Which Mass Should You Use in q = mcΔT? (HSC Chemistry)

Suppose you mix 50.0 mL of hydrochloric acid with 50.0 mL of sodium hydroxide in a coffee-cup calorimeter. The temperature rises by 6.5°C. You want to calculate the heat released using \(q = mc\Delta T\).

Which mass belongs in the equation: the mass of the acid, the mass of the sodium hydroxide, the mass that actually reacted, or the mass of the combined solution?

Use the mass of the material whose temperature change you are measuring. In this example, that is the mass of the combined solution, not the mass of the reactants that reacted.

Assuming both solutions have a density of \(1.00\ \text{g mL}^{-1}\), the mass is:

\[ m = (50.0+50.0)\times 1.00 =100.0\ \text{g} \]

This distinction is important because calorimetry involves two different calculations:

• \(q=mc\Delta T\) measures the heat gained or lost by the surroundings, such as the solution or water.
• The molar enthalpy calculation divides the reaction’s heat change by the amount of substance that actually reacted.

The mass in the first calculation and the amount in the second calculation usually come from different parts of the experiment.

01First identify what is being heated

Before choosing a mass, identify the system and its surroundings.

The system is the chemical reaction you are investigating. The surroundings are everything outside that reaction that can exchange energy with it.

In a typical acid-base neutralisation experiment:

• System: the reacting acid and base.
• Surroundings: the combined solution, the calorimeter, and the nearby environment.

The reaction releases or absorbs energy. The solution responds with a temperature change, which we measure.

For an exothermic reaction, the solution gains heat while the reaction loses heat.

Therefore:

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

This is the ideal energy-conservation model. In a real experiment, some energy may also escape to the cup or the surrounding air.

02Choose the mass by drawing the measurement boundary

The equation is:

\[ q=mc\Delta T \]

where:

• \(q\) is the heat gained or lost by the material being measured, in joules (J).
• \(m\) is the mass of that material, in grams (g).
• \(c\) is its specific heat capacity, in \(\text{J g}^{-1}\,^\circ\text{C}^{-1}\).
• \(\Delta T\) is its temperature change, calculated as final temperature minus initial temperature.

For dilute aqueous solutions, HSC Chemistry questions commonly use:

\[ c=4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1} \]

This approximates the specific heat capacity of water. It isn’t an exact value for every solution.

The important question is not which substance reacted. It’s which material experienced the temperature change you’re using in the equation.

Consider these different experimental arrangements.

Let’s look at the decisions behind this table.

Mixing two solutions: include both solutions

Suppose you mix hydrochloric acid and sodium hydroxide.

Both solutions are inside the cup. Both experience the temperature change, so both contribute to the heat absorbed by the solution.

Even if one reactant is in excess, its solution still counts.

Why? Because unreacted material doesn’t magically stop absorbing energy. It remains in the mixture and warms up along with everything else.

If the two solutions have the same density, the combined mass is:

\[ m_{\text{solution}}=m_{\text{acid}}+m_{\text{base}} \]

If only volumes are supplied, use:

\[ m=\rho V \]

where \(\rho\) is density and \(V\) is volume.

For the common HSC approximation of \(1.00\ \text{g mL}^{-1}\), the numerical value of the total volume in millilitres equals the approximate mass in grams.

If the densities differ, calculate each mass separately and add them.

Dissolving a solid: include the dissolved substance

Imagine adding potassium nitrate crystals to water.

The crystals dissolve, producing one solution. If the solution is treated as a single material with a specified or assumed specific heat capacity, its mass includes both the original water and the dissolved potassium nitrate.

For example, dissolving \(3.00\text{ g}\) of potassium nitrate in \(80.0\text{ g}\) of water gives:

\[ m_{\text{solution}}=80.0+3.00 =83.0\text{ g} \]

This assumes all the solid dissolves and the final solution is treated as having the stated specific heat capacity.

If the question explicitly tells you to use the mass of water alone, follow that model instead. The supplied assumptions determine the calculation.

Burning a fuel: use the water mass, not the fuel mass

Now consider a different experiment.

A spirit burner containing ethanol sits beneath a metal can of water. The ethanol burns, transferring energy to the can and its water.

The thermometer measures the water’s temperature change.

Therefore, the mass in the water’s \(q=mc\Delta T\) calculation is the mass of water.

The ethanol mass has a different job. It tells you how much fuel was consumed, which allows you to calculate the moles of ethanol burned.

This is a particularly important distinction:

• Water mass: used to calculate heat absorbed.
• Fuel mass: used to calculate the amount of fuel reacted.

You would not put the fuel mass into the water’s heat calculation simply because the fuel released the energy.

What about the calorimeter itself?

The cup, lid, thermometer, and other equipment can also absorb energy.

If the question supplies the calorimeter’s heat capacity, include it as a separate term.

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

Here, \(C_{\text{cal}}\) is the calorimeter’s heat capacity, measured in \(\text{J}\,^\circ\text{C}^{-1}\).

Notice that \(C_{\text{cal}}\) is not the same as \(c\):

• \(c\) describes the heat capacity per gram of a substance.
• \(C_{\text{cal}}\) describes the heat capacity of the entire calorimeter.

Do not add the physical mass of the cup to the solution mass when a calorimeter heat capacity is already supplied. Account for the cup using its separate heat-capacity term.

If the question says the calorimeter absorbs negligible heat, omit this correction.

03The mass heated is not the amount reacted

This is the second calculation you need to keep separate.

Once you’ve calculated the heat absorbed by the surroundings, you need to determine how much reaction occurred.

For a reaction at approximately constant atmospheric pressure, the molar enthalpy change is calculated using:

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

Here:

• \(\Delta H\) is the molar enthalpy change, in \(\text{kJ mol}^{-1}\).
• \(q_{\text{reaction}}\) is the heat transferred by the reaction, in kJ.
• \(n_{\text{reaction}}\) is the amount of reaction, in mol, defined according to the balanced equation and the quantity being reported.

The amount reacted is determined using stoichiometry, usually by identifying the limiting reactant.

For example, consider neutralisation:

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

The reaction has a \(1:1\) mole ratio between hydrochloric acid and sodium hydroxide.

If \(0.0400\text{ mol}\) of hydrochloric acid reacts completely with excess sodium hydroxide, then:

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

If the question asks for the enthalpy of neutralisation per mole of water formed, this is the denominator.

It is not the mass of the solution, and it is not the total amount of acid and base added together.

The two calculations answer different questions:

For a fuller explanation of limiting reactants and the correct denominator, see How to Find the Reacted Amount in Molar Enthalpy Calculations.

Don’t lose track of the sign

The temperature change also tells you whether the surroundings gained or lost heat.

Always calculate:

\[ \Delta T=T_{\text{final}}-T_{\text{initial}} \]

The sign of \(q\) depends on which substance you’re calculating it for.

For example, if the solution warms by \(5.0^\circ\text{C}\), its heat change is positive.

The reaction has released that energy, so its heat change is negative.

For an ideal insulated system:

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

In a real experiment, the accuracy of this relationship depends on whether you’ve accounted for the calorimeter and other important heat transfers.

04Worked examples

05A quick decision method

Before substituting numbers into \(q=mc\Delta T\), work through these decisions.

1. Identify the measured material. What substance or mixture has the temperature change?
2. Draw its boundary. Which materials are included in that measured body?
3. Find its mass. Include all material within the boundary, using supplied masses or density and volume.
4. Check for additional heat capacities. If the calorimeter has a supplied heat capacity, add its contribution separately.
5. Check the initial temperatures. If different parts start at different temperatures, calculate their heat changes separately rather than assuming one common \(\Delta T\).
6. Calculate the amount reacted separately. Use the balanced equation and limiting reactant.
7. Apply the correct sign. The reaction and surroundings have opposite heat changes when energy losses are accounted for.

A useful way to remember the whole method is:

The mass belongs to the temperature measurement. The moles belong to the chemical reaction.

For the full experimental treatment of calorimetry, including heat losses and temperature corrections, see Calorimetry and q = mcΔT: HSC Chemistry Guide.

06Questions and solutions

The following questions test whether you can choose the correct mass before starting the calculation. They gradually introduce additional complications, including calorimeter heat capacity, fuel combustion, and different initial temperatures.

Question 1

A student mixes \(35.0\text{ mL}\) of hydrochloric acid with \(25.0\text{ mL}\) of sodium hydroxide.

Both solutions have a density of \(1.00\text{ g mL}^{-1}\). The combined solution warms by \(4.50^\circ\text{C}\).

Assume:

• The volumes are additive.
• The combined solution has a specific heat capacity of \(4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}\).
• The calorimeter absorbs negligible heat.

Calculate:

a) The mass used in \(q=mc\Delta T\).

b) The heat absorbed by the solution.

c) The heat change of the reaction.

Solution 1

The correct mass is \(60.0\text{ g}\), and the reaction releases \(1.13\text{ kJ}\).

Part a: Find the solution mass.

Both solutions are mixed and experience the temperature rise.

\[ \begin{aligned} m&=(35.0+25.0)(1.00)\\ &=\boxed{60.0\text{ g}} \end{aligned} \]

Part b: Calculate the heat absorbed.

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=(60.0)(4.18)(4.50)\\ &=1129\text{ J}\\ &=\boxed{+1.13\text{ kJ}} \end{aligned} \]

The positive sign indicates that the solution gained heat.

Part c: Calculate the reaction heat.

The reaction supplied the energy absorbed by the solution.

\[ q_{\text{reaction}}=-1.13\text{ kJ} \]

The reaction is exothermic.

The common mistake is to use only one reactant’s mass. Both solutions contribute to the combined solution’s thermal energy change.

Question 2

A student dissolves \(5.00\text{ g}\) of ammonium chloride, \(\ce{NH4Cl}\), in \(100.0\text{ g}\) of water.

The temperature decreases from \(25.0^\circ\text{C}\) to \(22.2^\circ\text{C}\).

The following information is supplied:

• Specific heat capacity of the final solution: \(4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}\).
• Calorimeter heat capacity: \(50.0\ \text{J}\,^\circ\text{C}^{-1}\).
• Molar mass of \(\ce{NH4Cl}\): \(53.49\text{ g mol}^{-1}\).

Assume all the ammonium chloride dissolves and there is no heat exchange with the external environment.

Calculate the molar enthalpy of dissolution.

Solution 2

The molar enthalpy of dissolution is \(+14.9\text{ kJ mol}^{-1}\).

The solution cools, so the dissolution reaction must absorb energy.

Step 1: Calculate the mass of the final solution.

\[ m=100.0+5.00=105.0\text{ g} \]

The dissolved ammonium chloride contributes to the final solution mass.

Step 2: Calculate the temperature change.

\[ \Delta T=22.2-25.0=-2.8^\circ\text{C} \]

Step 3: Calculate the heat change of the solution.

\[ \begin{aligned} q_{\text{solution}} &=(105.0)(4.18)(-2.8)\\ &=-1229\text{ J} \end{aligned} \]

Step 4: Calculate the heat change of the calorimeter.

\[ \begin{aligned} q_{\text{cal}} &=C_{\text{cal}}\Delta T\\ &=(50.0)(-2.8)\\ &=-140\text{ J} \end{aligned} \]

Step 5: Calculate the heat change of the reaction.

Both the solution and calorimeter have lost energy.

\[ \begin{aligned} q_{\text{reaction}} &=-(q_{\text{solution}}+q_{\text{cal}})\\ &=-(-1229-140)\\ &=+1369\text{ J}\\ &=+1.369\text{ kJ} \end{aligned} \]

Step 6: Calculate the moles of ammonium chloride dissolved.

\[ \begin{aligned} n&=\frac{m}{M}\\ &=\frac{5.00}{53.49}\\ &=0.0935\text{ mol} \end{aligned} \]

Step 7: Calculate the molar enthalpy.

\[ \begin{aligned} \Delta H &=\frac{+1.369}{0.0935}\\ &=\boxed{+14.6\text{ kJ mol}^{-1}} \end{aligned} \]

The small rounding difference from using unrounded intermediate values is approximately \(14.9\text{ kJ mol}^{-1}\).

The reaction is endothermic.

The important trap is using \(100.0\text{ g}\), the original water mass, rather than \(105.0\text{ g}\), the final solution mass. The calorimeter also absorbs energy, so ignoring its heat capacity would underestimate the energy absorbed by the surroundings.

Question 3

A student uses a spirit burner containing methanol to heat water in a metal can.

The experimental measurements are:

The molar mass of methanol is \(32.04\text{ g mol}^{-1}\).

Assume complete combustion and that all energy released is absorbed by the water and metal can.

Calculate:

a) The heat absorbed by the water.

b) The total heat absorbed by the surroundings.

c) The experimental molar enthalpy of combustion of methanol.

Solution 3

The water absorbs \(6.27\text{ kJ}\), the total surroundings absorb \(6.42\text{ kJ}\), and the molar enthalpy of combustion is approximately \(-686\text{ kJ mol}^{-1}\).

Part a: Calculate the heat absorbed by the water.

The water is the material whose temperature is measured.

Therefore, use \(300.0\text{ g}\), not the mass of methanol.

\[ \begin{aligned} \Delta T&=25.0-20.0\\ &=5.0^\circ\text{C} \end{aligned} \]

\[ \begin{aligned} q_{\text{water}} &=mc\Delta T\\ &=(300.0)(4.18)(5.0)\\ &=6270\text{ J}\\ &=\boxed{+6.27\text{ kJ}} \end{aligned} \]

Part b: Include the metal can.

\[ \begin{aligned} q_{\text{can}} &=C_{\text{cal}}\Delta T\\ &=(30.0)(5.0)\\ &=150\text{ J} \end{aligned} \]

Therefore:

\[ \begin{aligned} q_{\text{surroundings}} &=6270+150\\ &=6420\text{ J}\\ &=\boxed{+6.42\text{ kJ}} \end{aligned} \]

Part c: Calculate the molar enthalpy of combustion.

The mass of methanol is used to calculate the moles burned.

\[ \begin{aligned} n&=\frac{0.300}{32.04}\\ &=0.00936\text{ mol} \end{aligned} \]

The reaction released the energy absorbed by the surroundings:

\[ q_{\text{reaction}}=-6.42\text{ kJ} \]

Therefore:

\[ \begin{aligned} \Delta H &=\frac{-6.42}{0.00936}\\ &=\boxed{-686\text{ kJ mol}^{-1}} \end{aligned} \]

The negative value indicates exothermic combustion.

Notice that the \(0.300\text{ g}\) of methanol never appeared in the water’s \(q=mc\Delta T\) calculation. It was only used to determine the amount of fuel consumed.

Question 4

Two different exothermic reactions are investigated using separate insulated solution calorimeters.

The following illustrative data are recorded:

Both solutions have a specific heat capacity of \(4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}\).

Assume there is no heat exchange with the external environment.

Calculate the molar enthalpy change for each reaction.

Then explain:

a) Which reaction released more total heat?

b) Which reaction released more heat per mole?

c) Why would comparing the temperature rises alone give the wrong conclusion?

Solution 4

Reaction B releases more total heat, but Reaction A releases more heat per mole.

This is a useful example of why temperature rise alone cannot tell you which reaction released more energy.

Step 1: Calculate the heat absorbed in Reaction A.

\[ \begin{aligned} q_{\text{solution,A}} &=(60.0)(4.18)(6.00)\\ &=1505\text{ J} \end{aligned} \]

The calorimeter absorbs:

\[ q_{\text{cal,A}}=(30.0)(6.00)=180\text{ J} \]

Therefore:

\[ \begin{aligned} q_{\text{surroundings,A}} &=1505+180\\ &=1685\text{ J}\\ &=1.685\text{ kJ} \end{aligned} \]

The reaction heat is:

\[ q_{\text{reaction,A}}=-1.685\text{ kJ} \]

Step 2: Calculate the molar enthalpy of Reaction A.

\[ \begin{aligned} \Delta H_A &=\frac{-1.685}{0.0200}\\ &=\boxed{-84.2\text{ kJ mol}^{-1}} \end{aligned} \]

Step 3: Calculate the heat absorbed in Reaction B.

\[ \begin{aligned} q_{\text{solution,B}} &=(180.0)(4.18)(4.00)\\ &=3010\text{ J} \end{aligned} \]

The calorimeter absorbs:

\[ q_{\text{cal,B}}=(30.0)(4.00)=120\text{ J} \]

Therefore:

\[ \begin{aligned} q_{\text{surroundings,B}} &=3010+120\\ &=3130\text{ J}\\ &=3.130\text{ kJ} \end{aligned} \]

The reaction heat is:

\[ q_{\text{reaction,B}}=-3.130\text{ kJ} \]

Step 4: Calculate the molar enthalpy of Reaction B.

\[ \begin{aligned} \Delta H_B &=\frac{-3.130}{0.0400}\\ &=\boxed{-78.2\text{ kJ mol}^{-1}} \end{aligned} \]

Part a: Which released more total heat?

Reaction B released \(3.130\text{ kJ}\), compared with \(1.685\text{ kJ}\) for Reaction A.

Therefore, Reaction B released more total heat.

Part b: Which released more heat per mole?

Reaction A has the more negative molar enthalpy:

\[ |-84.2|>|-78.2| \]

Therefore, Reaction A releases more energy per mole of reaction.

Part c: Why is comparing temperature rises misleading?

Reaction A had a larger temperature rise, but it heated a much smaller mass of solution.

Reaction B had a smaller temperature rise, but it heated three times as much solution.

The total energy absorbed depends on both mass and temperature change, as well as the calorimeter’s heat capacity.

There is another distinction: total heat released and heat released per mole are not interchangeable. The second calculation also depends on the amount of reaction that occurred.

Question 5

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

The two solutions begin at different temperatures.

The experimental arrangement and data are:

• \(60.0\text{ g}\) of hydrochloric acid solution at \(18.0^\circ\text{C}\), containing \(0.0400\text{ mol}\) of \(\ce{HCl}\).
• \(40.0\text{ g}\) of sodium hydroxide solution at \(32.0^\circ\text{C}\), containing \(0.0600\text{ mol}\) of \(\ce{NaOH}\).
• The calorimeter initially has a temperature of \(25.0^\circ\text{C}\).
• The calorimeter heat capacity is \(40.0\ \text{J}\,^\circ\text{C}^{-1}\).
• The final temperature after mixing is \(28.7^\circ\text{C}\).

Assume both solutions have a specific heat capacity of \(4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}\), and no energy is transferred to the external environment.

The reaction is:

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

A student argues:

“The combined solution has a mass of \(100.0\text{ g}\). Its average initial temperature is \(25.0^\circ\text{C}\). Therefore, I can use one \(mc\Delta T\) calculation with a temperature rise of \(3.7^\circ\text{C}\).”

Another student argues:

“The final temperature is lower than the initial temperature of the sodium hydroxide solution, so the reaction must be endothermic.”

Evaluate both arguments.

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

Solution 5

The reaction is exothermic, and its molar enthalpy of neutralisation is \(-57.0\text{ kJ mol}^{-1}\) per mole of water formed. Both students’ arguments are incorrect.

This question requires you to account for three different starting temperatures and two separate solution masses.

The key is that you cannot automatically combine everything into one \(mc\Delta T\) calculation.

Step 1: Identify the amount of reaction.

The balanced equation has a \(1:1\) ratio between hydrochloric acid and sodium hydroxide.

Initial amounts:

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

\[ n(\ce{NaOH})=0.0600\text{ mol} \]

Hydrochloric acid is the limiting reactant.

Therefore:

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

This is the amount used to calculate the molar enthalpy per mole of water formed.

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

The hydrochloric acid solution starts at \(18.0^\circ\text{C}\) and finishes at \(28.7^\circ\text{C}\).

\[ \begin{aligned} q_{\ce{HCl}} &=(60.0)(4.18)(28.7-18.0)\\ &=(60.0)(4.18)(10.7)\\ &=+2684\text{ J} \end{aligned} \]

The hydrochloric acid solution gains energy.

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

The sodium hydroxide solution starts at \(32.0^\circ\text{C}\) and finishes at \(28.7^\circ\text{C}\).

\[ \begin{aligned} q_{\ce{NaOH}} &=(40.0)(4.18)(28.7-32.0)\\ &=(40.0)(4.18)(-3.3)\\ &=-552\text{ J} \end{aligned} \]

The sodium hydroxide solution actually loses energy.

This is the first important surprise: not every part of the surroundings has to warm up.

Step 4: Calculate the heat change of the calorimeter.

The calorimeter begins at \(25.0^\circ\text{C}\) and finishes at \(28.7^\circ\text{C}\).

\[ \begin{aligned} q_{\text{cal}} &=C_{\text{cal}}\Delta T\\ &=(40.0)(28.7-25.0)\\ &=+148\text{ J} \end{aligned} \]

The calorimeter gains energy.

Step 5: Calculate the total heat absorbed by the surroundings.

Add the three heat changes:

\[ \begin{aligned} q_{\text{surroundings}} &=2684-552+148\\ &=2280\text{ J}\\ &=+2.280\text{ kJ} \end{aligned} \]

The surroundings have gained energy overall, even though the sodium hydroxide solution cooled.

Therefore:

\[ q_{\text{reaction}}=-2.280\text{ kJ} \]

Step 6: Calculate the molar enthalpy of neutralisation.

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

\[ \begin{aligned} \Delta H &=\frac{-2.280}{0.0400}\\ &=\boxed{-57.0\text{ kJ mol}^{-1}} \end{aligned} \]

The negative value confirms that neutralisation is exothermic.

Why the first student’s argument fails

The student used the ordinary arithmetic average of the two initial solution temperatures:

\[ \frac{18.0+32.0}{2}=25.0^\circ\text{C} \]

But the solutions have different masses.

The correct combined initial temperature, if treating the two solutions as one body with the same specific heat capacity, must be weighted by their masses:

\[ \begin{aligned} T_{\text{weighted}} &=\frac{(60.0)(18.0)+(40.0)(32.0)} {60.0+40.0}\\ &=23.6^\circ\text{C} \end{aligned} \]

The combined solution’s temperature rise relative to this weighted initial temperature is \(5.1^\circ\text{C}\), not \(3.7^\circ\text{C}\).

Using the weighted temperature gives:

\[ \begin{aligned} q_{\text{solution}} &=(100.0)(4.18)(28.7-23.6)\\ &=2132\text{ J} \end{aligned} \]

Adding the calorimeter contribution gives:

\[ q_{\text{surroundings}}=2132+148=2280\text{ J} \]

The same correct result is obtained.

The separate calculations are generally safer because they make each initial temperature explicit.

Why the second student’s argument fails

The sodium hydroxide solution cools, but the hydrochloric acid solution and calorimeter gain more energy than the sodium hydroxide solution loses.

The total surroundings gain \(2.280\text{ kJ}\).

Therefore, the reaction loses \(2.280\text{ kJ}\) and is exothermic.

The trap is judging the entire reaction using the temperature change of only one part of the surroundings.

The next decision after choosing the correct mass is choosing the correct mole denominator. Once you can separate the material being heated from the substance reacting, limiting reactants and balanced equations become much easier to apply to experimental enthalpy calculations.