How to Calculate Specific Heat Capacity from Energy and Temperature Data

An immersion heater supplies 3.0 kJ of energy to a cup containing a liquid. The liquid warms by 5.0°C. A student divides 3.0 kJ by the liquid’s mass and temperature change to calculate its specific heat capacity.

There is one problem: what if some of that energy heated the cup instead?

To find specific heat capacity correctly, you need the energy absorbed by the sample, not necessarily the total energy supplied to the apparatus.

The calculation itself is straightforward:

\[
\boxed{c=\frac{q}{m\Delta T}}
\]

Here, \(c\) is specific heat capacity, \(q\) is energy transferred to the sample, \(m\) is its mass, and \(\Delta T\) is its temperature change.

The important skill is identifying the correct energy value before substituting.

01The equation for specific heat capacity

Specific heat capacity describes how much energy is needed to raise the temperature of one gram of a substance by one degree Celsius.

For example, a substance with a specific heat capacity of \(4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}\) requires 4.18 J of energy to increase the temperature of 1 g of that substance by 1°C.

The relationship between energy, mass, temperature change, and specific heat capacity is:

\[
q=mc\Delta T
\]

Rearranging to find specific heat capacity:

\[
\boxed{c=\frac{q}{m\Delta T}}
\]

The symbols mean:

Calculate temperature change using:

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

Be careful with the order. Final temperature comes first.

A temperature increase of 25°C to 30°C gives:

\[
\Delta T=30-25=+5.0^{\circ}\mathrm C
\]

A temperature decrease from 25°C to 20°C gives:

\[
\Delta T=20-25=-5.0^{\circ}\mathrm C
\]

Temperature differences have the same numerical value in degrees Celsius and kelvin.

02How do you measure the energy accurately?

Imagine trying to heat a liquid in an ordinary glass beaker.

You place an immersion heater inside, record the initial temperature, switch on the heater, and measure the final temperature.

You know how much electrical energy the heater supplied. You also know the liquid’s mass and temperature change.

What should the calculation show?

If every joule supplied by the heater entered the liquid, then:

\[
E_{\mathrm{electrical}}=q_{\mathrm{liquid}}
\]

However, that prediction is not quite correct.

Some energy heats the glass beaker. Some escapes into the surrounding air. The energy supplied by the heater is therefore not necessarily the energy absorbed by the liquid.

Improving the experimental design

A better arrangement uses an insulated calorimeter, often made from polystyrene.

The apparatus contains:

• A measured quantity of liquid.
• An immersion heater connected to a power supply.
• A thermometer to measure temperature.
• A lid to reduce heat exchange with the surroundings.
• A stirrer to distribute energy evenly through the liquid.

The experiment follows a simple sequence:

1. Measure the mass of the liquid.
2. Record its initial temperature.
3. Switch on the heater for a measured time.
4. Record the voltage and current supplied to the heater.
5. Stir the liquid to maintain a uniform temperature.
6. Record its final temperature.

The electrical energy supplied is:

\[
E_{\mathrm{electrical}}=VIt
\]

But we must account for where that energy goes.

If the calorimeter absorbs some energy and some escapes to the surroundings, then:

\[
q_{\mathrm{liquid}}
=
E_{\mathrm{electrical}}
-q_{\mathrm{calorimeter}}
-q_{\mathrm{loss}}
\]

Here, \(q_{\mathrm{loss}}\) is the energy transferred out of the apparatus to the surroundings.

This is an energy balance. It is more accurate than automatically treating the electrical energy reading as the energy absorbed by the liquid.

An insulated calorimeter reduces energy loss, but it does not eliminate it completely.

03Applying specific heat capacity to chemical reactions

In HSC Chemistry, specific heat capacity is particularly useful in calorimetry experiments.

These experiments measure temperature changes during chemical reactions. The measured temperature change allows us to calculate the energy transferred between the reaction and its surroundings.

There is an important distinction between the two calculations:

• When finding specific heat capacity, we calculate the energy absorbed by the sample.
• When finding reaction enthalpy, we calculate the energy transferred by the reaction.

The signs of these energy transfers are opposite.

Understanding the system and surroundings

Before calculating reaction enthalpy, identify two things.

The system is the part of the experiment being investigated. For a chemical reaction, this is the reacting chemical substances.

The surroundings are everything outside the system that can exchange energy with it. In a typical calorimetry experiment, this includes the solution, calorimeter, thermometer, and potentially the surrounding air.

Imagine a reaction taking place inside an insulated cup.

The reaction releases energy, which transfers into the solution. The solution becomes warmer.

The reaction is the system, and the solution is part of the surroundings.

The sign convention for heat energy

In chemistry, we use the convention:

• Positive \(q\): energy enters the system.
• Negative \(q\): energy leaves the system.

For an ideal insulated calorimeter, energy is conserved:

\[
q_{\mathrm{system}}+q_{\mathrm{surroundings}}=0
\]

Therefore:

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

This gives us two important cases.

An exothermic reaction releases energy into the surroundings. The solution warms, so the reaction has a negative \(q\).

An endothermic reaction absorbs energy from the surroundings. The solution cools, so the reaction has a positive \(q\).

The temperature change tells you what happened to the solution. The sign of the reaction energy must then be determined separately.

Calculating energy absorbed by a solution

For a solution in a calorimeter:

\[
q_{\mathrm{solution}}=m_{\mathrm{solution}}c_{\mathrm{solution}}\Delta T
\]

The specific heat capacity of dilute aqueous solutions is often approximated as:

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

This is an assumption, not a universal value for every solution. Use a different value if the question provides one.

If the calorimeter itself absorbs a significant amount of energy, its contribution must also be included.

For a calorimeter with heat capacity \(C_{\mathrm{cal}}\):

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

If heat losses to the room are negligible:

\[
q_{\mathrm{reaction}}
=
-\left(q_{\mathrm{solution}}+q_{\mathrm{cal}}\right)
\]

If the calorimeter’s contribution is negligible, this simplifies to:

\[
\boxed{q_{\mathrm{reaction}}=-mc\Delta T}
\]

This is the relationship commonly used in introductory reaction calorimetry calculations.

04The assumptions that affect your answer

Calorimetry calculations use several simplifying assumptions. You need to recognise them because they affect the accuracy of your result.

Two limitations are particularly important.

Heat loss to the surroundings

Suppose an exothermic reaction releases energy, but some of that energy escapes into the surrounding air.

The solution will experience a smaller temperature rise than it would in a perfectly insulated calorimeter.

If you assume that all the energy stayed in the solution, you will underestimate the magnitude of the reaction’s enthalpy change.

For an exothermic reaction, your calculated \(\Delta H\) will be less negative than the actual value.

Using the wrong amount of reactant

The initial amount of a reactant is not necessarily the amount that reacts.

You must identify the limiting reagent using the balanced chemical equation.

For a reaction:

\[
\ce{A + 2B -> products}
\]

If you start with 0.030 mol of A and 0.040 mol of B, A is not the limiting reagent.

The reaction needs two moles of B for every mole of A. The available 0.040 mol of B can react with only:

\[
n(\mathrm A)=\frac{0.040}{2}
\]

\[
n(\mathrm A)=0.020\ \mathrm{mol}
\]

Therefore, B is limiting, and only 0.020 mol of reaction can occur.

Using the original 0.030 mol of A would overestimate the amount of reaction and produce an incorrect molar enthalpy.

05A quick method for choosing the correct calculation

There are two related calculations, but they answer different questions.

When solving a calorimetry question, follow this order:

1. Identify the system and surroundings.
2. Calculate the temperature change.
3. Determine the mass of the solution.
4. Calculate the energy absorbed or released by the surroundings.
5. Apply the correct sign to the reaction energy.
6. Identify the limiting reagent and calculate the amount of reaction.
7. Divide the reaction energy by the amount reacted to obtain molar enthalpy.

For a deeper explanation of the underlying temperature and energy relationship, see Specific Heat Capacity: Using Q = mcΔT in HSC Physics.

For more practice converting temperature changes into reaction enthalpy, see How to Calculate Enthalpy Change from a Temperature Rise.

06Questions and solutions

Question 1

A 75.0 g sample of liquid absorbs 1.35 kJ of energy. Its temperature increases from 16.0°C to 20.0°C.

Calculate the specific heat capacity of the liquid in J g⁻¹ °C⁻¹.

Solution 1

The specific heat capacity is 4.50 J g⁻¹ °C⁻¹.

Use:

\[
c=\frac{q}{m\Delta T}
\]

First, convert energy into joules:

\[
q=1.35\times1000=1350\ \mathrm J
\]

Calculate the temperature change:

\[
\Delta T=20.0-16.0=4.0^{\circ}\mathrm C
\]

Substitute:

\[
c=\frac{1350}{75.0\times4.0}
\]

\[
\boxed{c=4.50\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}}
\]

This means each gram of the liquid requires 4.50 J to increase its temperature by 1°C.

Question 2

A student uses an electrical heater to determine the specific heat capacity of an unknown liquid.

The following data are recorded:

• Mass of liquid: 100.0 g
• Electrical energy supplied: 3.00 kJ
• Energy absorbed by the calorimeter: 0.250 kJ
• Energy lost to the surrounding air: 0.150 kJ
• Initial temperature: 19.0°C
• Final temperature: 25.0°C

Calculate the specific heat capacity of the liquid.

Explain why using the full electrical energy supplied would produce an incorrect answer.

Solution 2

The specific heat capacity is 4.33 J g⁻¹ °C⁻¹.

The key is to calculate the energy absorbed by the liquid, rather than using the total electrical energy.

Step 1: Calculate the energy absorbed by the liquid.

The total energy supplied is 3.00 kJ.

Some energy heats the calorimeter, and some escapes into the surrounding air.

Therefore:

\[
q_{\mathrm{liquid}}
=3.00-0.250-0.150
\]

\[
q_{\mathrm{liquid}}=2.60\ \mathrm{kJ}
\]

Convert to joules:

\[
q_{\mathrm{liquid}}=2600\ \mathrm J
\]

Step 2: Calculate the temperature change.

\[
\Delta T=25.0-19.0
\]

\[
\Delta T=6.0^{\circ}\mathrm C
\]

Step 3: Calculate specific heat capacity.

\[
c=\frac{q}{m\Delta T}
\]

\[
c=\frac{2600}{100.0\times6.0}
\]

\[
\boxed{c=4.33\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}}
\]

Using the full 3.00 kJ would incorrectly assume that every joule supplied entered the liquid.

The energy balance shows that only 2.60 kJ was absorbed by the liquid. The remaining 0.40 kJ was transferred elsewhere.

Question 3

Zinc reacts with copper(II) sulfate solution according to the equation:

\[
\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}
\]

A student adds 0.650 g of zinc to 25.0 mL of 0.300 mol L⁻¹ copper(II) sulfate solution.

The temperature increases from 20.5°C to 23.7°C.

Assume:

• The solution has a mass of 25.0 g.
• Its specific heat capacity is 4.18 J g⁻¹ °C⁻¹.
• The calorimeter absorbs negligible energy.
• No heat is lost to the surroundings.
• The reaction goes to completion.
• The heat capacity of the added zinc and deposited copper is negligible.

Calculate the enthalpy change per mole of reaction. Identify the limiting reagent.

Solution 3

The enthalpy change is −44.6 kJ mol⁻¹, and copper(II) sulfate is the limiting reagent.

Work through the energy calculation and stoichiometry separately.

Step 1: Calculate the temperature change.

\[
\Delta T=23.7-20.5
\]

\[
\Delta T=3.2^{\circ}\mathrm C
\]

The solution warms, so it absorbs energy. The reaction must release energy.

Therefore:

\[
q_{\mathrm{solution}}>0
\]

\[
q_{\mathrm{reaction}}<0
\]

Step 2: Calculate the energy absorbed by the solution.

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

\[
q_{\mathrm{solution}}
=25.0\times4.18\times3.2
\]

\[
q_{\mathrm{solution}}=334.4\ \mathrm J
\]

Therefore:

\[
q_{\mathrm{reaction}}=-334.4\ \mathrm J
\]

Convert to kilojoules:

\[
q_{\mathrm{reaction}}=-0.3344\ \mathrm{kJ}
\]

Step 3: Calculate the initial amount of zinc.

The molar mass of zinc is 65.4 g mol⁻¹.

\[
n(\mathrm{Zn})=\frac{0.650}{65.4}
\]

\[
n(\mathrm{Zn})=0.00994\ \mathrm{mol}
\]

Step 4: Calculate the initial amount of copper(II) sulfate.

Convert the volume into litres:

\[
V=25.0\ \mathrm{mL}=0.0250\ \mathrm L
\]

Use:

\[
n=CV
\]

\[
n(\ce{CuSO4})=0.300\times0.0250
\]

\[
n(\ce{CuSO4})=0.00750\ \mathrm{mol}
\]

Step 5: Identify the limiting reagent.

The balanced equation has a 1:1 mole ratio between zinc and copper(II) sulfate.

The initial amounts are:

• Zinc: 0.00994 mol
• Copper(II) sulfate: 0.00750 mol

Copper(II) sulfate is present in the smaller stoichiometric amount, so it is the limiting reagent.

Therefore:

\[
n_{\mathrm{reaction}}=0.00750\ \mathrm{mol}
\]

Step 6: Calculate the enthalpy change.

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

\[
\Delta H=\frac{-0.3344}{0.00750}
\]

\[
\boxed{\Delta H=-44.6\ \mathrm{kJ\,mol^{-1}}}
\]

The negative sign confirms that the reaction is exothermic.

A common mistake would be dividing by the initial amount of zinc. However, not all the zinc reacts because copper(II) sulfate is the limiting reagent.

Question 4

An ionic solid dissolves in water in an insulated calorimeter.

During the process:

• The final mass of the solution is 80.0 g.
• The specific heat capacity of the solution is 4.18 J g⁻¹ °C⁻¹.
• The initial temperature is 24.0°C.
• The final temperature is 20.8°C.
• The amount of solid dissolved is 0.0400 mol.

Assume the calorimeter absorbs negligible energy, no energy escapes to the surroundings, and the dissolution is complete.

Calculate:

(a) The energy transferred to or from the solution.

(b) The energy transferred to or from the dissolving process.

(c) The molar enthalpy change of dissolution.

Explain why the sign of the enthalpy change is important.

Solution 4

The molar enthalpy change of dissolution is +26.8 kJ mol⁻¹. The dissolution process is endothermic.

The key observation is that the solution cools. It loses energy to the dissolving process.

Step 1: Calculate the temperature change.

\[
\Delta T=20.8-24.0
\]

\[
\Delta T=-3.2^{\circ}\mathrm C
\]

The negative temperature change indicates that the solution has lost energy.

Step 2: Calculate the energy transferred to or from the solution.

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

\[
q_{\mathrm{solution}}
=80.0\times4.18\times(-3.2)
\]

\[
\boxed{q_{\mathrm{solution}}=-1070\ \mathrm J}
\]

The solution loses approximately 1.07 kJ of energy.

Step 3: Calculate the energy transferred by the dissolution process.

The solution is the surroundings, while the dissolving process is the system.

Therefore:

\[
q_{\mathrm{dissolution}}=-q_{\mathrm{solution}}
\]

\[
q_{\mathrm{dissolution}}=+1070\ \mathrm J
\]

Convert to kilojoules:

\[
q_{\mathrm{dissolution}}=+1.07\ \mathrm{kJ}
\]

Step 4: Calculate the molar enthalpy change.

\[
\Delta H=\frac{q_{\mathrm{dissolution}}}{n}
\]

\[
\Delta H=\frac{1.070}{0.0400}
\]

\[
\boxed{\Delta H=+26.8\ \mathrm{kJ\,mol^{-1}}}
\]

The positive sign means that the dissolution process absorbs energy.

The tempting mistake is to see a temperature decrease and automatically assign a negative enthalpy change. However, the temperature decrease describes the surroundings, not the process itself.

The solution loses energy, but the dissolving process gains it. That is why the signs are opposite.

Question 5

A student wants to determine the specific heat capacity of an aqueous solution using a calorimeter.

Before testing the solution, the student calibrates the calorimeter using 100.0 g of water.

During calibration:

• An electrical heater supplies 2.00 kJ of energy.
• The water and calorimeter temperature increases by 4.00°C.
• The specific heat capacity of water is 4.18 J g⁻¹ °C⁻¹.
• No energy is lost to the surroundings.

The student then uses the same calorimeter to investigate the following reaction:

\[
\ce{A + 2B -> products}
\]

The initial amounts of reactants are:

\[
n(\mathrm A)=0.0300\ \mathrm{mol}
\]

\[
n(\mathrm B)=0.0400\ \mathrm{mol}
\]

The reaction enthalpy is independently known to be:

\[
\Delta H=-60.0\ \mathrm{kJ\,mol^{-1}}
\]

This value is per mole of reaction as written.

During the reaction experiment:

• The final mass of the solution is 80.0 g.
• The temperature increases by 3.00°C.
• The reaction goes to completion.
• The calorimeter has the same heat capacity as during calibration.
• No energy is lost to the surroundings.

Determine the specific heat capacity of the aqueous solution.

In your explanation, address why both of the following approaches are incorrect:

• Using the full reaction energy as the energy absorbed by the solution.
• Using the initial amount of A as the amount of reaction.

Solution 5

The specific heat capacity of the aqueous solution is 3.98 J g⁻¹ °C⁻¹.

This problem requires three connected calculations:

1. Determine the heat capacity of the calorimeter.
2. Calculate the amount of reaction that actually occurs.
3. Separate the energy absorbed by the solution from the energy absorbed by the calorimeter.

Step 1: Calculate the heat capacity of the calorimeter.

During calibration, both the water and calorimeter absorb energy.

The total energy supplied is:

\[
q_{\mathrm{total}}=2000\ \mathrm J
\]

The water absorbs:

\[
q_{\mathrm{water}}=mc\Delta T
\]

\[
q_{\mathrm{water}}
=100.0\times4.18\times4.00
\]

\[
q_{\mathrm{water}}=1672\ \mathrm J
\]

Therefore, the calorimeter absorbs:

\[
q_{\mathrm{cal}}=2000-1672
\]

\[
q_{\mathrm{cal}}=328\ \mathrm J
\]

The heat capacity of the calorimeter is:

\[
C_{\mathrm{cal}}=\frac{q_{\mathrm{cal}}}{\Delta T}
\]

\[
C_{\mathrm{cal}}=\frac{328}{4.00}
\]

\[
\boxed{C_{\mathrm{cal}}=82.0\ \mathrm{J\,^{\circ}C^{-1}}}
\]

This value describes the energy required to increase the temperature of the entire calorimeter assembly by 1°C.

Step 2: Identify the limiting reagent and amount of reaction.

The balanced equation is:

\[
\ce{A + 2B -> products}
\]

Compare the initial amount of each reactant with its stoichiometric coefficient.

For A:

\[
\frac{0.0300}{1}=0.0300
\]

For B:

\[
\frac{0.0400}{2}=0.0200
\]

B has the smaller stoichiometric amount, so B is the limiting reagent.

The amount of reaction is:

\[
\boxed{n_{\mathrm{reaction}}=0.0200\ \mathrm{mol}}
\]

This is the maximum amount of reaction that can occur with the available reactants.

Step 3: Calculate the total energy released by the reaction.

The reaction enthalpy is −60.0 kJ per mole of reaction.

Therefore:

\[
q_{\mathrm{reaction}}=n\Delta H
\]

\[
q_{\mathrm{reaction}}
=0.0200\times(-60.0)
\]

\[
\boxed{q_{\mathrm{reaction}}=-1.20\ \mathrm{kJ}}
\]

The reaction releases 1.20 kJ of energy.

Because the calorimeter is insulated, this energy is absorbed by the surroundings:

\[
q_{\mathrm{surroundings}}=+1200\ \mathrm J
\]

Step 4: Calculate the energy absorbed by the calorimeter.

The temperature increases by 3.00°C.

Use the heat capacity found during calibration:

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

\[
q_{\mathrm{cal}}=82.0\times3.00
\]

\[
\boxed{q_{\mathrm{cal}}=246\ \mathrm J}
\]

Step 5: Calculate the energy absorbed by the solution.

The surroundings consist of the solution and calorimeter.

Therefore:

\[
q_{\mathrm{surroundings}}
=q_{\mathrm{solution}}+q_{\mathrm{cal}}
\]

Rearrange:

\[
q_{\mathrm{solution}}
=q_{\mathrm{surroundings}}-q_{\mathrm{cal}}
\]

Substitute:

\[
q_{\mathrm{solution}}=1200-246
\]

\[
\boxed{q_{\mathrm{solution}}=954\ \mathrm J}
\]

Step 6: Calculate the specific heat capacity of the solution.

Now we have the energy absorbed by the solution, its mass, and its temperature change.

Use:

\[
c=\frac{q_{\mathrm{solution}}}{m\Delta T}
\]

\[
c=\frac{954}{80.0\times3.00}
\]

\[
\boxed{c=3.98\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}}
\]

Why the two tempting approaches fail

The first tempting approach is to use the full 1200 J as the energy absorbed by the solution:

\[
c=\frac{1200}{80.0\times3.00}
\]

This gives 5.00 J g⁻¹ °C⁻¹.

However, 246 J was absorbed by the calorimeter. Only 954 J entered the solution, so the larger result is incorrect.

The second tempting approach is to use 0.0300 mol of A as the amount of reaction.

That would suggest that the reaction released:

\[
q=0.0300\times(-60.0)
\]

\[
q=-1.80\ \mathrm{kJ}
\]

But this amount of reaction would require 0.0600 mol of B, and only 0.0400 mol is available.

The reaction cannot proceed that far. B is limiting, so only 0.0200 mol of reaction occurs.

The important idea is that energy accounting and stoichiometry must agree with the actual physical system. You cannot assign energy to the solution that was absorbed by the calorimeter, or calculate energy for a quantity of reaction that the available reactants cannot support.

Finding specific heat capacity is ultimately an energy-accounting problem. You need the correct mass, temperature change, and energy transferred to the substance itself.

Once you can distinguish the reaction from its surroundings, the same measurements become useful for calculating reaction enthalpy, comparing exothermic and endothermic processes, and evaluating the reliability of calorimetry experiments.