Cambridge AS & A Level9702

Temperature

Physics 9702 Chapter Notes

What this chapter covers

Temperature - Thermal equilibriumTemperature - Temperature scalesTemperature - Specific heat capacity and specific latent heat
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Temperature notes

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In this chapter

  1. 1Temperature and Thermal Equilibrium
  2. 2The Kelvin Scale and Absolute Zero
  3. 3Internal Energy
  4. 4The First Law of Thermodynamics

1. Temperature and Thermal Equilibrium

Temperature is a fundamental concept in physics that indicates the 'hotness' or 'coldness' of an object. On a microscopic level, it's related to the average kinetic energy of the particles in a substance. However, its most important role is determining the direction of energy transfer. When two objects are in thermal contact, energy will spontaneously flow from the object at a higher temperature to the object at a lower temperature. This flow of energy due to a temperature difference is called thermal energy, or more commonly, heat. The transfer continues until both objects reach the same temperature. At this point, there is no longer any net flow of energy between them. They are said to be in thermal equilibrium. A thermometer works on this principle: it comes into thermal equilibrium with the object it's measuring and then displays its own temperature, which we take to be the temperature of the object.

Key term

Thermal Equilibrium: The state in which there is no net flow of thermal energy between two objects in thermal contact, meaning they are at the same temperature.

Examiner insight

Examiners award marks for clearly stating that energy flows from a higher temperature to a lower temperature, not just from a 'hot' object to a 'cold' one.

Worked example 14 marks

A small, hot block of iron at 90 °C is dropped into a large, insulated beaker of water at 20 °C. Describe the energy transfers that occur and the final state of the system.

  1. 1

    Step 1: Identify the initial temperature difference. The iron block (90 °C) is at a higher temperature than the water (20 °C).

  2. 2

    Step 2: State the direction of energy transfer. Because there is a temperature difference, thermal energy will flow from the hotter object (the iron block) to the colder object (the water).

  3. 3

    Step 3: Describe the consequences of the energy transfer. The iron block will cool down as it loses internal energy, and the water will warm up as it gains internal energy.

  4. 4

    Step 4: Describe the final state. The energy transfer will continue until the iron block and the water reach the same temperature. At this point, they are in thermal equilibrium, and the net flow of energy between them stops. The final temperature will be somewhere between 20 °C and 90 °C.

Recap

  • Temperature determines the direction of thermal energy transfer.
  • Thermal energy flows from a region of higher temperature to a region of lower temperature.
  • Two objects are in thermal equilibrium when they are at the same temperature.
  • When in thermal equilibrium, there is no net flow of energy between objects in thermal contact.

Quick check

  1. If object A is in thermal equilibrium with object B, and object B is in thermal equilibrium with object C, what can be said about objects A and C?2 marks

2. The Kelvin Scale and Absolute Zero

While the Celsius scale (°C) is common in daily life, it is based on the properties of a specific substance—water (0 °C for freezing, 100 °C for boiling). This can be problematic as these points change with pressure. The thermodynamic, or Kelvin, scale (K) is an absolute temperature scale that is not dependent on the properties of any particular substance. Its zero point, 0 K, is called absolute zero. This is the lowest possible temperature, where the particles of a substance have their minimum possible internal energy. It's impossible to remove any more energy from a system at absolute zero. The size of one unit on the Kelvin scale is the same as one degree on the Celsius scale, which makes conversion straightforward. A change of 1 K is identical to a change of 1 °C.

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.

Common pitfall

Forgetting that while absolute temperatures must be converted, a *change* in temperature is the same value in both Kelvin and Celsius (e.g., a ΔT of 50 °C is also a ΔT of 50 K).

Fun fact

The coldest place in the known universe is the Boomerang Nebula, with a temperature of about 1 K (–272 °C). This is even colder than the background temperature of empty space (2.7 K)!

Worked example 12 marks

The body temperature of a healthy human is approximately 37 °C. What is this temperature in kelvin?

  1. 1

    Step 1: Recall the conversion formula: T (K) = θ (°C) + 273.15.

  2. 2

    Step 2: Substitute the Celsius temperature into the formula.

  3. 3

    T = 37 + 273.15

  4. 4

    Step 3: Calculate the result.

  5. 5

    T = 310.15 K

Worked example 22 marks

A chemical reaction needs to be cooled to a temperature of 80 K. What is this temperature in degrees Celsius?

  1. 1

    Step 1: Rearrange the conversion formula: θ (°C) = T (K) - 273.15.

  2. 2

    Step 2: Substitute the Kelvin temperature into the formula.

  3. 3

    θ = 80 - 273.15

  4. 4

    Step 3: Calculate the result.

  5. 5

    θ = -193.15 °C

Worked example 32 marks

A block of aluminium is heated from 25 °C to 125 °C. Calculate the temperature change in kelvin.

  1. 1

    Step 1: Calculate the temperature change in Celsius.

  2. 2

    Δθ = 125 °C - 25 °C = 100 °C

  3. 3

    Step 2: Recognise that the size of a kelvin is the same as the size of a degree Celsius.

  4. 4

    Therefore, a change in temperature of 100 °C is equal to a change in temperature of 100 K.

  5. 5

    Alternatively, convert start and end temperatures to Kelvin: T_initial = 25 + 273.15 = 298.15 K. T_final = 125 + 273.15 = 398.15 K. ΔT = 398.15 K - 298.15 K = 100 K.

Recap

  • The thermodynamic temperature scale (Kelvin) is an absolute scale.
  • Absolute zero (0 K) is the lowest possible temperature.
  • At 0 K, a substance has its minimum internal energy.
  • To convert from Celsius to Kelvin, add 273.15.
  • A change in temperature has the same numerical value in Kelvin and Celsius.

Quick check

  1. Convert a room temperature of 20 °C to the Kelvin scale.1 mark
  2. What is the key advantage of the Kelvin scale over the Celsius scale?1 mark

3. Internal Energy

Any object, whether it's a solid, liquid, or gas, is made of a huge number of molecules that are in constant, random motion. The internal energy (symbol U) of a system is the total energy of its constituent particles. It is defined as the sum of the random distribution of kinetic and potential energies of the molecules in the system.

  1. Kinetic Energy: The molecules are moving and vibrating randomly. This motion gives them kinetic energy. The average kinetic energy of the molecules is directly proportional to the object's temperature in kelvin. So, when you heat an object and its temperature rises, you are increasing the average kinetic energy of its molecules.
  1. Potential Energy: There are electrostatic forces of attraction between molecules. The energy stored in these intermolecular bonds is the potential energy. This energy depends on the separation of the molecules. When a substance changes phase (e.g., melting or boiling), energy is used to overcome these forces and move the molecules further apart. This increases their potential energy, even while the temperature (and thus kinetic energy) remains constant.

Key term

Internal Energy (U): The sum of the random distribution of the kinetic and potential energies of the molecules within a system.

Examiner insight

High-scoring answers clearly distinguish between the kinetic energy component of internal energy (linked to temperature) and the potential energy component (linked to intermolecular forces and phase).

Worked example 14 marks

Compare the internal energy of 1 kg of water at 100 °C and 1 kg of steam at 100 °C. Explain your reasoning in terms of molecular energies.

  1. 1

    Step 1: State the relationship between temperature and kinetic energy. Since both the water and the steam are at the same temperature (100 °C), the average kinetic energy of the molecules in both phases is the same.

  2. 2

    Step 2: Consider the phase change. To turn water at 100 °C into steam at 100 °C, energy must be supplied (the latent heat of vaporisation).

  3. 3

    Step 3: Relate the added energy to potential energy. This energy does not increase the temperature, so it does not increase the kinetic energy of the molecules. Instead, it does work to overcome the intermolecular forces holding the water molecules together in the liquid state, moving them much further apart in the gaseous state.

  4. 4

    Step 4: Conclude by comparing internal energies. This work increases the potential energy of the molecules. Therefore, the internal energy (the sum of kinetic and potential energies) of 1 kg of steam at 100 °C is significantly greater than the internal energy of 1 kg of water at 100 °C.

Recap

  • Internal energy is the sum of the random kinetic and potential energies of molecules.
  • The kinetic energy component is related to the temperature of the system.
  • The potential energy component is related to the intermolecular forces and separation (i.e., the phase).
  • An increase in temperature causes an increase in the system's internal energy.
  • A change of phase at constant temperature also causes a change in internal energy.

Quick check

  1. When a solid melts at a constant temperature, which component of its internal energy increases: kinetic or potential?1 mark
  2. What happens to the internal energy of a gas if its temperature is decreased?1 mark

4. The First Law of Thermodynamics

The first law of thermodynamics is a statement of the principle of conservation of energy applied to thermal systems. It states that the change in the internal energy of a system (ΔU) is equal to the thermal energy supplied to the system(q) plus the work done on the system (W).

The law is expressed by the equation: ΔU = q + W.

Let's break down the terms and their signs, which is crucial: • ΔU: The change in internal energy. An increase in U is positive, a decrease is negative. • q: The thermal energy (heat) transferred. If heat is added *to* the system, q is positive. If heat is removed *from* the system, q is negative. • W: The work done. If work is done *on* the system (e.g., a gas is compressed), W is positive. If work is done *by* the system (e.g., a gas expands and pushes a piston), W is negative.

For a gas expanding or being compressed at a constant pressure `p`, the work done *on* the gas is given by `W = -pΔV`, where `ΔV` is the change in volume (`V_final - V_initial`). Notice the negative sign:

  • If a gas is compressed, `ΔV` is negative, making `W` positive (work is done on the gas).
  • If a gas expands, `ΔV` is positive, making `W` negative (work is done by the gas).

ΔU = q + W

W = -pΔV

Key term

First Law of Thermodynamics: The principle 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).

Examiner insight

Examiners specifically test the sign conventions for the first law. Always write down the formula and clearly state the values and signs of q and W before calculating.

Common pitfall

Getting the signs of q and W wrong. Remember: energy *into* the system (heat added, work done on) is positive; energy *out of* the system (heat lost, work done by) is negative.

Worked example 13 marks

In a thermodynamic process, a system absorbs 400 J of heat and has 150 J of work done on it. What is the change in the internal energy of the system?

  1. 1

    Step 1: State the first law of thermodynamics: ΔU = q + W.

  2. 2

    Step 2: Identify the values of q and W from the question, paying attention to the signs.

  3. 3

    Heat is absorbed by the system, so q = +400 J.

  4. 4

    Work is done on the system, so W = +150 J.

  5. 5

    Step 3: Substitute these values into the equation.

  6. 6

    ΔU = (+400 J) + (+150 J)

  7. 7

    Step 4: Calculate the change in internal energy.

  8. 8

    ΔU = +550 J. The internal energy increases by 550 J.

Worked example 24 marks

A gas in a cylinder expands at a constant pressure of 2.0 × 10⁵ Pa from a volume of 5.0 × 10⁻³ m³ to 7.5 × 10⁻³ m³. During the expansion, it absorbs 800 J of thermal energy. Calculate the change in the internal energy of the gas.

  1. 1

    Step 1: Calculate the change in volume, ΔV.

  2. 2

    ΔV = V_final - V_initial = (7.5 × 10⁻³ m³) - (5.0 × 10⁻³ m³) = 2.5 × 10⁻³ m³.

  3. 3

    Step 2: Calculate the work done on the gas, W, using W = -pΔV.

  4. 4

    W = -(2.0 × 10⁵ Pa) × (2.5 × 10⁻³ m³) = -500 J.

  5. 5

    The negative sign indicates that the gas did 500 J of work on its surroundings as it expanded.

  6. 6

    Step 3: State the first law of thermodynamics: ΔU = q + W.

  7. 7

    Step 4: Substitute the values for q and W. The gas absorbs heat, so q = +800 J.

  8. 8

    ΔU = (+800 J) + (-500 J) = 300 J.

  9. 9

    Step 5: State the final answer. The internal energy of the gas increases by 300 J.

Recap

  • The first law of thermodynamics is ΔU = q + W.
  • ΔU is the change in internal energy of the system.
  • q is positive for heat supplied to the system, negative for heat removed.
  • W is positive for work done on the system, negative for work done by the system.
  • For a gas at constant pressure, the work done on it is W = -pΔV.

Quick check

  1. If the internal energy of a system decreases by 100 J and 40 J of heat is supplied to it, calculate the work done, W. Was work done on or by the system?3 marks

End-of-chapter exercise

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

  1. Convert a temperature of 50 °C to kelvin.1 mark
  2. Define the term 'thermal equilibrium'.2 marks
  3. Explain why the internal energy of 10 g of steam at 100 °C is greater than the internal energy of 10 g of water at 100 °C. Refer to the energy of the molecules.3 marks
  4. A gas is held in a cylinder with a movable piston at a constant pressure of 1.8 x 10⁵ Pa. The gas is cooled and its volume decreases from 2.1 x 10⁻³ m³ to 1.6 x 10⁻³ m³. Calculate the work done on the gas.3 marks
  5. The internal energy of a system decreases by 320 J. If 180 J of thermal energy is removed from the system, calculate the work done, stating whether it was done on or by the system.3 marks
  6. A block of lead is hammered forcefully. Its temperature is observed to rise. Explain this observation in terms of the first law of thermodynamics and internal energy.4 marks
  7. A student claims that a thermometer based on the expansion of alcohol is an 'absolute' measure of temperature. Explain why this claim is incorrect and why the thermodynamic Kelvin scale is preferred for scientific work.3 marks
  8. In one cycle of a heat engine, a gas absorbs 2500 J of heat from a high-temperature source and does 700 J of work. Calculate the change in the internal energy of the gas and the heat released to the cold source.4 marks
  9. What is the temperature of absolute zero in degrees Celsius (to 2 decimal places) and in kelvin?2 marks
  10. A perfectly insulated cylinder contains a gas. The gas is compressed by a piston, which does 500 J of work on the gas. State the heat transferred, q, and calculate the change in the internal energy of the gas, ΔU. What will happen to the temperature of the gas?4 marks

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