Analytical Chemistry Technology  ·  Level 6
Physics Principles
Chapter 2: Apply principles of thermodynamics
📚 5 Topics
⚑ What you will be able to do

By the end of this chapter, you will be able to:

  • measure heat energy accurately by following the physics laboratory thermodynamics manual
  • calculate heat energy correctly using thermodynamic principles
  • determine thermodynamic processes accurately by applying the physics laboratory manual guidelines
  • calculate thermodynamic quantities correctly following the procedures in the physics laboratory manual

Mastering these skills will help you understand how energy works in real-world systems, making you confident in applying physics principles in your trade.

2.1 Heat energy

Heat energy is energy that flows from a hotter body to a colder one because of the temperature difference between them. It is measured in joules (J), the same unit as all other forms of energy. When heat is added to a substance, it either raises the temperature (the particles move faster) or changes the state (the particles are pulled apart without a rise in temperature). In an analytical laboratory these ideas explain how water baths, ovens, muffle furnaces, distillation units and freeze dryers work, and how much energy each one uses. Example: the quality control laboratory of a sugar factory in Katito works out how much steam heat its juice heaters need from the specific heat capacity of cane juice.

Two kinds of heat are distinguished:

  • Sensible heat: Heat that changes the temperature of a body and can be sensed with a thermometer. It is calculated with Q = mcΔθ.
  • Latent heat: Heat that changes the state of a body at constant temperature. It is calculated with Q = mL.

2.1.1 Heat capacity

The heat capacity (C) of a body is the quantity of heat needed to raise the temperature of the whole body by one kelvin (or one degree Celsius, since a change of 1 K equals a change of 1 °C). It depends on both the material and the amount of material, so a large copper calorimeter has a bigger heat capacity than a small one.

Formula: C = Q ÷ Δθ

Unit: joule per kelvin (J/K)

Worked example: an aluminium block absorbs 4,500 J and its temperature rises by 20 K. Find its heat capacity.

C = Q ÷ Δθ

C = 4,500 ÷ 20

C = 225 J/K

Heat capacity matters when a calorimeter, beaker or heating block takes up part of the heat in an experiment: its heat capacity must be added to that of the sample, or the result will be too low.

2.1.2 Specific heat capacity

The specific heat capacity (c) of a substance is the quantity of heat needed to raise the temperature of 1 kg of the substance by 1 K. It is a property of the material only, so it can be used to identify materials and to size heating and cooling equipment.

Formula: Q = mcΔθ, so c = Q ÷ (mΔθ), and heat capacity C = mc

Unit: joule per kilogram per kelvin (J/kg K)

Substance Specific heat capacity (J/kg K)
Water 4,200
Ice 2,100
Ethanol 2,400
Aluminium 900
Glass 670
Iron 450
Copper 390
Mercury 140

Worked example: how much heat is needed to raise 0.5 kg of water from 20 °C to 80 °C?

Q = mcΔθ

Q = 0.5 × 4,200 × (80 − 20)

Q = 0.5 × 4,200 × 60

Q = 126,000 J = 126 kJ

The high specific heat capacity of water is why it is used in water baths, cooling jackets and car radiators: it absorbs a lot of heat for a small temperature rise.

2.1.3 Latent heat

Latent heat is the heat absorbed or released when a substance changes state at constant temperature. The word latent means hidden: the heat goes into breaking (or forming) the bonds between particles, not into making them move faster, so a thermometer shows no rise during melting or boiling. The specific latent heat (L) is the heat needed to change the state of 1 kg of the substance without a change in temperature.

Formula: Q = mL

Unit: joule per kilogram (J/kg)

  • Absorbed on heating: Melting, boiling and sublimation take in latent heat from the surroundings, which is why evaporating ether or acetone cools the skin.
  • Released on cooling: Freezing, condensation and deposition give out the same amount of latent heat, which is why steam scalds more severely than boiling water at the same temperature.
  • Shown on a heating curve: A graph of temperature against time for ice heated steadily has flat parts at 0 °C and 100 °C where latent heat is being absorbed.

2.1.4 Latent heat of fusion

The specific latent heat of fusion (Lf) is the heat needed to change 1 kg of a solid at its melting point into liquid at the same temperature. For ice, Lf = 3.34 × 10⁵ J/kg. The same amount of heat is released when 1 kg of the liquid freezes.

Worked example: find the heat needed to melt 0.2 kg of ice at 0 °C.

Q = mLf

Q = 0.2 × 3.34 × 10⁵

Q = 66,800 J = 66.8 kJ

Ice packs and ice baths in sample transport keep samples at about 0 °C because the ice absorbs heat while it melts, without its temperature rising.

2.1.5 Latent heat of vaporization

The specific latent heat of vaporization (Lv) is the heat needed to change 1 kg of a liquid at its boiling point into vapour at the same temperature. For water at 100 °C, Lv = 2.26 × 10⁶ J/kg, nearly seven times its latent heat of fusion, because the particles must be separated completely.

Worked example: find the heat needed to boil away 0.05 kg of water already at 100 °C.

Q = mLv

Q = 0.05 × 2.26 × 10⁶

Q = 113,000 J = 113 kJ

Latent heat of vaporization is used in distillation and rotary evaporation (heat supplied at the flask is removed again at the condenser), in steam sterilisation in autoclaves, and in evaporative cooling.

2.1.6 Latent heat of sublimation

Sublimation is the direct change from solid to vapour without passing through the liquid state. The specific latent heat of sublimation (Ls) is the heat needed to change 1 kg of a solid directly into vapour at the same temperature. For a substance at the same temperature, Ls = Lf + Lv, because sublimation does the work of melting and boiling in one step. For ice near 0 °C, Ls ≈ 3.34 × 10⁵ + 2.50 × 10⁶ ≈ 2.83 × 10⁶ J/kg.

Worked example: dry ice (solid carbon dioxide) has Ls ≈ 5.71 × 10⁵ J/kg at −78.5 °C. Find the heat absorbed when 0.1 kg sublimes.

Q = mLs

Q = 0.1 × 5.71 × 10⁵

Q = 57,100 J = 57.1 kJ

Sublimation is used in freeze drying (lyophilisation) of biological and food samples, in purifying iodine, naphthalene and camphor, and in dry ice for shipping frozen samples.

Laboratory applications of heat energy

  • Calorimetry: Measuring heats of reaction, neutralisation and combustion (bomb calorimeter for fuel calorific value).
  • Thermal analysis: Differential scanning calorimetry (DSC) measures the heat absorbed at melting points to identify and check the purity of compounds.
  • Drying to constant mass: Ovens at 105 °C supply latent heat to drive off moisture in moisture content tests.
  • Temperature control: Thermostatic baths use water's high specific heat capacity to keep samples steady during viscosity, density and enzyme tests.
  • Energy costing: Knowing Q = mcΔθ lets a laboratory estimate the electricity used by heaters and furnaces.
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🔒2.2 Determination of Specific heat capacity

The specific heat capacity of a solid or liquid is found in the laboratory by supplying a known quantity of heat to a known mass and measuring the temperature change, then using c = Q ÷ (mΔθ). Two methods are used at this level: the method of mixtures, where h…

🔒2.3 Determination of specific latent heats of fusion, vaporization and sublimation

The specific latent heat of a substance is found by measuring the heat needed to change the state of a known mass at constant temperature, then using L = Q ÷ m. As with specific heat capacity, the heat can be supplied electrically (Q = VIt or Pt) or found from…

🔒2.4 Thermodynamic processes

A thermodynamic process is any change in the state of a system, such as a gas in a cylinder, in which its pressure, volume or temperature changes. All processes obey the first law of thermodynamics, which is the law of conservation of energy applied to heat an…

🔒2.5 Thermodynamic quantities

Thermodynamic quantities are the measurable properties that describe the state of a system at any moment. For a fixed mass of gas, three of them, pressure, volume and temperature, are enough to fix its state, and they are linked by the ideal gas equation PV =…

Am I competent?

At the start of this chapter we promised you would be able to:

  • measure heat energy accurately by following the physics laboratory thermodynamics manual
  • calculate heat energy correctly using thermodynamic principles
  • determine thermodynamic processes accurately by applying the physics laboratory manual guidelines
  • calculate thermodynamic quantities correctly following the procedures in the physics laboratory manual

Tick each one you can genuinely do.

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