Cambridge AS & A Level9702

Thermodynamics

Physics 9702 Chapter Notes

What this chapter covers

Thermodynamics - Internal energyThermodynamics - The first law of thermodynamics
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In this chapter

  1. 1Internal Energy and Temperature
  2. 2The First Law of Thermodynamics
  3. 3Temperature, Heat, and Equilibrium
  4. 4Specific Heat Capacity
  5. 5Specific Latent Heat

1. Internal Energy and Temperature

Every substance is made of atoms and molecules in constant, random motion. The internal energy (symbol U) of a system is the total energy stored within it. It is the sum of the random kinetic and potential energies of all its constituent particles. The kinetic energy comes from the particles' movement (vibrating, rotating, and translating). The potential energy comes from the forces (bonds) between the particles. Temperature is a measure of the average kinetic energy of the particles. Therefore, if you increase the temperature of an object, you increase the average kinetic energy of its particles, which in turn increases its total internal energy. For example, the molecules in a hot cup of tea are moving faster and have more internal energy than the molecules in a cold glass of water.

Key term

Internal Energy (U): The sum of the random distribution of kinetic and potential energies of the atoms or molecules that make up a system.

Examiner insight

Examiners look for a clear distinction between the kinetic and potential energy components of internal energy, especially when explaining temperature changes versus phase changes.

Common pitfall

Stating that internal energy is just the kinetic energy of molecules. It is crucial to remember the potential energy component, which becomes especially important during phase changes.

Worked example 14 marks

A 1 kg block of iron is heated from 293 K to 323 K. Describe the changes that occur to the internal energy of the iron block, referencing its atomic structure.

  1. 1
    1. Identify the change: The temperature of the iron block increases.
  2. 2
    1. Relate temperature to energy: Temperature is a measure of the average kinetic energy of the particles in a substance.
  3. 3
    1. Describe the particle behaviour: As the iron is heated, energy is transferred to its atoms. The iron atoms, which are fixed in a lattice structure, vibrate more vigorously about their fixed positions.
  4. 4
    1. Link to internal energy components: This increased vibration means the average kinetic energy of the atoms has increased. While the potential energy due to intermolecular forces doesn't change significantly (as the iron remains solid), the overall internal energy (sum of kinetic and potential energies) increases due to the large increase in kinetic energy.
  5. 5
    1. Conclusion: The internal energy of the iron block has increased because the average kinetic energy of its vibrating atoms has increased.

Recap

  • Internal energy (U) is the sum of the random kinetic and potential energies of a system's molecules.
  • The kinetic energy component is related to the temperature of the system.
  • The potential energy component is related to the forces between molecules and the state of matter.
  • Increasing the temperature of an object increases its internal energy by increasing the average kinetic energy of its particles.

Quick check

  1. What are the two types of energy that contribute to the internal energy of a substance?2 marks

2. The First Law of Thermodynamics

The first law of thermodynamics is a fundamental principle of energy conservation. It states that the change in the internal energy of a system (ΔU) is equal to the heat added to the system(q) plus the work done on the system (W). The equation is ΔU = q + W. It's crucial to understand the sign convention: 'q' is positive if heat is added to the system, and negative if heat is removed. 'W' is positive if work is done ON the system (e.g., compression of a gas), and negative if work is done BY the system (e.g., expansion of a gas). When a gas changes volume at a constant pressure 'p', the work done ON the system is given by W = -pΔV, where ΔV is the change in volume (V_final - V_initial).

ΔU = q + W

W = -pΔV (work done ON a gas at constant pressure)

Key term

First Law of Thermodynamics: The principle that the change in internal energy (ΔU) of a system is equal to the heat added to the system (q) plus the work done on the system (W).

Examiner insight

Examiners frequently test the sign convention of the first law. Always clearly state whether work is done 'on' or 'by' the system and whether heat is 'added' or 'removed' to justify your signs.

Common pitfall

Confusing the sign for work done. Remember: if the gas expands, it does work, so W (work done on the system) is negative. If it's compressed, work is done on it, so W is positive.

Worked example 13 marks

A gas in a cylinder is supplied with 2500 J of heat. The gas expands, doing 1500 J of work on its surroundings. Calculate the change in the internal energy of the gas.

  1. 1
    1. State the First Law of Thermodynamics: ΔU = q + W
  2. 2
    1. Identify the given values with correct signs: Heat is supplied TO the gas, so q = +2500 J.
  3. 3
    1. The gas does work on its surroundings (expands), so work is done BY the gas. This means the work done ON the gas is negative: W = -1500 J.
  4. 4
    1. Substitute the values into the equation: ΔU = (+2500 J) + (-1500 J)
  5. 5
    1. Calculate the result: ΔU = 1000 J. The internal energy of the gas increases by 1000 J.

Worked example 24 marks

A gas is compressed at a constant pressure of 2.0 x 10^5 Pa from a volume of 8.0 x 10^-3 m^3 to 3.0 x 10^-3 m^3. During this process, 1200 J of heat is removed from the gas. Calculate the change in its internal energy.

  1. 1
    1. First, calculate the work done ON the gas. State the formula: W = -pΔV
  2. 2
    1. Calculate the change in volume: ΔV = V_final - V_initial = (3.0 x 10^-3) - (8.0 x 10^-3) = -5.0 x 10^-3 m^3
  3. 3
    1. Substitute values to find W: W = -(2.0 x 10^5 Pa) * (-5.0 x 10^-3 m^3) = +1000 J. The positive sign indicates work was done on the gas.
  4. 4
    1. Now use the First Law of Thermodynamics: ΔU = q + W
  5. 5
    1. Identify q: Heat is removed from the gas, so q = -1200 J.
  6. 6
    1. Substitute q and W: ΔU = (-1200 J) + (+1000 J) = -200 J.
  7. 7
    1. Conclusion: The internal energy of the gas decreases by 200 J.

Recap

  • The first law of thermodynamics is ΔU = q + W.
  • ΔU is the change in internal energy.
  • q is the heat added TO the system (positive) or removed FROM it (negative).
  • W is the work done ON the system (positive) or BY the system (negative).
  • For a gas at constant pressure, the work done ON it is W = -pΔV.

Quick check

  1. If the internal energy of a system increases by 50 J while it does 20 J of work, was heat added or removed? How much?2 marks

3. Temperature, Heat, and Equilibrium

Heat is not something an object 'has'; it is energy in transit. Thermal energy naturally flows from a region of higher temperature to a region of lower temperature. This process continues until both regions reach the same temperature. At this point, they are in thermal equilibrium, and there is no longer any net flow of energy between them. While the Celsius scale (°C) is common, the absolute thermodynamic scale is the Kelvin scale (K). Its zero point, 0 K, is called absolute zero. This is the coldest possible temperature, where particles have the minimum possible internal energy. The size of one kelvin is the same as one degree Celsius, which makes conversion simple: T (in K) = θ (in °C) + 273.15. Because temperature differences drive energy flow, the Kelvin scale is essential for many laws in thermodynamics.

T / K = θ / °C + 273.15

Key term

Absolute Zero: The lowest possible temperature (0 K or –273.15 °C) at which a system has its minimum possible internal energy.

Examiner insight

Examiners reward students who can articulate that absolute zero is the point of minimum, not zero, internal energy for a system.

Fun fact

The coldest temperature ever achieved in a laboratory is just 38 picokelvin (0.000000000038 K) above absolute zero, far colder than the average temperature of deep space (about 2.7 K).

Worked example 12 marks

The surface temperature of the Sun is approximately 5778 K. What is this temperature in degrees Celsius?

  1. 1
    1. State the conversion formula: T / K = θ / °C + 273.15
  2. 2
    1. Rearrange for degrees Celsius: θ / °C = T / K - 273.15
  3. 3
    1. Substitute the value in Kelvin: θ / °C = 5778 - 273.15
  4. 4
    1. Calculate the result: θ = 5504.85 °C. To an appropriate number of significant figures, this is 5505 °C.

Worked example 23 marks

An aluminium block at 80 °C is placed in thermal contact with a copper block at 20 °C in an insulated container. Describe the energy transfer and the final state of the system.

  1. 1
    1. Identify the temperature difference: The aluminium block is at a higher temperature than the copper block.
  2. 2
    1. State the direction of energy flow: Thermal energy will flow from the hotter object (aluminium block) to the colder object (copper block).
  3. 3
    1. Describe the consequence: The internal energy of the aluminium block will decrease, and its temperature will fall. The internal energy of the copper block will increase, and its temperature will rise.
  4. 4
    1. Describe the final state: This net energy transfer will continue until both blocks reach the same temperature. At this point, they are in thermal equilibrium, and the net flow of energy between them becomes zero.

Recap

  • Thermal energy flows from a region of higher temperature to a region of lower temperature.
  • Thermal equilibrium is reached when connected objects are at the same temperature, with no net energy flow.
  • The Kelvin scale is the absolute temperature scale, where 0 K is absolute zero.
  • To convert from Celsius to Kelvin, add 273.15.

Quick check

  1. What is the boiling point of water (100 °C) in Kelvin?1 mark

4. Specific Heat Capacity

Different materials require different amounts of energy to heat up. The specific heat capacity (symbol 'c') of a substance is a measure of this property. It is formally defined as the energy required to raise the temperature of 1 kg of the substance by 1 K (or 1 °C). The formula to calculate the heat energy(q) transferred is q = mcΔT, where 'm' is the mass and 'ΔT' is the temperature change. Substances with a high specific heat capacity, like water (c ≈ 4200 J kg⁻¹ K⁻¹), can absorb a lot of heat for a small temperature rise. Substances with a low specific heat capacity, like copper (c ≈ 385 J kg⁻¹ K⁻¹), heat up and cool down very quickly.

q = mcΔT

Key term

Specific Heat Capacity (c): The energy required per unit mass of a substance to raise its temperature by one Kelvin (or one degree Celsius).

Examiner insight

When using q = mcΔT, marks are often awarded for correctly calculating the temperature change (ΔT) and for using the correct units for mass (kg) and energy (J).

Common pitfall

Forgetting to convert mass from grams to kilograms before using the formula q = mcΔT. All units must be in their standard SI form.

Worked example 13 marks

Calculate the heat energy required to raise the temperature of a 0.50 kg aluminium block from 20 °C to 100 °C. The specific heat capacity of aluminium is 910 J kg⁻¹ K⁻¹.

  1. 1
    1. State the relevant formula: q = mcΔT
  2. 2
    1. Identify the given values: m = 0.50 kg, c = 910 J kg⁻¹ K⁻¹.
  3. 3
    1. Calculate the temperature change, ΔT. Note that a change in °C is the same as a change in K. ΔT = 100 °C - 20 °C = 80 °C = 80 K.
  4. 4
    1. Substitute the values into the formula: q = (0.50 kg) * (910 J kg⁻¹ K⁻¹) * (80 K)
  5. 5
    1. Calculate the result: q = 36400 J or 36.4 kJ.

Worked example 24 marks

A 200 W electric heater is used to heat a 1.5 kg block of an unknown material. In 5.0 minutes, the temperature of the block increases from 25 °C to 65 °C. Assuming no heat loss, calculate the specific heat capacity of the material.

  1. 1
    1. First, calculate the total energy supplied by the heater. Energy = Power × time. q = P × t.
  2. 2
    1. Convert time to SI units: t = 5.0 min * 60 s/min = 300 s.
  3. 3
    1. Calculate q: q = 200 W * 300 s = 60000 J.
  4. 4
    1. State the formula for specific heat capacity and rearrange it for 'c': q = mcΔT => c = q / (mΔT)
  5. 5
    1. Calculate the temperature change: ΔT = 65 °C - 25 °C = 40 °C = 40 K.
  6. 6
    1. Substitute all values into the rearranged formula: c = 60000 J / (1.5 kg * 40 K)
  7. 7
    1. Calculate the result: c = 60000 / 60 = 1000 J kg⁻¹ K⁻¹.

Recap

  • Specific heat capacity (c) is the energy needed to heat 1 kg of a substance by 1 K.
  • The formula for heat energy transfer is q = mcΔT.
  • A high 'c' means a substance heats up and cools down slowly (e.g., water).
  • A low 'c' means a substance heats up and cools down quickly (e.g., metals).
  • Always ensure mass is in kg and temperature change is calculated correctly.

Quick check

  1. Substance A has c = 400 J kg⁻¹ K⁻¹ and Substance B has c = 800 J kg⁻¹ K⁻¹. If 1 kg of each absorbs 800 J of energy, which substance has the greater temperature rise?2 marks

5. Specific Latent Heat

When a substance changes state (e.g., solid to liquid or liquid to gas), it absorbs or releases energy without any change in its temperature. This energy is called latent heat. The specific latent heat (L) of a substance is the energy required to change the state of 1 kg of the substance at a constant temperature. The formula is q = mL. There are two types:

  1. Specific Latent Heat of Fusion (L_f): Energy needed to melt (solid to liquid) or released when freezing (liquid to solid).
  2. Specific Latent Heat of Vaporisation (L_v): Energy needed to boil (liquid to gas) or released when condensing (gas to liquid).

During a phase change, the energy supplied goes into increasing the potential energy of the molecules (breaking bonds) rather than their kinetic energy, which is why the temperature remains constant.

q = mL

Key term

Specific Latent Heat (L): The energy required per unit mass of a substance to change its state without any change in temperature.

Examiner insight

Candidates must be clear that latent heat involves a change of state at a constant temperature, which is a common point of confusion with specific heat capacity.

Common pitfall

Trying to include a temperature change (ΔT) in the latent heat calculation. The formula q = mL applies only when the temperature is constant during the phase change.

Fun fact

The specific latent heat of vaporisation for water is very large. This is why steam burns are so severe: a large amount of energy is released onto the skin as the steam condenses back to liquid water.

Worked example 12 marks

How much energy is required to completely melt a 2.5 kg block of ice at 0 °C into water at 0 °C? The specific latent heat of fusion of ice is 3.34 x 10^5 J kg⁻¹.

  1. 1
    1. Identify the process: This is a phase change (melting) at constant temperature.
  2. 2
    1. State the relevant formula: q = mL_f
  3. 3
    1. Identify the given values: m = 2.5 kg, L_f = 3.34 x 10^5 J kg⁻¹.
  4. 4
    1. Substitute the values into the formula: q = (2.5 kg) * (3.34 x 10^5 J kg⁻¹)
  5. 5
    1. Calculate the result: q = 835000 J or 8.35 x 10^5 J (or 835 kJ).

Worked example 24 marks

A 2.0 kW kettle contains water at 100 °C. It is left on and boils dry in 6.5 minutes. Assuming all the energy from the kettle is transferred to the water, calculate the mass of water that was in the kettle. (Specific latent heat of vaporisation of water = 2.26 x 10^6 J kg⁻¹)

  1. 1
    1. First, calculate the total energy supplied by the kettle. Energy = Power × time. q = P × t.
  2. 2
    1. Convert power and time to SI units: P = 2.0 kW = 2000 W. t = 6.5 min * 60 s/min = 390 s.
  3. 3
    1. Calculate q: q = 2000 W * 390 s = 780000 J.
  4. 4
    1. State the formula for latent heat and rearrange it for mass 'm': q = mL_v => m = q / L_v
  5. 5
    1. Substitute the energy and latent heat values: m = 780000 J / (2.26 x 10^6 J kg⁻¹)
  6. 6
    1. Calculate the result: m ≈ 0.345 kg (or 345 g).

Recap

  • Specific latent heat (L) relates to energy transfer during a phase change at constant temperature.
  • The formula is q = mL.
  • Latent heat of fusion (L_f) is for melting/freezing.
  • Latent heat of vaporisation (L_v) is for boiling/condensing.
  • During a phase change, internal energy increases due to potential energy gain, not kinetic energy.

Quick check

  1. What happens to the temperature of a puddle of water as it evaporates?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define internal energy and explain why the internal energy of a block of lead increases when its temperature rises.3 marks
  2. The temperature on a cold day is -5.0 °C. What is this temperature on the absolute (Kelvin) scale?1 mark
  3. A fixed mass of gas is compressed, and 800 J of work is done on it. In the process, its internal energy increases by 500 J. According to the first law of thermodynamics, was heat added to or removed from the gas, and how much?3 marks
  4. A 400 g copper pan is heated from 20 °C to 110 °C. Calculate the thermal energy absorbed by the pan. (Specific heat capacity of copper = 385 J kg⁻¹ °C⁻¹).3 marks
  5. Explain, in terms of molecules, why energy is needed to boil water without a change in temperature.3 marks
  6. A bicycle pump's barrel becomes hot during use. Using the first law of thermodynamics, explain this phenomenon. Assume heat loss to the surroundings is negligible at first.3 marks
  7. A gas expands at a constant pressure of 1.5 x 10^5 Pa, and its volume increases by 4.0 x 10^-3 m^3. During this expansion, 1000 J of heat is supplied to the gas. Calculate the change in the internal energy of the gas.4 marks
  8. Calculate the total energy required to turn 500 g of ice at -10 °C into steam at 100 °C. (c_ice = 2100 J kg⁻¹K⁻¹, L_f = 3.34x10^5 J kg⁻¹, c_water = 4200 J kg⁻¹K⁻¹, L_v = 2.26x10^6 J kg⁻¹).6 marks
  9. Distinguish between specific heat capacity and specific latent heat. Your answer should include definitions and a comment on temperature change.4 marks
  10. An insulated container holds 2.0 kg of water at 80 °C. A 0.5 kg block of an unknown metal at 20 °C is dropped into the water. The system reaches thermal equilibrium at a final temperature of 78 °C. Calculate the specific heat capacity of the metal. (c_water = 4200 J kg⁻¹K⁻¹).5 marks

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