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Pearson Edexcel PhysicsParticle model of matter

Internal energy and specific heat capacity

Link heating to temperature change and energy transfer.

Start here

The key idea

Internal energy is the total kinetic and potential energy of the particles in a system.

Equation to know

change in thermal energy = mass x specific heat capacity x temperature change

Internal Energy And Specific Heat Capacity
energy intemp risesE = m c delta T

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

1

Internal energy

Internal energy is the total energy stored by the particles of a substance, made up of their kinetic energy (from movement) and the potential energy of the bonds between them.Heating a substance increases its internal energy.This either raises the temperature, increasing particle kinetic energy, or causes a change of state, changing the potential energy of the particles.

2

Specific heat capacity

The specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1 degree C.Change in thermal energy = mass x specific heat capacity x temperature change, written E = m x c x change in temperature, with energy in J, mass in kg, c in J/kg degrees C and temperature in degrees C.A high specific heat capacity means a substance heats up and cools down slowly.

3

Specific latent heat

The specific latent heat is the energy needed to change the state of 1 kg of a substance with no change in temperature.Energy for a change of state = mass x specific latent heat, written E = m x L.There is specific latent heat of fusion for melting or freezing, and specific latent heat of vaporisation for boiling or condensing.During a change of state the temperature stays constant.

4

Heating and cooling graphs

On a temperature-time graph for heating, sloping sections show the temperature rising as kinetic energy increases.Flat (horizontal) sections show a change of state, where energy is being supplied but the temperature stays constant because it is breaking bonds (increasing potential energy).The flat section for boiling is usually longer than for melting because the latent heat of vaporisation is larger.

Key terms

Definitions to learn

Internal energy

The total kinetic and potential energy of a substance's particles.

Specific heat capacity

The energy to raise 1 kg of a substance by 1 degree C.

Specific latent heat

The energy to change the state of 1 kg with no temperature change.

Latent heat of fusion

The energy to melt or freeze 1 kg of a substance.

Latent heat of vaporisation

The energy to boil or condense 1 kg of a substance.

Worked example

Calculate the energy needed to heat 2.0 kg of water by 15 degrees Celsius. The specific heat capacity is 4200 J/kg degrees Celsius.

1

Use E = mc delta T.

2

Substitute 2.0 × 4200 × 15.

Final answer

126,000 J

Exam habit

Quote the specific heat capacity equation: E = mcΔT.State c (with units J/kg°C), m (kg) and ΔT (°C) before substituting.If temperature stays constant, explain that energy is breaking bonds, not raising temperature.

Watch out

Do not forget the mass or the temperature change.

Examiner tips

How to score full marks

  • 1During a change of state the temperature stays constant — flat lines on a heating graph mean melting or boiling.
  • 2Use the temperature change (difference), not the final temperature, in E = m x c x change in temperature.
  • 3Convert mass to kg before substituting into specific heat or latent heat equations.
Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1A 0.50 kg metal block receives 9000 J and warms by 60 degrees Celsius. Find its specific heat capacity.
Mark scheme
  1. 1.Rearrange c = E / (m delta T).
300 J/kg degrees Celsius
2Explain why water is useful in central-heating systems.
Mark scheme
  1. 1.Use the idea of specific heat capacity.
Water has a high specific heat capacity, so it can transfer a large amount of energy for a given temperature change.
3Why does temperature stay constant while a pure substance melts?
Mark scheme
  1. 1.Think about energy and particle bonds.
Energy is used to overcome forces between particles rather than increasing their kinetic energy.
4Define specific heat capacity.[1 mark]
Mark scheme
  1. 1.Recall the standard definition for 1 kg and 1 degree C.
The energy needed to raise the temperature of 1 kg of a substance by 1 degree C (1)
5Calculate the energy needed to heat 2.0 kg of water from 20 degrees C to 70 degrees C. Specific heat capacity of water = 4200 J/kg degrees C.[3 marks]
Mark scheme
  1. 1.Find the temperature change: 70 - 20.
  2. 2.Use E = m x c x change in temperature.
  3. 3.Substitute m = 2.0, c = 4200.
change in temperature = 70 - 20 = 50 degrees C (1); E = m x c x change in temperature = 2.0 × 4200 × 50 (1) = 420000 J (1)
6Explain why the temperature stays constant while a substance is boiling.[2 marks]
Mark scheme
  1. 1.Link energy supplied to bonds.
  2. 2.Link to particle potential energy.
The energy supplied is used to break the bonds between particles, not to increase their kinetic energy (1); so the potential energy increases but the temperature does not change until all the liquid has boiled (1)
7Calculate the energy needed to melt 0.50 kg of ice. Specific latent heat of fusion of ice = 334000 J/kg.[2 marks]
Mark scheme
  1. 1.Use E = m x L.
  2. 2.Substitute m = 0.50 and L = 334000.
E = m x L (1) = 0.50 × 334000 = 167000 J (1)
8A 0.30 kg metal block is supplied with 5400 J of energy and its temperature rises by 40 degrees C. Calculate the specific heat capacity of the metal and explain why the measured value might be higher than the true value.[3 marks]
Mark scheme
  1. 1.Rearrange E = m x c x change in temperature to c = E / (m x change in temperature).
  2. 2.Substitute the values.
  3. 3.Consider energy lost to the surroundings.
c = E / (m x change in temperature) = 5400 / (0.30 × 40) (1) = 5400 / 12 = 450 J/kg degrees C (1); the measured value may be too high because some energy is lost to the surroundings, so more energy than calculated is needed per degree rise (1)
9A 1.5 kg aluminium pan contains 2.0 kg of water. Both start at 15 degrees C and are heated to 100 degrees C. Calculate the total energy needed. Specific heat capacity of aluminium = 900 J/kg degrees C; specific heat capacity of water = 4200 J/kg degrees C.[4 marks]
Mark scheme
  1. 1.Temperature change = 100 - 15 = 85 degrees C for both.
  2. 2.Energy for pan = m x c x delta T = 1.5 × 900 × 85.
  3. 3.Energy for water = 2.0 × 4200 × 85.
  4. 4.Total = sum of both.
delta T = 85 degrees C (1); E_pan = 1.5 × 900 × 85 = 114750 J (1); E_water = 2.0 × 4200 × 85 = 714000 J (1); E_total = 114750 + 714000 = 828750 J (accept 829 kJ) (1)
10How much energy is released when 0.20 kg of steam at 100 degrees C condenses to water at 100 degrees C? Specific latent heat of vaporisation of water = 2 260 000 J/kg.[3 marks]
Mark scheme
  1. 1.Condensation releases the same energy as boiling requires.
  2. 2.Use E = m x L.
  3. 3.Substitute m = 0.20 and L = 2 260 000.
E = m x L (1) = 0.20 × 2 260 000 (1) = 452 000 J = 452 kJ (1)
11Explain why burns from steam at 100 degrees C are typically more severe than burns from boiling water at 100 degrees C.[4 marks]
Mark scheme
  1. 1.Both are at the same temperature.
  2. 2.Steam has extra latent heat of vaporisation.
  3. 3.When steam condenses on skin it releases this extra energy.
  4. 4.More total energy transferred to skin causes greater damage.
Both steam and boiling water are at 100 degrees C, so they transfer the same amount of energy as they cool the same amount (1); however, when steam condenses on the skin it must first release its latent heat of vaporisation (1); this releases a very large additional amount of energy (about 2 260 000 J per kg) directly into the skin tissue (1); the total energy transferred to the skin is therefore much greater from steam, causing more severe burns (1)
12A heating experiment gives the following data: 0.50 kg of an unknown substance is heated at a constant rate of 250 W. The temperature rises from 20 degrees C at t = 0, reaches 80 degrees C at t = 120 s, then remains constant until t = 280 s before rising again. Calculate the specific heat capacity of the substance and the specific latent heat of its phase change.[4 marks]
Mark scheme
  1. 1.Energy supplied in first section = P x t = 250 × 120.
  2. 2.c = E / (m x delta T).
  3. 3.Energy for phase change = P x time for flat section = 250 × (280 - 120).
  4. 4.L = E / m.
Energy in sloping section = 250 × 120 = 30000 J (1); c = 30000 / (0.50 × 60) = 30000 / 30 = 1000 J/kg degrees C (1); time for flat section = 280 - 120 = 160 s; energy for phase change = 250 × 160 = 40000 J (1); L = E / m = 40000 / 0.50 = 80000 J/kg (1)
13Explain, using ideas about particles and internal energy, why it takes much more energy to boil 1 kg of water than to melt 1 kg of ice, even though the temperature does not change in either case. Refer to the specific latent heats of fusion and vaporisation in your answer.[4 marks]
Mark scheme
  1. 1.Melting: breaking solid bonds, particles stay close.
  2. 2.Boiling: completely separating particles, overcoming all intermolecular attractions.
  3. 3.Link to values: L_vap >> L_fus.
  4. 4.Both involve potential energy not kinetic energy increasing.
In both cases the energy supplied goes into increasing the potential energy of the particles, not their kinetic energy, so temperature stays constant (1); when ice melts, the bonds between particles are weakened enough for them to move past each other, but the particles remain close together — this requires relatively little energy (latent heat of fusion ≈ 334 000 J/kg) (1); when water boils, the particles must be completely separated from one another, overcoming all the attractive forces between them; because these forces are strong and must be fully overcome, a much larger amount of energy is needed (latent heat of vaporisation ≈ 2 260 000 J/kg, about 6.8 times greater) (1); the difference reflects the much greater increase in particle separation and potential energy during boiling compared with melting (1)
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