT is the temperature change.
For an aqueous solution, the specific heat may be treated as a value close to that of water, but the actual specific heat of the solution is not necessarily exactly the same as that of water.
In a report, use the specific heat specified in the laboratory manual.

Determining Exothermic and Endothermic Reactions

In calorimetry, a temperature increase is considered to indicate an exothermic reaction, while a temperature decrease is considered to indicate an endothermic reaction.
In an exothermic reaction, the solution receives heat generated by the reaction and its temperature rises.
In an endothermic reaction, the reaction takes heat from the surrounding solution, so the temperature decreases.

However, the sign of the heat of reaction and the sign of the heat received by the solution must be considered oppositely.
If the solution receives heat, the reaction has released that heat.
In a report, it is important to handle the sign according to the instructions in the laboratory manual.

What Is Heat of Neutralization?

Heat of neutralization is the heat generated when an acid and a base react to form water.
In the neutralization of a strong acid and strong base, the main reaction is the reaction of hydrogen ions and hydroxide ions to form water.
Therefore, for combinations of strong acids and strong bases, the heat of neutralization when 1 mol of water is formed is considered to be close to a nearly constant value.

H⁺ + OH⁻ → H₂O

On the other hand, when a weak acid or weak base is used, heat associated with ionization is also involved, so the value may differ from the heat of neutralization of a strong acid and strong base.

How to Determine the Heat of Neutralization

To determine the heat of neutralization, first calculate the heat received by the solution from the temperature rise caused by mixing.
Next, divide by the amount of water formed during neutralization or the amount of acid or base that reacted to determine the heat of neutralization per 1 mol.
It is also important to check whether the acid or base is the limiting reagent.

Heat of neutralization = Heat received by solution ÷ Amount of water formed

When calculating the heat of neutralization, the total mass of the solution after mixing is often used.
Determine the amount of substance from the concentration and volume and confirm the amount that actually participated in neutralization.

Why the Heat of Neutralization May Be Smaller Than the Literature Value

One reason the experimental heat of neutralization may be smaller than the literature value is that heat escaped to the surroundings.
Ideally, all heat generated by the reaction would be used to raise the temperature of the solution, but in reality, heat is transferred to the container, thermometer, air, desk, and other surroundings.
As a result, the measured temperature rise becomes smaller than the true value, and the calculated heat of neutralization also becomes smaller.

In addition, if mixing or stirring is insufficient, the maximum temperature may not be measured correctly and the temperature change may be underestimated.

What Is Heat of Dissolution?

Heat of dissolution is the heat generated or absorbed when a substance dissolves in a solvent.
During dissolution, energy required to break the crystal lattice and energy released by hydration or solvation are involved.
As a result, the overall process may be exothermic or endothermic.

If the temperature rises during dissolution, the dissolution is considered exothermic, while if the temperature falls, it is considered endothermic.
However, when the temperature change is small, the relative effects of heat loss and the resolution of the thermometer become larger.

How to Determine the Heat of Dissolution

To determine the heat of dissolution, calculate the heat received or lost by the solution from the temperature change before and after dissolution.
Divide this heat by the amount of substance dissolved to determine the heat of dissolution per 1 mol.

Heat of dissolution = Heat quantity ÷ Amount of solute

In exothermic dissolution, the temperature rises, while in endothermic dissolution, the temperature falls.
The treatment of signs depends on whether the value is expressed as the heat of reaction or as the heat received by the solution, so organize it according to the laboratory manual.

What Is Heat Loss?

Heat loss is the escape of heat generated by a reaction or dissolution to places other than the solution whose heat is intended to be measured.
Even when an insulated container is used, heat cannot be completely confined and is transferred to the container, thermometer, air, surrounding desk, and other surroundings.
Therefore, the temperature change observed experimentally tends to be smaller than under ideal conditions.

In an exothermic reaction, heat loss causes the observed temperature rise to become smaller and may lead to underestimation of the heat of reaction.
In an endothermic process, if heat flows in from the surroundings, the observed temperature decrease becomes smaller and the amount of absorbed heat may be underestimated.

Effect of the Heat Capacity of the Container

In calorimetry, heat generated by a reaction warms not only the solution but also the container and thermometer.
If the heat capacity of the container is ignored, the heat received by the container is not included in the calculation, and the amount of heat generated may be underestimated.
For more accurate determination, the heat capacity of the calorimeter is corrected.

q = mcΔT + C_calΔT

Here, C_cal is the heat capacity of the calorimeter.
In student experiments, the heat capacity of the container may be ignored for simplification, but in that case it can be discussed as a source of error.

Reading the Maximum and Minimum Temperatures

In calorimetry, it is important to read the maximum or minimum temperature after mixing correctly.
In an exothermic reaction, the temperature rises after mixing and then gradually decreases because of heat loss.
In an endothermic process, the temperature decreases and may then gradually return because of heat inflow from the surroundings.

If the maximum or minimum temperature is missed, the temperature change is underestimated.
Therefore, temperature changes may be recorded over time and corrected using a temperature-time graph.

Discussion of a Temperature-Time Graph

A temperature-time graph makes it possible to visually confirm the temperature before mixing, the temperature change after mixing, and the temperature decrease caused by heat loss.
In an exothermic reaction, the temperature rises immediately after mixing and then gradually decreases.
In an endothermic process, the temperature decreases and may then gradually increase because of heat inflow from the surroundings.

Using a temperature-time graph, it may be possible to estimate a corrected temperature that takes heat loss into account.
In a report, rather than simply reading the maximum temperature, discuss the trend in temperature change in relation to heat loss.

Error Caused by Insufficient Stirring

In calorimetry, stirring is important for making the temperature of the entire solution uniform.
If stirring is insufficient, only the area around the thermometer may become locally hotter or colder, and the average temperature of the entire solution cannot be measured correctly.
In addition, the reaction or dissolution may not proceed uniformly, affecting the reading of the maximum or minimum temperature.

Errors in Reading the Thermometer

Errors are also included in reading the scale or digital display of the thermometer.
With an analog thermometer, parallax may be a problem, while with a digital thermometer, display delay and response time may be problematic.
If the temperature change caused by a reaction or dissolution occurs rapidly, a slow thermometer response may prevent accurate measurement of the maximum or minimum temperature.

Error Caused by Assuming the Specific Heat Is the Same as That of Water

In heat calculations for aqueous solutions, the specific heat may be treated as the same as that of water.
However, the specific heat of a solution containing solute is not exactly the same as that of pure water.
For high-concentration solutions or depending on the type of solute, differences in specific heat may affect the heat calculation.

Error Caused by Assuming the Density of the Solution Is the Same as That of Water

When determining the mass of a solution, it may be calculated from volume and density.
In student experiments, the density of an aqueous solution may be assumed to be 1.0 g/mL.
However, the actual density of a solution varies with concentration and solute.
If this assumption is inaccurate, the mass m becomes inaccurate and the heat q also shifts.

Effect of a Temperature Difference Before Mixing

When two solutions are mixed in a heat-of-neutralization experiment, if their temperatures before mixing are different, heat transfer occurs between the solutions independently of the heat of reaction.
Therefore, if the observed temperature change is treated as being caused only by the reaction heat, an error occurs.
If possible, it is desirable to adjust the solutions to the same temperature before mixing.

Error Caused by Undissolved Sample

In a heat-of-dissolution experiment, if the sample does not dissolve completely, the amount of substance that actually dissolves is smaller than the amount used in the calculation.
As a result, the heat of dissolution per 1 mol cannot be determined correctly.
In addition, if undissolved material remains, the temperature change also tends to become smaller, causing the heat of dissolution to be underestimated.

Discussion When the Reaction Is Not Complete

In calculations of heat of neutralization and heat of reaction, the reaction may be assumed to proceed completely.
However, because of insufficient mixing, concentration errors, insufficient reaction time, or ionization of weak acids and weak bases, the reaction may not proceed ideally.
If the reaction is incomplete, the amount of heat generated becomes smaller than the theoretical value.

When the Heat Quantity Is Larger Than the Literature Value

If the heat quantity is larger than the literature value, possible causes include reading the temperature change as too large, underestimating the amount of substance that reacted, or including another exothermic process.
For example, if a thermometer-reading error occurs or heat transfer caused by a temperature difference before mixing is incorrectly included as reaction heat, the heat quantity may be overestimated.

When the Heat Quantity Is Smaller Than the Literature Value

The most common reason the heat quantity is smaller than the literature value is heat loss to the surroundings.
If part of the generated heat escapes to the container or air, the temperature change of the solution becomes smaller.
Incomplete reaction, missing the maximum temperature, and inaccurate assumptions about the specific heat or mass of the solution may also cause the heat quantity to be underestimated.

Calculation of Error Rate

When experimentally determined heats of neutralization or dissolution are compared with literature values, the error rate can be calculated to express the difference quantitatively.
The error rate is obtained by dividing the difference between the experimental value and literature value by the literature value and expressing it as a percentage.

Error rate (%) = |Experimental value − Literature value| ÷ Literature value × 100

Simply calculating the error rate is not sufficient as a discussion.
Explain which factors, such as heat loss, temperature measurement, the specific-heat assumption, heat capacity of the container, and insufficient stirring, shifted the value and in which direction.

When the Results Can Be Considered Good

Calorimetry results can be considered good when a clear temperature change is observed, the trend in the temperature-time graph agrees with an exothermic or endothermic process, and the determined heat quantity does not greatly contradict the literature value.
It is also important that mixing and stirring be sufficient and that the maximum or minimum temperature be read correctly.

Example Discussion When the Experiment Did Not Go Well

If calorimetry does not go well, possible causes are considered from results such as a small temperature change, a large difference from the literature value, unstable temperature, inability to read the maximum temperature, undissolved sample, or insufficient stirring.
It is easier to organize the discussion by separately considering heat loss, heat capacity of the container, thermometer, specific-heat assumption, density assumption, and incomplete reaction.

How to Write Points for Improvement

In a discussion of calorimetry, including points for improvement as well as sources of error makes the report easier to organize.
Improvements can be organized by dividing them into insulation, temperature measurement, mixing, and calculation conditions.

Improvements to Reduce Heat Loss

• Use a highly insulating container
• Cover the container to reduce heat exchange with the surroundings
• Perform measurements quickly
• Correct for the heat capacity of the container
• Correct for heat loss using a temperature-time graph

Improvements to Temperature Measurement

• Check the response of the thermometer
• Do not miss the maximum or minimum temperature
• Record the temperature at regular intervals
• Keep the thermometer-reading position constant
• Equalize the temperatures of the solutions before mixing

Improvements to Mixing and Sample Handling

• Mix the solutions quickly
• Stir sufficiently
• Dissolve the sample completely
• Determine the amount of reactant accurately
• Check the assumptions for solution mass and specific heat

Difference Between a Superficial Discussion and a Good Discussion

In a discussion of calorimetry, simply writing that “the temperature increased” or “heat escaped” results in a superficial discussion.
Relating the temperature change, q = mcΔT, heat loss, heat capacity of the container, and differences from literature values produces a more persuasive discussion.

Examples of Expressions That Can Be Used in Reports

The following expressions can be used when writing the results and discussion of calorimetry.
Adjust the necessary parts according to your own experimental results.

• Because the temperature increased after mixing, this reaction is considered exothermic.
• Because a temperature decrease was observed, the dissolution process can be judged to be endothermic.
• The heat received by the solution was determined using q = mcΔT.
• The heat of reaction per 1 mol was calculated by dividing the heat quantity by the amount of substance that reacted.
• Because of heat loss, the observed temperature change may have been smaller than the actual value.
• Because the container and thermometer absorbed heat, the amount of heat generated cannot be completely evaluated from the temperature rise of the solution alone.
• If the maximum temperature is missed, ΔT is underestimated and the heat of reaction is underestimated.
• Insufficient stirring may have caused a nonuniform temperature distribution, so the thermometer reading may not have reflected the average temperature of the entire solution.
• Assuming that the specific heat and density of the solution are the same as those of water may be a source of error in the heat calculation.
• When a weak acid or weak base is used, heat associated with ionization is involved, so the value differs from the heat of neutralization of a strong acid and strong base.

Points to Check When Discussing Calorimetry

Checking the following points before writing the report makes the discussion easier to write.

• Have you calculated the temperature change ΔT correctly?
• Have you checked each value and unit in q = mcΔT?
• Have you determined whether the reaction is exothermic or endothermic from the temperature change?
• Have you correctly determined the amount of substance that reacted?
• Have you converted the heat quantity to a value per 1 mol?
• Have you compared the result with a literature or theoretical value?
• Have you considered the effect of heat loss?
• Have you discussed the effect of neglecting the heat capacity of the container?
• Have you considered errors in reading the maximum or minimum temperature?
• Have you discussed insufficient stirring and temperature nonuniformity?
• Have you considered assumptions about specific heat and density as sources of error?
• Do the points for improvement correspond to the sources of error?

Summary

In calorimetry, the heat quantity is determined from the temperature change associated with a reaction or dissolution.
Basically, q = mcΔT is used to calculate the heat received or lost by the solution.
For heat of neutralization, the heat generated by the reaction between an acid and base is divided by the amount of water formed or a similar quantity to determine the heat per 1 mol.
For heat of dissolution, the heat associated with dissolution per 1 mol is determined from the temperature change before and after dissolution and the amount of solute.

One of the most important sources of error in calorimetry is heat loss.
If part of the generated heat escapes to the container, thermometer, air, and other surroundings, the observed temperature change becomes smaller and the heat of reaction or dissolution may be underestimated.
The heat capacity of the container, reading of the maximum or minimum temperature, insufficient stirring, and assumptions about specific heat and density also affect the results.

In a report, do not simply write that “the temperature increased” or “heat escaped.”
Explain the temperature change, heat calculation, heat per amount of substance, difference from the literature value, and sources of error in relation to one another.
A more persuasive discussion of calorimetry can be produced by explaining the direction in which heat loss shifts the experimental value.