 Heat Capacity Calculations Chemistry Tutorial

Key Concepts

• Specific Heat Capacity of a substance is the amount of heat required to raise the temperature of 1 gram of the substance by 1°C (or by 1 K). 1
• Cg is the symbol used for specific heat capacity.
• The S.I. units2 of specific heat capacity are :

joules per degree celsius per gram, J °C-1 g-1

or

joules per Kelvin per gram, J K-1 g-1

• We can express the relationship between specific heat capacity (Cg), energy in joules (q), mass in grams (m) and temperature change (ΔT) as a mathematical equation:
 Cg = q     m × ΔT = J     g × °C
• This equation can be rearranged to find the amount of heat energy (q) gained or lost by a substance given its specific heat capacity (Cg), mass in grams (g) and the change in temperature (ΔT)

q = m × Cg × ΔT

• Molar Heat Capacity of a substance is the amount of heat required to raise the temperature of 1 mole of the substance by 1°C (or by 1 K).
• Cn is the symbol used for molar heat capacity.
• The units of molar heat capacity are:

joules per degree Celsius per mole, J °C-1 mol-1

or

joules per Kelvin per mole, J K-1 mol-1

• We can express the relationship between molar heat capacity (Cn), energy in joules (q), amount of substance in moles (n) and temperature change (ΔT) as a mathematical equation:
 Cn = q     n × ΔT = J     mol × °C
• This equation can be rearranged to calculate the amount of heat energy gained or lost by a substance given its molar heat capacity (Cn), the amount of the substance in moles (n), and the temperature change (ΔT):

q = n × Cn × ΔT

• Specific heat capacity or molar heat capacity can be used to determine the energy change of a chemical reaction in aqueous solution (the Heat of Reaction or Enthalpy of Reaction) in the school laboratory using a simple calorimeter made up of a polystyrene cup fitted with a lid and a thermometer.
• Specific heat capacity or molar heat capacity can also be used to detemine the energy change of a combustion reaction (see Heat of Combustion of Fuels and Heat of Combustion of Food)

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Specific Heat Capacity

If you heat some water gently using a heat source like a bunsen burner, the temperature of the water increases.
The energy supplied by the bunsen burner causes the water molecules to move faster, increasing their kinetic energy.
We can measure the result of this increased kinetic energy as an increase in temperature.
The amount of energy absorbed by the water molecules to increase their kinetic energy is referred to as the "heat energy".3
Heat energy of the water particles, q, is proportional to temperature change, ΔT.
ΔT = final temperature - initial temperature

q ∝ ΔT

This means that if you use the same mass of water but double the heat energy (q) then the temperature change (ΔT) will also double.
Similarly, if you halve the heat energy (q) then the temperature change (ΔT) will also be halved.

You can could also heat the "cold" water by adding some "hot" water to it.

Imagine you have a beaker of water containing 100 g of water at a temperature of 25.0°C.
What would happen to the temperature of the water if you added 10 g of boiling water (100°C)?
Heat will flow from the hot water to the cold water.4
The kinetic energy of the "hot" water molecules will decrease, and the kinetic energy of the "cold" water molecules will increase, until all the water molecules have the same average kinetic energy.5
Because temperature is a measure of the average kinetic energy of all the water molecules, we find that the temperature of the water will become constant.
In this example, a constant temperature6 of 31.8°C will be achieved.
The temperature change, ΔT is
ΔT = final temperature - initial temperature = 31.8 - 25.0 = 6.8°C

Now, imagine repeating the experiment, but this time using 20 g of boiling water.
What will be the final temperature of the water?
Once again heat will flow from the hot water to the cold water, the hot water cools and the cold water heats up until a constant temperature is achieved everywhere in the volume of water.
But, this time the temperature will be higher, 37.5°C.
The temperature change, ΔT is
ΔT = final temperature - initial temperature = 37.5 - 25.0 = 12.5°C

Adding a greater mass of hot water to the same mass of cool water raises the temperature more.
This tells us that the amount of heat energy that can be transferred from a hot substance to a cold substance is dependent on the mass of the substance used.
The heat energy (q) is proportional to the mass of substance used (m) and to the change in temperature (ΔT):

q ∝ m × ΔT

We could turn this relationship into a mathematical equation by using a constant of proportionality.
Let C be the constant of proportionality, then:

q = C × m × ΔT

Let's see what happens to this constant of proportionality, C, when we change the substance used to heat the water.

What would happen to the temperature of 100 g of water initially at 25.0°C if we added 20 g of a different substance instead of water, say, 20 g of copper metal at 100 °C?
Heat will flow from the hot copper to the cooler water, the copper will cool down and the water will heat up until a constant temperature is achieved.
The final temperature of the water is only 26.5°C, less than the temperature when 20 g of water was added!
The temperature change, ΔT is
ΔT = final temperature - initial temperature = 26.5 - 25.0 = 1.5°C

For equal masses of hot water and hot copper at the same temperature the hot water can transfer more heat energy to the cold water than hot copper metal can.7
That is, the value of the constant of proportionality, C, for water is greater than that for copper.
The term that is used to describe this ability (or capacity) to transfer heat energy is "heat capacity".
When comparisons are made using the mass in grams of substances, this "heat capacity" is referred to as the specific heat capacity.
So, the specific heat capacity of water is greater than the specific heat capacity of copper.
Specific heat capacity has been given the symbol Cg (think "g" for grams, that is, mass).

Now we can replace the constant of proportionality (C) in our mathematical equation above with specific heat capacity (Cg):

q = Cg × m × ΔT

We can rearrange this equation by dividing both sides of the equation by m × ΔT:

 q     m × ΔT = Cg × m × ΔT m × ΔT q     m × ΔT = Cg

Now, if I want to compare the specific heat capacities of various substances I would need to keep the mass constant, say 1 gram, and I would use just enough heat energy to produce a temperature change of 1°C (or 1K),
Substituting these values into the equation:

 q     1 × 1 = Cg q = Cg

That is, the specific heat capacity of a substance is the energy (q) required to raise the temperature of 1 gram of the substance by 1°C (or 1K)!

Different substances have different specific heat capacities. The specific heat capacity of some substances is given in the table below:8

Specific Heat Capacities of Some Substances
ElementsCg
(J K-1 g-1
or J °C-1 g-1)
CompoundsCg
(J K-1 g-1
or J °C-1 g-1)
aluminiumCg = 0.90 water (liquid)Cg = 4.18
carbonCg = 0.72 ethanol (liquid)Cg = 2.44
copperCg = 0.39 sulfuric acid (liquid)Cg = 1.42
leadCg = 0.13 sodium chloride (solid)Cg = 0.85
mercury (liquid)Cg = 0.14 potassium hydroxide (solid)Cg = 1.18

From the table above we see that the specific heat capacity of copper is 0.39 J °C-1 g-1 while the specific heat capacity of water is much higher, 4.18 J °C-1 g-1.
It requires 0.39 J of energy to change the temperature of 1 gram of copper metal by 1°C (or 1 K).
It requires 4.18 J of energy to change the temperature of 1 gram of liquid water by 1°C (or 1 K).

Specific heat capacity, Cg, as described above is useful because we can easily measure the mass of many substances.
However, when we look at the table of values some of these values seem counter-intuitive.
Why should it require 0.13 J of energy to raise the temperature of 1 g of lead 1°C, but almost 7 times as much energy to raise the temperature of 1 g of aluminium by 1°C?
And why would carbon have a higher heat capacity than metallic copper or lead?

Perhaps comparisons based on mass are not the best option available.....

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Molar Heat Capacity

Equal masses of different substances contain different numbers of "particles" (atoms, ions, or molecules).
Chemists use the "mole" as a measure of the "amount" of substance because a mole of a pure substance always contains the same number of particles (Avogadro's number, NA = 6.02 × 1023).

The mass of 1 mole of a pure substance is equal to its relative molecular mass expressed in grams:

mass of 1 mole = relative molecular mass in grams

Recall that specific heat capacity is the energy required to raise the temperature of 1 gram of substance by 1°C (or 1 K).

example: Cg for metallic copper, Cu(s), is 0.39 J °C-1 g-1

If we want to find the heat capacity of 1 mole of substance, we need to multiply the specific heat capacity, Cg, by the relative molecular mass (Mr) or molar mass (M) of the substance:

heat capacity of 1 mole = Mr × C(g)
or
heat capacity of 1 mole = M × C(g)

The quantity "M × Cg" is referred to as the molar heat capacity and is given the symbol Cn (n is the symbol used for moles).

The molar heat capacity of a substance is the energy required to raise the temperature of 1 mole of substance by 1°C (or 1K).

For example, the specific heat capacity of copper metal: Cg = 0.39 J°C-1 g-1
The relative atomic mass of copper from the Periodic Table: Mr = 63.55
Molar heat capacity of copper metal = Cg × Mr = 0.39 × 63.55 = 24.8 J °C-1 mol-1

You could perform this calculation yourself for each of the substances listed in the table of specific heat capacities above.
You can check your calculations against those listed in the table of molar heat capacities given below:

Molar Heat Capacities of Some Substances
ElementsCn
(J K-1 mol-1
or J °C-1 mol-1)
CompoundsCn
(J K-1 mol-1
or J °C-1 mol-1)
mercuryCn = 28.1 sulfuric acid (liquid)Cn = 139
leadCn = 27.0 waterCn = 75
copperCn = 24.8 potassium hydroxide (solid)Cn = 66
aluminiumCn = 24.3 sodium chloride (solid)Cn = 50
carbonCn = 8.6 ethanol (ethyl alcohol)Cn = 22

This table allows us to compare the heat capacities of the same number of particles, that is, 1 mole, of different substances.
We find that the molar heat capacities of the metals is quite similar while the molar heat capacity of carbon is much lower.
It requires about 25 J of energy to raise the temperature of 1 mole of metal by 1°C (or 1 K), but it only needs about 9 J of energy to raise the temperature of 1 mole of carbon by 1°C (or 1 K).

We could write a new equation for calculating the amount of heat required (q) to raise the temperature (ΔT) of an amount of substance in moles (n):

q = Cn × n × ΔT

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Heat Capacity Calculations Worked Examples

Question 1: Calculate the quantity of heat in joules needed to increase the temperature of 250 g of water from 20°C to 56°C.
Specific heat capacity of water is 4.18 J°C-1g-1.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate the quantity of heat in joules
q = ? J

2. What data (information) have you been given in the question?

Extract the data from the question:

m = mass of water = 250 g

Ti = initial temperature = 20°C

Tf = final temperature = 56°C

Cg = specific heat capacity of water = 4.18 J°C-1g-1

3. What is the relationship between what you know and what you need to find out?
Write the equation to find heat energy:

q = m × Cg × ΔT

(a) Calculate the change in temperature, ΔT:

ΔT = Tf - Ti

(b) Substitute the values for m, Cg and ΔT into the equation and solve for q

4. Perform the calculations
(a) Calculate the change in temperature, ΔT:

ΔT = Tf - Ti
ΔT = 56 - 20 = 36°C

(b) Substitute the values for m, Cg and ΔT into the equation and solve for q

q = m × Cg × ΔT
q = 250 × 4.18 × 36 = 37 620 J

Think about what the term "specific heat capacity" means. It is the energy required to raise the temperature of 1 g of substance (water) by 1°C.
Therefore, since we have 250 g, we will need 250 times the "specific heat capacity of water (4.18)":
that is, we need 250 × 4.18 = 1045 J
But, we also raised the temperature by 36°C not 1°C, so we will need 36 times this amount of energy:
that is, we need 36 × 1045 = 37 620 J

Since this answer, 37 620 J, is the same as our calculation above, we are reasonably confident that our answer is plausible.

6. State your solution to the problem "quantity of heat in joules":

q = 37 620 J

Question 2: Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15 g of copper from 25°C to 60°C.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate the specific heat capacity of copper
Cg(Cu) = ? J°C-1g-1

2. What data (information) have you been given in the question?

Extract the data from the question:

q = heat energy = 204.75 J

m = mass of copper = 15 g

Ti = initial temperature = 25°C

Tf = final temperature = 60°C

3. What is the relationship between what you know and what you need to find out?
Write the equation to find heat energy:

q = m × Cg × ΔT

(a) Calculate the change in temperature, ΔT:

ΔT = Tf - Ti

(b) Substitute the values for q, m and ΔT into the equation and solve for Cg

4. Perform the calculations
(a) Calculate the change in temperature, ΔT:

ΔT = Tf - Ti
ΔT = 60 - 25 = 35°C

(b) Substitute the values for q, m and ΔT into the equation and solve for Cg

 q = m × Cg × ΔT 204.75 = 15 × Cg × 35 204.75 = Cg × 525 204.75 525 = Cg × 525 525 0.39 = Cg

Specific heat capacity is the energy required to raise the temperature of 1 g of substance (copper) by 1°C
204.75 J of heat energy was used to raise the temperature of 15 g of copper, so this value of heat energy is 15 times too high, so we need to divide the heat energy by 15:
204.75 J ÷ 15 g = 13.65 J g-1
Note that the temperature increase was not 1°C, it was 35°C, that is, the value of heat energy will be 35 times too high so we need to divide it by 35:
13.65 J g-1 ÷ 35°C = 0.39 J g-1 °C-1
That is, 0.39 J of heat energy is required to raise the temperature of 1 g of copper by 1°C.
The heat energy required to raise the temperature of 1 g of substance by 1°C is known as its specific heat capacity.
Therefore the specific heat capacity of copper is 0.39 J°C-1g-1

Since this result is the same as the one we calculated above we are reasonably confident that our answer is plausible.

6. State your solution to the problem " specific heat capacity of copper":

Cg(Cu) = 0.39 J°C-1g-1

Question 3: 216 J of energy is required to raise the temperature of aluminium from 15°C to 35°C.
Calculate the mass in grams of aluminium.
(Specific Heat Capacity of aluminium is 0.90 J°C-1g-1).

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate the mass in grams of aluminium
m(Al) = ? g

2. What data (information) have you been given in the question?

Extract the data from the question:

q = heat energy = 216 J

Ti = initial temperature = 15°C

Tf = final temperature = 35°C

Cg(Al) = 0.90 J°C-1g-1

3. What is the relationship between what you know and what you need to find out?
Write the equation to find heat energy:

q = m × Cg × ΔT

(a) Calculate the change in temperature, ΔT:

ΔT = Tf - Ti

(b) Substitute the values for q, Cg and ΔT into the equation and solve for m

4. Perform the calculations
(a) Calculate the change in temperature, ΔT:

ΔT = Tf - Ti
ΔT = 35 - 15 = 20°C

(b) Substitute the values for q, Cg and ΔT into the equation and solve for m

 q = m × Cg × ΔT 216 = m × 0.90 × 20 216 = m × 18 216 18 = m × 18 18 12 = m

Specific heat capacity is the energy required to raise the temperature of 1 g of substance (Al) by 1°C.
The specific heat capacity of aluminium is given as 0.90 J°C-1g-1.
The temperature change is not 1°C, it is 20°C, so we need to multiply the specific heat capacity by 20 in order to determine the amount of energy that would be required to raise the temperature of 1g of Al by 20°C:
heat energy = 20 °C × 0.90 J °C-1g-1 = 18 J g-1
We are told that the heat energy required was 216 J which is greater than 18 J so the mass of Al used must have been greater than 1 g.
Divide the heat energy given (216 J) by 18 J g-1 to determine the mass of Al used:
216 J ÷ 18 J g-1 = 12 g

Since this result is the same as the one we calculated above we are reasonably confident that our answer is plausible.

6. State your solution to the problem "mass in grams of aluminium":

m(Al) = 12 g

Question 4: The initial temperature of 150 g of ethanol was 22.0°C.
What will be the final temperature in degrees celsius of the ethanol if 3240 J was needed to raise the temperature of the ethanol?
(Specific heat capacity of ethanol is 2.44 J°C-1g-1)

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate the final temperature of the ethanol
Tf = ? °C

2. What data (information) have you been given in the question?

Extract the data from the question:

m = mass of ethanol = 150 g

Ti = initial temperature = 22.0°C

q = heat energy needed to raise temperature = 3240 J

Cg = heat capacity of ethanol = 2.44 J°C-1g-1

3. What is the relationship between what you know and what you need to find out?
Write the equation to find heat energy:

q = m × Cg × ΔT

(a) Substitute the values for q, m and Cg into the equation and solve for ΔT

(b) Calculate the final temperature of ethanol:

ΔT = Tf - Ti
Tf = ΔT + Ti

4. Perform the calculations
(a) Substitute the values for q, m and Cg into the equation and solve for ΔT

 q = m × Cg × ΔT 3240 = 150 × 2.44 × ΔT 3240 = 366 × ΔT 3240 366 = 366 × ΔT 366 8.85 = ΔT

(b) Calculate the final temperature of ethanol:

Tf = ΔT + Ti
Tf = 22.0 + 8.85 = 30.85°C = 30.9°C

Sepcific heat capacity is the energy required to raise the temperature of 1 g of substance (ethanol) by 1°C.
2.44 J of heat energy is needed to raise the temperature of 1 g of ethanol by 1°C.
150 g of ethanol was used, so the heat energy required will be 150 times the specific heat capacity of ethanol:
150 g × 2.44 J°C-1g-1 = 366 J°C-1
The heat energy actually used was greater than 366 J, it was 3240 J, that is, the heat energy used was 3240 ÷ 366 = 8.85 times greater than that required to raise the temperature by 1°C, therefore the temperature change must be 1 × 8.85 = 8.85°C.
The initial temperature was 22.0°C, so the final temperature must be 8.85°C higher than this, that is, the final temperature must be 22.0 + 8.85 = 30.85 ≈ 30.9°C.

Since this result is the same as the one we calculated above we are reasonably confident that our answer is plausible.

6. State your solution to the problem "final temperature of the ethanol":

Tf = 30.9°C

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Footnotes

1. Since the graduations in the scale of both the celsius and Kelvin temperature scales are the same, and because we are not concerned here with either the initial or final temperature but only in the difference between the two, it can been seen that a difference of 1°C is the same as a difference of 1 K.
Careful experiments show that the specific heat of a substance is itself a function of temperature, so, in the nineteenth century a standard was set, that is, heat capacity is the heat required to raise the temperature of 1 g of water from 14.5°C to 15.5°C.

2. In 1960 the General Conference of Weights and Measures agreed upon a unified version of the metric system. The units in this system are known as SI units (Systèm International d'Unités). Seven base units constitute the foundation of the SI system:

Physical quantityName of UnitSymbol
masskilogramkg
lengthmeterm
timeseconds
electric currentampereA
temperaturekelvinK
luminous intensitycandelacd
quantity of substancemolemol

Derived units are based on the above SI units.
The unit of force is the newtown (N), it is a derived unit, 1 N = 1 kg m s-2
The unit for energy is also a derived unit, the joule (J), 1 J = 1 N m = 1 kg m2 s-2
Electrical measurements are capable of greater precision than calorimetric measurements as described in this discussion, so the joule can also be defined as a volt coulomb.

3. Heat, or heat energy, is the energy directly transferred from one object to another.
Heat is energy in transit, a substance like water at a constant temperature does not have a "heat content", but it does have an "energy content".
The energy content of a substance is made up of the kinetic energy (movement) of its particles and potential energy such as the stored chemical potential energy in its chemical bonds.
Temperature is a measure of the average kinetic energy of the particles.

4. Heat always flows from "hot" to "cold".
In 1803, and 4 years after his death, Joseph's Black work on calorimetry (the measurement of heat changes) was published. In it he showed that equality of temperature did not imply that there was also an "equality of heat" in different substances. He investigated the capacity for heat or the amount of heat needed to increase the temperature of different bodies by a given number of degrees. In explaining his experiments, he treated heat like a substance which could flow from one body to another.

5. The particles will not have exactly the same kinetic energy. There is a distribution of kinetic energies for the particles, so we refer to the "average" kinetic energy of the particles in the system.

6. This is known as thermal equilibrium.

7. It is more accurate to say that the heat capacity is the ability of a substance to transfer heat to another substance since heat is energy in transit.
That is, heat capacity is the capacity or ability of a substance to transfer heat to another substance.
But, since the word capacity is usually thought of as "containment", eg, a 250 mL volumetric flask has a capacity of 250 mL, we often think of the heat capacity of a substance being its ability to contain heat energy.
We cannot really refer to "heat" as being stored, that is, heat can be absorbed by the molecules to increase their kinetic energy but it is not "stored" because it has done work to speed up the particles. Heat energy could be "stored" as potential energy in chemical bonds if a chemical reaction takes place, but this is not the case in these examples.

8. Values of specific heat capacities refer to conditions of constant atmospheric pressure.