Physics · 2 Thermal physics

2.2.2 Specific heat capacity

Before you start

The particle model: particles in a solid vibrate about fixed positions, and particles in a liquid or a gas move about from place to place (2.1).
Kinetic energy is the energy an object has because it moves, and energy can be held in an internal (thermal) store (1.7.1).
Power is energy transferred per unit time, so the energy from a heater is $Δ E = P t$ , with $P$ in W and $t$ in s (1.7.4).
Converting grams to kilograms, kJ to J, and minutes to seconds.
Rearranging an equation of the form $a = \frac{b}{c d}$ to make $b$ , $c$ or $d$ the subject.

Syllabus objectives covered (4)

Know that a rise in the temperature of an object increases its internal energy
Describe an increase in temperature of an object in terms of an increase in the average kinetic energies of all of the particles in the object
Define specific heat capacity as the energy required per unit mass per unit temperature increase; recall and use the equation c = ΔE / (mΔθ)
Describe experiments to measure the specific heat capacity of a solid and a liquid

Part 1 of 8

Heating an object raises its internal energy

Every object is made of particles, and those particles store energy. The total of that energy is the object's internal energy. It is the kinetic energy the particles have because they move, plus the potential energy stored in the forces between them.
In the language of energy stores, internal energy is the internal (thermal) store. When you heat an object, energy is transferred into that store.
The fact this objective asks you to know: when the temperature of an object rises, its internal energy increases. A pan of water at $60$ °C has more internal energy than the same pan of water at $20$ °C.
It works the other way too. When an object cools, its temperature falls and its internal energy decreases, because energy is transferred out of it to the surroundings.
Do not turn the fact round. A rise in temperature always means more internal energy, but more internal energy does not always mean a rise in temperature. When a solid melts, its internal energy increases while its temperature stays the same. That is the next subtopic.

Part 2 of 8

A higher temperature means faster particles, on average

Now look inside the object. In every state of matter the particles are moving, so every particle has kinetic energy. In a solid they vibrate about fixed positions. In a liquid or a gas they move from place to place.
When you heat a solid, its particles vibrate faster. When you heat a liquid or a gas, its particles move about faster. Faster particles have more kinetic energy.
The particles in an object do not all have the same speed. At any moment some are fast and some are slow, and they keep passing energy to one another. So the quantity that describes the whole object is the average kinetic energy of all its particles.
That gives the description this objective asks for: an increase in the temperature of an object is an increase in the average kinetic energy of all of the particles in the object.
Two words carry that sentence. 'Average', because single particles speed up and slow down all the time. 'Kinetic', because it is the energy of the particles' motion that rises. 'The particles get hotter' does not describe the particles: one particle on its own does not have a temperature.
So temperature and internal energy are different quantities. Temperature goes with the average kinetic energy per particle. Internal energy is the total for all the particles, so it also depends on how many particles there are. $2$ kg of water at $50$ °C has twice as many particles as $1$ kg at $50$ °C, with the same average kinetic energy, so it has twice the internal energy.
For the same reason, warming $2$ kg of water by $10$ °C needs twice the energy that warming $1$ kg by $10$ °C needs. That is why the mass appears in the equation in the next section.

Part 3 of 8

Specific heat capacity: energy per kilogram per degree

Give the same energy to $1$ kg of water and to $1$ kg of aluminium. The temperature of the aluminium rises more than four times as much. Different materials need different amounts of energy to warm up by the same amount.
Specific heat capacity, $c$ , measures this. Learn the definition in the syllabus words: specific heat capacity is the energy required per unit mass per unit temperature increase.
'Per unit mass' means for each kilogram. 'Per unit temperature increase' means for each degree Celsius. So $c$ is the energy needed to raise the temperature of $1$ kg of the substance by $1$ °C.
The unit comes from the definition: joules, divided by kilograms, divided by degrees Celsius. It is written J/(kg °C). You may also see J/(kg K). A change of $1$ K is the same size as a change of $1$ °C, so the two units are equal.
Read a value as a sentence. The specific heat capacity of water is about $4200$ J/(kg °C): it takes $4200$ J to raise the temperature of $1$ kg of water by $1$ °C. For aluminium it is about $900$ J/(kg °C), so $900$ J does the same job for $1$ kg of aluminium.
A large $c$ means a substance needs a lot of energy for each degree. With the same mass and the same rate of heating, it warms up more slowly. As it cools, it also gives out more energy for each degree it falls.
The syllabus does not ask you to recall a value of $c$ for any material. A question that needs one gives it to you.

Part 4 of 8

The equation, and how to rearrange it

The definition turns straight into the equation. Divide the energy by the mass, and then by the temperature change.
$c = \frac{Δ E}{m Δ θ}$
$Δ E$ is the energy transferred, in J. $m$ is the mass, in kg. $Δ θ$ is the change in temperature, in °C. $θ$ (theta) is the symbol for temperature, and $Δ$ means 'change in'.
On Paper 4 nothing gives you this equation, so you must recall it. On the practical papers, 5 and 6, a formula you need is given in the question.
To find the energy, multiply both sides by $m Δ θ$ . You multiply because $m Δ θ$ is dividing $Δ E$ , and doing the same to both sides keeps the equation balanced.
$Δ E = m c Δ θ$
To find the temperature change, start from $Δ E = m c Δ θ$ and divide both sides by $m c$ , because $m c$ is multiplying $Δ θ$ . To find the mass, divide both sides by $c Δ θ$ instead: $m = Δ E / (c Δ θ)$ .
$Δ θ = \frac{Δ E}{m c}$
The trap here: $Δ θ$ is a change, final temperature minus starting temperature. Water heated from $20$ °C to $80$ °C has $Δ θ = 60$ °C. Putting $80$ into the equation gives the wrong answer.
If a question asks for the final temperature, find $Δ θ$ first, then add it to the starting temperature.
The same equation gives the energy an object releases as it cools. Then $Δ θ$ is the size of the fall in temperature.
Put the mass in kg and the energy in J before you substitute. $250$ g is $0.250$ kg, and $12$ kJ is $12 000$ J. A mass left in grams makes the answer wrong by a factor of $1000$ .

Part 5 of 8

When the energy comes from a heater

A question may give you the power of a heater and a time, and no energy. Then the calculation has two steps: find $Δ E$ first, then use it in the specific heat capacity equation.
The energy comes from your work on power: $Δ E = P t$ . A $40$ W heater supplies $40$ J every second, so in $300$ s it supplies $40 \times 300 = 12 000$ J.
Written as one line, the two steps combine like this.
$c = \frac{P t}{m Δ θ}$
Put the time in seconds before you multiply. A watt is a joule per second, so a time left in minutes gives an energy $60$ times too small.
A question may tell you to assume that all the energy from the heater goes into the object. That assumption is what lets you use the heater's energy as $Δ E$ . In a real kettle or block, some energy also goes into the container and the surroundings, so the real temperature rise is smaller.
Write the equation in symbols, substitute, then calculate. Under this marking system a C mark can be scored for correct working that is seen, even when the final answer is wrong.

Part 6 of 8

Measuring c for a solid

To measure $c$ you need three things: the mass of the object, its temperature rise, and the energy you gave it. Every step of the method either measures one of those, or helps make sure the energy you measure really goes into the object.
Use a metal block, for example of mass $1$ kg, with two holes drilled in it. An electric heater goes in one hole and a thermometer in the other.
apparatus diagram — not drawn yet
A cylindrical metal block wrapped in insulating lagging and standing on an insulating mat, with two vertical holes in its top: an electric heater in one hole, connected by leads through a joulemeter to a low-voltage power supply, and a thermometer in the other hole with a little oil around its bulb.
Energy in from the joulemeter, temperature rise from the thermometer, mass from a balance: those three measurements give c.
Measure the mass of the block on a balance, and convert it to kg.
Put a few drops of oil in the thermometer hole. The oil fills the air gap around the bulb, so the thermometer reads the temperature of the metal and not of the air in the hole.
Wrap the block in insulation (lagging) and stand it on an insulating mat. This reduces the energy transferred to the surroundings.
Record the starting temperature. Connect the heater to the power supply through a joulemeter, which reads the energy supplied in joules. Without a joulemeter, use a heater of known power and time it with a stopwatch ( $Δ E = P t$ ), or measure the current and the voltage with an ammeter and a voltmeter ( $Δ E = I V t$ ).
Switch on for a measured time, for example $10$ minutes, then switch off. Keep watching the thermometer. The heater is still hotter than the block, so the temperature goes on rising for a short time. Record the highest temperature reached.
Work out $Δ θ$ as the highest temperature minus the starting temperature, and $Δ E$ as the final joulemeter reading minus the starting reading. Then calculate $c = Δ E / (m Δ θ)$ .

Part 7 of 8

Measuring c for a liquid

The method for a liquid uses the same three measurements. What changes is how you find the mass, and how you make the liquid all one temperature.
Find the mass by difference. Weigh the empty beaker, then the beaker with the liquid in it, and subtract. The mass of the beaker is not part of $m$ .
Lower an immersion heater into the liquid so that its heating element is completely covered. If part of it is in the air, some of its energy goes into the air and not into the liquid.
apparatus diagram — not drawn yet
A beaker of liquid standing inside an insulating jacket and covered by a lid; an immersion heater passes through the lid with its element fully under the liquid and is connected through a joulemeter to a low-voltage power supply; a thermometer and a stirrer also pass through the lid, with the thermometer bulb in the liquid away from the heater.
Element fully covered, a stirrer to spread the energy, and a lid and insulation to cut losses.
Stir the liquid gently while it is heated. The liquid next to the heater is hotter than the rest. Stirring spreads the energy, so the thermometer reads the temperature of all the liquid.
Stand the beaker in an insulating jacket, or use a polystyrene cup, and cover it with a lid. The lid reduces the energy lost from the surface by evaporation and convection.
Then record the starting temperature, the energy supplied and the highest temperature, and calculate $c$ exactly as for the solid.

Part 8 of 8

Why your measured value comes out too large

In a real experiment, not all the energy the heater supplies stays in the block or the liquid. Some is transferred to the surroundings, and some warms the heater, the thermometer and the container.
So the temperature rise you measure is smaller than it would be if all the measured energy stayed in the sample.
Look at the equation $c = Δ E / (m Δ θ)$ . $Δ E$ is the full energy from the joulemeter, but $Δ θ$ is smaller than it should be. Dividing by a number that is too small gives a $c$ that is too large.
Get the direction right: energy losses make the measured specific heat capacity larger than the true value, not smaller.
Each precaution in the method works against this. Insulation and a lid reduce losses to the surroundings. Oil in the hole, or stirring, makes the thermometer read the true temperature of the sample. Reading the highest temperature after switching off catches the energy still flowing out of the heater.

Worked examples

Example 1. Calculate the energy needed to raise the temperature of $0.50$ kg of water from $20$ °C to $80$ °C. The specific heat capacity of water is $4200$ J/(kg °C).

Show the working

You want an energy, so use the equation in the form $Δ E = m c Δ θ$ .
Find the temperature change first. It is the rise, not the final temperature.
$Δ θ = 80 - 20 = 60^{\circ} C$
Substitute. The mass is already in kg.
$Δ E = m c Δ θ = 0.50 \times 4200 \times 60 = 126 000 J$
Sense check: $1$ kg needs $4200$ J for each degree, so $0.50$ kg needs $2100$ J for each degree, and $60$ degrees needs $60 \times 2100 = 126 000$ J.

Answer: $126 000$ J ( $126$ kJ, or $1.3 \times 10^{5}$ J to 2 significant figures).

Example 2. A student heats an aluminium block of mass $1.00$ kg with a $40$ W electric heater for $6.0$ minutes. The temperature of the block rises from $18.0$ °C to $33.0$ °C. Calculate the specific heat capacity of aluminium given by this experiment.

Show the working

The energy is not given, so there are two steps: the energy from the heater first, then $c$ .
Convert the time to seconds, because a watt is a joule per second.
$t = 6.0 \times 60 = 360 s$
Now the energy supplied.
$Δ E = P t = 40 \times 360 = 14 400 J$
The temperature change.
$Δ θ = 33.0 - 18.0 = 15.0^{\circ} C$
Substitute into the equation for $c$ .
$c = \frac{Δ E}{m Δ θ} = \frac{14 400}{1.00 \times 15.0} = 960 J/(kg^{\circ} C)$
The accepted value for aluminium is about $900$ J/(kg °C). The experiment gives a larger value, which is the direction energy losses push it: some of the $14 400$ J went into the surroundings, so the measured temperature rise was smaller than it would otherwise have been.

Answer: $960$ J/(kg °C).

Example 3. An electric kettle of power $2.0$ kW contains $1.5$ kg of water at $15$ °C. The kettle is switched on for $2.0$ minutes. Assume that all the energy supplied goes into the water. Calculate the final temperature of the water. The specific heat capacity of water is $4200$ J/(kg °C).

Show the working

Plan: the energy from $Δ E = P t$ , then $Δ θ$ from the rearranged equation, then add $Δ θ$ to the starting temperature.
Power in W and time in s: $2.0$ kW $= 2000$ W and $2.0$ min $= 120$ s.
$Δ E = P t = 2000 \times 120 = 240 000 J$
Rearrange $Δ E = m c Δ θ$ by dividing both sides by $m c$ .
$Δ θ = \frac{Δ E}{m c} = \frac{240 000}{1.5 \times 4200} = 38.1^{\circ} C$
That is the rise. The question asks for the final temperature, so add the starting temperature.
$θ_{final} = 15 + 38.1 = 53.1^{\circ} C$
Sense check: the answer is below $100$ °C, so the water has not reached its boiling point and the equation applies for the whole time.

Answer: $53$ °C (to 2 significant figures).

Example 4. A student measures the specific heat capacity of water. Her results: mass of empty beaker $= 125$ g; mass of beaker and water $= 425$ g; joulemeter reading at the start $= 1200$ J; joulemeter reading at the end $= 29 400$ J; starting temperature $= 19.5$ °C; highest temperature $= 41.5$ °C. Calculate the specific heat capacity of water from her results, and comment on the value. The accepted value is about $4200$ J/(kg °C).

Show the working

Find the mass of the water by difference, then convert it to kg.
$m = 425 - 125 = 300 g = 0.300 kg$
The energy supplied is the change in the joulemeter reading, not the final reading.
$Δ E = 29 400 - 1200 = 28 200 J$
The temperature change.
$Δ θ = 41.5 - 19.5 = 22.0^{\circ} C$
Substitute, and round to 3 significant figures to match the data.
$c = \frac{28 200}{0.300 \times 22.0} = 4272.7 \dots \approx 4270 J/(kg^{\circ} C)$
Comment: her value is larger than the accepted value. That is what energy losses cause. Some of the energy warmed the beaker and the surroundings, so the temperature rise was smaller than it would have been if all $28 200$ J had stayed in the water.

Answer: $4270$ J/(kg °C) to 3 significant figures. It is larger than the accepted value because some of the energy supplied went into the beaker and the surroundings, not the water.

Where people go wrong

Mistake: Using the final temperature as $Δ θ$ , for example putting $80$ into the equation for water heated from $20$ °C to $80$ °C.

Instead: $Δ θ$ is the change: final temperature minus starting temperature, here $60$ °C.

Specific heat capacity is energy per unit temperature increase, so it is the increase that goes into the equation.

Mistake: Leaving the mass in grams.

Instead: Convert to kilograms first: divide by $1000$ .

The unit J/(kg °C) is energy per kilogram. A mass in grams makes the answer wrong by a factor of $1000$ .

Mistake: Treating temperature and internal energy as the same thing.

Instead: Temperature goes with the average kinetic energy of the particles. Internal energy is the total energy of all the particles.

A larger mass of the same substance at the same temperature has more particles with the same average kinetic energy, so it has more internal energy at the same temperature.

Mistake: Writing that heating makes 'the particles get hotter', or that 'every particle moves faster'.

Instead: Write that the average kinetic energy of all the particles increases.

One particle does not have a temperature, and the particles have a range of speeds that keep changing. The objective describes the average over all the particles.

Mistake: Saying that energy losses make the measured specific heat capacity too small.

Instead: Energy losses make it too large.

The joulemeter counts all the energy supplied, but the temperature rise is smaller than it would be with no losses. Dividing $Δ E$ by a $Δ θ$ that is too small gives a $c$ that is too large.

Check yourself

The temperature of a metal spoon rises from $20$ °C to $70$ °C. State what happens to its internal energy.
Answer
Its internal energy increases.
Describe what happens to the particles of a gas when its temperature increases.
Answer
The particles move faster on average: the average kinetic energy of all the particles increases.
Define specific heat capacity.
Answer
The energy required per unit mass per unit temperature increase. In other words, the energy needed to raise the temperature of $1$ kg of the substance by $1$ °C.
Calculate the energy needed to raise the temperature of $2.0$ kg of aluminium by $15$ °C. The specific heat capacity of aluminium is $900$ J/(kg °C).
Answer
$Δ E = m c Δ θ = 2.0 \times 900 \times 15 = 27 000$ J.
A copper block of mass $0.40$ kg absorbs $3120$ J of energy. The specific heat capacity of copper is $390$ J/(kg °C). Calculate the rise in its temperature.
Answer
$Δ θ = Δ E / (m c) = 3120 / (0.40 \times 390) = 3120 / 156 = 20$ °C.
The same energy is given to $1$ kg of water ( $c = 4200$ J/(kg °C)) and to $1$ kg of aluminium ( $c = 900$ J/(kg °C)). Deduce which has the greater temperature rise.
Answer
The aluminium. $Δ θ = Δ E / (m c)$ , so with the same $Δ E$ and the same $m$ , the smaller $c$ gives the larger rise.
In an experiment to measure the specific heat capacity of a liquid, explain why the liquid is stirred.
Answer
The liquid next to the heater is hotter than the rest. Stirring spreads the energy, so the thermometer reads the temperature of all the liquid.
In the experiment with a metal block, explain why a few drops of oil are put in the thermometer hole.
Answer
The oil fills the air gap around the thermometer bulb, so the thermometer reads the temperature of the metal and not of the air in the hole.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A student measures the specific heat capacity of a metal. She has a metal block with two holes drilled in it, an electric heater of known power, a thermometer, a stopwatch, a balance and some insulation.

(a) Define specific heat capacity. 1 mark No calculator

Mark scheme

B1 energy required per unit mass per unit temperature increase, owtte; accept energy needed to raise the temperature of 1 kg by 1 °C; energy to heat 1 kg with no mention of temperature change is insufficient

(b) Describe how she uses this apparatus to determine the specific heat capacity of the metal. Include the measurements she takes. 4 marks No calculator

Mark scheme

B1 measures the mass of the block with the balance
B1 heater in one hole and thermometer in the other; measures the starting temperature and the highest (final) temperature
B1 switches on the heater for a measured time using the stopwatch; energy supplied = power × time
B1 c = ΔE / (mΔθ) with Δθ = final temperature − starting temperature

(c) The heater has a power of $50$ W and is switched on for $8.0$ minutes. The mass of the block is $1.2$ kg. The temperature of the block rises from $21.0$ °C to $42.5$ °C. Calculate the specific heat capacity of the metal. 3 marks

Mark scheme

C1 ΔE = Pt = 50 × 480 = 24 000 J
C1 c = ΔE / (mΔθ) or 24 000 / (1.2 × 21.5) seen; ecf for ΔE
A1 930 J/(kg °C) (accept 930.2 J/(kg °C), or the same value in J/(kg K)); unit required

(d) The accepted value of the specific heat capacity of this metal is $900$ J/(kg °C). Explain why the student's value is larger than this. 2 marks No calculator

Mark scheme

B1 some energy is transferred to the surroundings, or heats the heater or the thermometer
B1 so the temperature rise is smaller than it would be for the energy supplied, which makes ΔE / (mΔθ) larger, owtte

Question 2. An electric kettle contains $0.80$ kg of water at $20$ °C. The specific heat capacity of water is $4200$ J/(kg °C).

(a) State what happens to the internal energy of the water as its temperature rises. 1 mark No calculator

Mark scheme

B1 it increases

(b) Describe how the motion of the water particles changes as the temperature of the water rises. 2 marks No calculator

Mark scheme

B1 the particles move faster or have more kinetic energy
B1 reference to the average: the average kinetic energy of all the particles increases, or the particles move faster on average, owtte; 'the particles get hotter' is insufficient

(c) Calculate the energy needed to raise the temperature of the water to $100$ °C. 2 marks

Mark scheme

C1 ΔE = mcΔθ or 0.80 × 4200 × 80 seen
A1 270 000 J or 2.7 × 10⁵ J (accept 268 800 J); unit required

(d) The power of the kettle is $2.4$ kW. Assume that all the energy supplied goes into the water. Calculate the time taken for the water to reach $100$ °C. 2 marks

Mark scheme

C1 t = ΔE / P or 268 800 / 2400 seen; ecf from (c)
A1 112 s (accept 110 s or 113 s, or 1.9 min); unit required

Question 3. A student measures the specific heat capacity of cooking oil. She heats the oil in a beaker with an immersion heater and reads its temperature with a thermometer.

(a) Explain why she stirs the oil before she reads the thermometer. 2 marks No calculator

Mark scheme

B1 the oil next to the heater is hotter than the rest, or the energy is not spread evenly
B1 stirring spreads the energy so the thermometer reads the temperature of all the oil, owtte

(b) Suggest one way she could reduce the energy transferred from the oil to the surroundings. 1 mark No calculator

Mark scheme

B1 insulate the beaker, or use a polystyrene cup, or put a lid on the beaker; ignore repeat the experiment

How this is examined

Describing how to measure c for a solid block or for a liquid: a four- to six-mark Describe, which may come with a diagram of the apparatus Each measurement named with its instrument: the mass on a balance; the starting temperature and the highest temperature with a thermometer; the energy supplied from a joulemeter, or from power × time with a stopwatch. Then the calculation, $c = Δ E / (m Δ θ)$ with $Δ θ$ = highest minus starting temperature. Add a precaution with its reason: insulation to reduce energy losses, stirring a liquid so its temperature is even, oil in the thermometer hole of a block for good thermal contact.
Working out c when the energy comes from a heater: a three-mark Calculate that gives a heater's power, a time in minutes, a mass and two temperatures $Δ E = P t$ with $t$ in seconds; then $c = Δ E / (m Δ θ)$ with $Δ θ$ as the change; then the answer with the unit J/(kg °C). A wrong energy used correctly in the second step can still earn the mark for that step.
Rearranging the equation to find an energy, a temperature rise or a final temperature: a two- or three-mark Calculate with the value of c given in the question The equation recalled and written in symbols; $Δ θ$ treated as a change, and added to the starting temperature if the final temperature is asked for; the mass in kg; the unit (J or °C) on the answer.
Why a measured value of c is larger than the accepted value: a two-mark Explain after an experimental calculation, which may be followed by a one-mark Suggest for an improvement 'Energy is transferred to the surroundings (or heats the container), so the temperature rise is smaller than it would be, so the calculated c is larger.' For an improvement: 'insulate the block or beaker' or 'put a lid on the beaker'. 'Repeat the experiment' does not reduce the energy lost.
What a rise in temperature does to the internal energy and to the particles: a one-mark State and a two-mark Describe about an object being heated 'The internal energy increases' and 'the average kinetic energy of the particles increases'. Use the words 'average' and 'kinetic energy', which the objective itself uses.
The definition of specific heat capacity: a one-mark Define 'The energy required per unit mass per unit temperature increase', or 'the energy needed to raise the temperature of 1 kg of a substance by 1 °C'. Leaving out either the mass or the temperature increase describes a different quantity.

Words to know

Internal energy: The total energy of the particles in an object: the kinetic energy of their motion plus the potential energy stored in the forces between them. It increases when the temperature of the object rises.
Temperature (particle view): A rise in the temperature of an object is a rise in the average kinetic energy of all of its particles.
Specific heat capacity, c: The energy required per unit mass per unit temperature increase: $c = Δ E / (m Δ θ)$ . Measured in J/(kg °C).
$Δ θ$ (temperature change): Final temperature minus starting temperature, in °C.
Joulemeter: A meter connected between a power supply and a heater that reads the energy supplied, in joules.
Immersion heater: An electric heater placed in a liquid, or in a hole in a solid block, so that its energy goes into the material around it.
Thermal insulation (lagging): Material wrapped round apparatus to reduce the energy transferred to the surroundings.

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