Water

Biology AS · Biological molecules. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A water molecule is polar. The oxygen atom pulls the shared electrons towards itself, so it has a small negative charge (δ−) and each hydrogen atom has a small positive charge (δ+).

A δ+ hydrogen of one molecule is attracted to the δ− oxygen of another. This attraction is a hydrogen bond. Each one is weak, but there are very many of them, and together they hold water molecules to each other.

Almost every property of water in the syllabus follows from this. Water dissolves ions and polar molecules, it needs a lot of energy to warm up, and it needs a lot of energy to evaporate.

Rules to remember

Solvent action: polar water molecules surround ions and polar molecules (glucose, amino acids), so these dissolve and can be transported and react in solution
Non-polar molecules such as triglycerides do not dissolve in water
High specific heat capacity (4.2 J g⁻¹ °C⁻¹): much energy is needed to raise the temperature, because energy is used to break hydrogen bonds; cells and habitats keep a stable temperature
High latent heat of vaporisation (about 2400 J g⁻¹ at body temperature): evaporation removes a lot of heat, so sweating and transpiration cool the organism
energy to change temperature = mass × specific heat capacity × temperature change
energy to evaporate = mass × latent heat of vaporisation

Common mistake

Mixing up the two heat properties. Specific heat capacity is about changing the temperature of liquid water; latent heat of vaporisation is about turning liquid into vapour at the same temperature.

Worked example

How much energy is needed to raise the temperature of 500 g of water by 3 °C? What mass of water would the same energy evaporate? (Specific heat capacity 4.2 J g⁻¹ °C⁻¹; latent heat of vaporisation 2400 J g⁻¹.)

energy = mass × specific heat capacity × temperature change = 500 × 4.2 × 3.
500 × 4.2 = 2100, and 2100 × 3 = 6300 J.
mass evaporated = energy ÷ latent heat = 6300 ÷ 2400 = 2.6 g (to 2 significant figures).

Answer: 6300 J (6.3 kJ), which would evaporate only about 2.6 g of water.

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