Cambridge IGCSE0972

Thermal properties and temperature

Physics 0972 Chapter Notes

What this chapter covers

Thermal properties and temperature - Thermal expansion of solids, liquids and gasesThermal properties and temperature - Specific heat capacityThermal properties and temperature - Melting, boiling and evaporation
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In this chapter

  1. 1Temperature, Scales and Absolute Zero
  2. 2Specific Heat Capacity
  3. 3Changes of State and Latent Heat
  4. 4Evaporation and Boiling
  5. 5Thermal Expansion

1. Temperature, Scales and Absolute Zero

Temperature is a measure of the average kinetic energy of the particles in a substance. The hotter an object is, the faster its particles are moving or vibrating. We measure temperature using two main scales: Celsius (°C) and Kelvin (K). The Celsius scale is based on the properties of water, with 0 °C as the freezing point and 100 °C as the boiling point. The Kelvin scale is an absolute scale, meaning its zero point, 0 K, is the coldest possible temperature, known as absolute zero. At absolute zero (-273 °C), particles have the minimum possible kinetic energy. A change of 1 °C is the same size as a change of 1 K.

Temperature in Kelvin (K) = Temperature in Celsius (°C) + 273

Key term

Absolute Zero: The lowest possible temperature (0 K or -273 °C) at which particles have their minimum possible kinetic energy.

Examiner insight

Examiners expect you to be able to convert between Celsius and Kelvin fluently. Marks are often lost for simple arithmetic errors or for using the wrong sign (+/-) during conversion.

Common pitfall

When calculating a temperature *change*, remember that the value is the same in both Kelvin and Celsius. For example, a rise from 10 °C to 30 °C is a change of 20 °C, which is also a change of 20 K.

Worked example 12 marks

The average human body temperature is 37 °C. What is this temperature in Kelvin?

  1. 1

    Start with the conversion formula: T(K) = T(°C) + 273

  2. 2

    Substitute the Celsius value: T(K) = 37 + 273

  3. 3

    Calculate the result: T(K) = 310 K

Worked example 22 marks

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

  1. 1

    Rearrange the conversion formula: T(°C) = T(K) - 273

  2. 2

    Substitute the Kelvin value: T(°C) = 5778 - 273

  3. 3

    Calculate the result: T(°C) = 5505 °C

Recap

  • Temperature measures the average kinetic energy of particles.
  • The Celsius scale uses 0 °C for freezing water and 100 °C for boiling water.
  • The Kelvin scale starts at absolute zero (0 K), the coldest possible temperature.
  • To convert from Celsius to Kelvin, add 273.
  • A change in temperature has the same value in both Celsius and Kelvin.

Quick check

  1. What is the boiling point of water in Kelvin?1 mark
  2. What is meant by the term 'absolute zero'?1 mark

2. Specific Heat Capacity

Different materials require different amounts of energy to heat up. Specific heat capacity, 'c', is the amount of thermal energy needed to raise the temperature of 1 kg of a substance by 1 °C (or 1 K). Substances with a high specific heat capacity, like water (c = 4200 J/(kg °C)), can absorb a lot of heat energy with only a small rise in temperature. This is why water is used in car cooling systems. Metals, like copper (c = 400 J/(kg °C)), have a low specific heat capacity and heat up very quickly.

Energy transferred (Q) = mass (m) × specific heat capacity (c) × temperature change (ΔT)

Key term

Specific Heat Capacity (c): The amount of thermal energy required to raise the temperature of 1 kg of a substance by 1 °C.

Examiner insight

Show your working clearly. Write down the formula, show the substitution of values, and then give the final answer with the correct units (J or kJ). Marks are often awarded for each of these steps.

Common pitfall

Confusing 'specific heat capacity' (energy per kg per °C) with 'thermal capacity' (energy per °C for the whole object).

Fun fact

The high specific heat capacity of water is a major reason why coastal areas have milder climates than inland areas. The sea takes a long time to heat up and cool down, moderating the temperature.

Worked example 13 marks

A 0.5 kg block of aluminium is heated from 20 °C to 100 °C. Calculate the thermal energy absorbed by the block. The specific heat capacity of aluminium is 900 J/(kg °C).

  1. 1

    Identify the formula: Q = mcΔT

  2. 2

    Identify the values: m = 0.5 kg, c = 900 J/(kg °C)

  3. 3

    Calculate the temperature change: ΔT = 100 °C - 20 °C = 80 °C

  4. 4

    Substitute the values into the formula: Q = 0.5 × 900 × 80

  5. 5

    Calculate the final answer: Q = 36,000 J or 36 kJ

Worked example 24 marks

A kettle transfers 168,000 J of energy to 0.5 kg of water, initially at 10 °C. What is the final temperature of the water? The specific heat capacity of water is 4200 J/(kg °C).

  1. 1

    Start with the formula: Q = mcΔT

  2. 2

    Rearrange to find the temperature change, ΔT: ΔT = Q / (mc)

  3. 3

    Substitute the values: ΔT = 168,000 / (0.5 × 4200)

  4. 4

    Calculate the temperature change: ΔT = 168,000 / 2100 = 80 °C

  5. 5

    Calculate the final temperature: Final Temp = Initial Temp + ΔT = 10 °C + 80 °C = 90 °C

Recap

  • Specific heat capacity is the energy needed to heat 1 kg of a substance by 1 °C.
  • The formula for energy transfer is Q = mcΔT.
  • Water has a very high specific heat capacity, making it good for cooling.
  • Metals have low specific heat capacities, so they heat up and cool down quickly.
  • The unit for specific heat capacity is Joules per kilogram per degree Celsius (J/(kg °C)).

Quick check

  1. State the units for specific heat capacity.1 mark
  2. Substance A has a higher specific heat capacity than substance B. Which will heat up faster if they have the same mass and are supplied with energy at the same rate?1 mark

3. Changes of State and Latent Heat

When a substance is heated, its temperature rises until it reaches its melting or boiling point. At these points, the energy being supplied is used to break the bonds holding the particles together, causing a change of state (e.g. solid to liquid). During a change of state, the temperature of the substance does not change. The energy required to change the state of 1 kg of a substance without changing its temperature is called the 'specific latent heat'.

Energy for state change (Q) = mass (m) × specific latent heat (L)

For melting/freezing: Q = mLf (where Lf is the specific latent heat of fusion)

For boiling/condensing: Q = mLv (where Lv is the specific latent heat of vaporisation)

Key term

Specific Latent Heat (L): The amount of thermal energy required to change the state of 1 kg of a substance at a constant temperature.

Examiner insight

Examiners often test the understanding that temperature does not change during a phase change by showing heating curves and asking students to identify the melting and boiling points.

Common pitfall

The most common mistake is to include a temperature change (ΔT) in the latent heat formula. Remember, latent heat calculations apply only when the temperature is constant.

Fun fact

Steam burns are often more severe than burns from boiling water at the same temperature (100 °C). This is because of the huge amount of energy (the specific latent heat of vaporisation) released when the steam condenses into liquid on the skin.

Worked example 12 marks

Calculate the energy needed to melt 2.5 kg of ice at its melting point of 0 °C. The specific latent heat of fusion of ice is 334,000 J/kg.

  1. 1

    Identify the correct formula: Q = mLf

  2. 2

    Identify the values: m = 2.5 kg, Lf = 334,000 J/kg

  3. 3

    Substitute the values: Q = 2.5 × 334,000

  4. 4

    Calculate the final answer: Q = 835,000 J or 835 kJ

Worked example 22 marks

A kettle boils away 0.2 kg of water into steam at 100 °C. How much energy was used to do this? The specific latent heat of vaporisation of water is 2,260,000 J/kg.

  1. 1

    Identify the correct formula: Q = mLv

  2. 2

    Identify the values: m = 0.2 kg, Lv = 2,260,000 J/kg

  3. 3

    Substitute the values: Q = 0.2 × 2,260,000

  4. 4

    Calculate the final answer: Q = 452,000 J or 452 kJ

Recap

  • During a change of state, the temperature of the substance remains constant.
  • Energy supplied during a state change is called latent heat.
  • Specific latent heat is the energy needed to change the state of 1 kg of a substance.
  • The formula is Q = mL.
  • Latent heat of fusion (Lf) is for melting/freezing.
  • Latent heat of vaporisation (Lv) is for boiling/condensing.

Quick check

  1. What happens to the temperature of water as it boils?1 mark
  2. What is the difference between latent heat of fusion and latent heat of vaporisation?2 marks

4. Evaporation and Boiling

Both evaporation and boiling are processes where a liquid turns into a gas (vaporisation), but they happen in different ways. Boiling occurs at a specific temperature, the boiling point, and happens throughout the entire volume of the liquid. It is a fast process that requires a continuous external heat source. Evaporation, however, can occur at any temperature below the boiling point. It happens only at the surface of the liquid. During evaporation, the most energetic particles escape from the surface, leaving behind the less energetic particles. This lowers the average kinetic energy of the remaining liquid, causing its temperature to drop. This is known as the cooling effect of evaporation.

Key term

Evaporation: The process by which a liquid turns into a gas at a temperature below its boiling point, occurring only at the liquid's surface.

Examiner insight

For questions asking you to explain cooling by evaporation, marks are specifically awarded for mentioning that the 'most energetic' or 'fastest' particles escape, leading to a decrease in the 'average' kinetic energy of the remaining particles.

Common pitfall

Stating that boiling and evaporation are the same thing. They are both types of vaporisation, but the conditions under which they occur are very different.

Worked example 14 marks

Explain, in terms of particle motion, why sweating helps to cool the body down.

  1. 1

    Sweat is mostly water, which is secreted onto the skin.

  2. 2

    The water absorbs thermal energy from the skin.

  3. 3

    The most energetic water molecules gain enough energy to escape from the surface of the liquid into the air (evaporation).

  4. 4

    As the most energetic particles leave, the average kinetic energy of the remaining water on the skin decreases.

  5. 5

    This means the temperature of the water drops, and since it is in contact with the skin, it cools the body.

Worked example 23 marks

State three differences between boiling and evaporation.

  1. 1

    Temperature: Boiling occurs at a fixed temperature (the boiling point), whereas evaporation occurs over a range of temperatures.

  2. 2

    Location: Boiling occurs throughout the liquid (bubbles form inside), whereas evaporation occurs only at the surface.

  3. 3

    Speed: Boiling is a rapid process, while evaporation is a much slower process.

Recap

  • Evaporation occurs at the surface of a liquid at any temperature.
  • Boiling occurs throughout a liquid at a fixed temperature (the boiling point).
  • Evaporation causes cooling because the most energetic particles escape.
  • The rate of evaporation can be increased by increasing temperature, surface area, or air flow (wind).

Quick check

  1. State two factors that can increase the rate of evaporation.2 marks

5. Thermal Expansion

Most materials expand when they are heated and contract when they are cooled. This is called thermal expansion. When a substance is heated, its particles gain kinetic energy and vibrate or move more vigorously. This increased movement causes them to push each other further apart, so the object as a whole expands. Gases expand the most, followed by liquids, and then solids. This effect can be useful, as seen in liquid-in-glass thermometers or bimetallic strips used in thermostats. However, it can also cause problems, such as bridges buckling on a hot day or railway lines bending. Engineers must account for thermal expansion by including expansion gaps in structures.

Key term

Thermal Expansion: The tendency of matter to increase in volume in response to an increase in temperature.

Fun fact

The Eiffel Tower in Paris is about 15 cm taller in the summer than in the winter due to the thermal expansion of its iron structure.

Worked example 13 marks

Large concrete bridges are often built with metal expansion joints (gaps) at regular intervals. Explain why these are necessary.

  1. 1

    On a hot day, the temperature of the bridge increases.

  2. 2

    The concrete and steel materials of the bridge gain thermal energy, causing their particles to vibrate more.

  3. 3

    This increased vibration causes the materials to expand (thermal expansion).

  4. 4

    The expansion joints provide a space for the bridge sections to expand into without pushing against each other.

  5. 5

    Without these joints, the immense forces from expansion could cause the bridge to buckle, crack, or become damaged.

Worked example 23 marks

A bimetallic strip is made of brass and steel bonded together. When heated, it bends. Brass expands more than steel for the same temperature rise. Draw a simple diagram to show which way the strip will bend when heated.

  1. 1

    Draw a straight strip labelled 'Brass' on top and 'Steel' on the bottom, marked 'Cold'.

  2. 2

    Draw a second diagram marked 'Hot'. This diagram should show the strip curved upwards.

  3. 3

    Explain the reasoning: Since brass expands more than steel, the brass side becomes longer than the steel side when heated.

  4. 4

    To accommodate this difference in length while remaining bonded, the strip must curve with the longer material (brass) on the outside of the curve.

Recap

  • Heating causes particles to move more and push each other apart, causing expansion.
  • Gases expand more than liquids, and liquids expand more than solids for the same temperature change.
  • Thermal expansion is used in thermometers and thermostats (bimetallic strips).
  • Engineers must allow for thermal expansion in structures like bridges and railway lines by using expansion gaps.

Quick check

  1. Why is there a small gap in the top of a bottle of a fizzy drink?1 mark
  2. Which state of matter expands the most when heated?1 mark

End-of-chapter exercise

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

  1. A 2 kg block of copper is at an initial temperature of 25 °C. It is heated by a 500 W heater for 2 minutes. Calculate the final temperature of the copper block. Assume all energy from the heater is transferred to the block. (Specific heat capacity of copper = 400 J/(kg °C))5 marks
  2. Explain in detail the differences between boiling and evaporation. You should state at least three distinct points of comparison.3 marks
  3. A student wants to determine the specific latent heat of fusion of ice. They melt 0.050 kg of ice at 0 °C and find that it requires 16,700 J of energy. (a) Calculate the specific latent heat of fusion of ice from these results. (b) The accepted value is 334,000 J/kg. Suggest one reason why the student's result might be different.4 marks
  4. The temperature on a cold day is -5 °C. Convert this temperature to Kelvin.2 marks
  5. Explain, using the kinetic theory of particles, why a railway track is laid with small gaps between sections of the rail.3 marks
  6. A 1.5 kg mass of water at 100 °C is converted into steam at 100 °C. Calculate the thermal energy required. (Specific latent heat of vaporisation of water = 2.26 x 10^6 J/kg)2 marks
  7. Describe the principle of a liquid-in-glass thermometer, explaining how it uses a thermal property to measure temperature.3 marks
  8. A 200 g block of an unknown material absorbs 9.6 kJ of energy, and its temperature rises from 20 °C to 140 °C. Calculate the specific heat capacity of the material and suggest what the material might be, using the table: Water = 4200, Aluminium = 900, Copper = 400 J/(kg °C).5 marks
  9. Explain why a puddle of water on the ground disappears even on a cool, windy day.3 marks
  10. Calculate the total energy required to turn 0.5 kg of ice at -10 °C into water at 20 °C. (c_ice = 2100 J/(kg °C), Lf_ice = 334,000 J/kg, c_water = 4200 J/(kg °C))6 marks

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