The shape of the lesson
By the end of the lesson, learners can
- Polar molecules, hydrogen bonds: Rotate the cluster of water molecules and follow the dashed hydrogen bonds.
- Three properties that matter: For each property, give the molecular reason and then the biological role.
- Polar: δ− O, δ+ H, bent molecule.
- Solvent: ions and polar molecules dissolve; transport and reactions.
- Specific heat capacity: temperature changes slowly.
- Latent heat of vaporisation: evaporation cools.
One picture every idea returns to
Oxygen attracts the shared electrons in each O-H bond more strongly than hydrogen, so a water molecule is polar: the O is slightly negative (δ−) and each H slightly positive (δ+). A hydrogen bond is the attraction between a δ+ hydrogen of one molecule and a lone pair on the δ− oxygen of another. Because water is polar it is an excellent solvent for ions and polar molecules. Because hydrogen bonds must be broken first, water has a high specific heat capacity (its temperature changes slowly) and a high latent heat of vaporisation (evaporation removes a lot of heat).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Small, bent and sticky.. Water is a tiny molecule that should be a gas at room temperature. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Polar molecules, hydrogen bonds | Rotate the cluster of water molecules and follow the dashed hydrogen bonds. Project tab 6 of the lab (Biological molecules lab, water) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Three properties that matter | For each property, give the molecular reason and then the biological role. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Three properties that matter, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Do not mix up the two thermal properties. | Specific heat capacity is about how much energy it takes to raise the temperature, so it explains stable temperatures. Latent heat of vaporisation is about the energy needed to evaporate water, so it explains cooling by sweating and transpiration. |
Water: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Explain how hydrogen bonds form between water molecules.
Explain why water is a good solvent and give two roles of this property in living organisms.
The specific heat capacity of water is 4.2 J g−1 °C−1. Calculate the energy needed to raise the temperature of 500 g of water by 3.0 °C.
During a long run an athlete evaporates 800 g of sweat. The latent heat of vaporisation of water at skin temperature is about 2.4 kJ g−1. Calculate the heat removed from the body.
Explain why the high specific heat capacity of water is important to organisms.
Over 24 hours the air above a lake varies from 6 °C to 24 °C, while the water 1 m below the surface varies from 15 °C to 16 °C. Describe and explain the difference.
Answer key and marking notes
- Oxygen is more electronegative than hydrogen, so electrons in the O-H bonds are pulled towards O.
- The O gains a partial negative charge (δ−) and each H a partial positive charge (δ+); the molecule is polar.
- A δ+ H of one molecule is attracted to a lone pair on the δ− O of another molecule: a hydrogen bond.
- Water is polar: δ− O is attracted to cations and δ+ H to anions, separating and surrounding the ions; polar molecules form hydrogen bonds with water.
- Role: a transport medium, for example glucose and ions in blood plasma or mineral ions in xylem.
- Role: metabolic reactions take place in solution in the cytoplasm (also accept removal of urea in urine).
- Energy = mass × specific heat capacity × temperature change.
- = 500 × 4.2 × 3.0 = 6300 J (6.3 kJ).
- Heat removed = mass evaporated × latent heat of vaporisation.
- = 800 × 2.4 = 1920 kJ (about 1.9 MJ).
- A lot of energy is needed to raise the temperature of water because hydrogen bonds must be broken first.
- So the temperature of cells and bodies, which are mostly water, changes only slowly, keeping conditions stable for enzymes.
- Large bodies of water have stable temperatures, providing stable aquatic habitats.
- The air temperature range is 18 °C; the water temperature range is only 1 °C.
- Water has a high specific heat capacity: much of the energy absorbed is used to break hydrogen bonds, so the temperature rises little.
- So the lake is a thermally stable habitat for aquatic organisms.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Why is a water molecule polar?
- Aoxygen is more electronegative than hydrogen, and the molecule is bent, so the partial charges do not cancel
- Bwater is made of H+ and OH− ions held together by ionic bonds
- Chydrogen is more electronegative than oxygen, so the electrons are pulled towards the hydrogen atoms
- Dthe molecule is linear, with the oxygen atom in the middle carrying a full negative charge
Which statement about the hydrogen bonds between water molecules in liquid water is correct?
- Aeach one is weak, but the very large number of them has a large combined effect
- Beach one is a strong covalent bond between an oxygen and a hydrogen atom
- Cthey form only when water freezes, holding the molecules in a fixed lattice
- Dthey form between the δ+ hydrogen atoms of two neighbouring molecules
Which pair of substances would both dissolve readily in water?
- Aglucose and potassium chloride
- Bglucose and a triglyceride
- Ccellulose and sodium chloride
- Dstarch and a triglyceride
The latent heat of vaporisation of water at body temperature is about 2.4 kJ g−1. What mass of sweat must evaporate to remove 120 kJ of heat from the body?
- A50 g
- B288 g
- C28.6 g
- D0.020 g
Cells are mostly water. Why is the high specific heat capacity of water important for the enzyme-catalysed reactions inside cells?
- Athe temperature inside cells changes only slowly, so enzyme activity stays fairly stable
- Bwater evaporates from inside the cells, keeping the enzymes cool
- Cthe heat released by respiration is lost from the cells very quickly
- Denzymes dissolve better in water that is at a high temperature
Equal masses (100 g) of water and of a plant oil are heated by identical heaters, and their temperatures are recorded. Using the graph, calculate the specific heat capacity of the oil. (Specific heat capacity of water = 4.2 J g−1 °C−1)
- A2.0 J g−1 °C−1
- B8.8 J g−1 °C−1
- C4.2 J g−1 °C−1
- D0.48 J g−1 °C−1
The specific heat capacity of water is 4.2 J g−1 °C−1 and its latent heat of vaporisation is about 2400 J g−1. Evaporating 1.0 g of sweat removes about how many times as much heat as is needed to warm 1.0 g of water from 20 °C to 37 °C?
- Aabout 34
- Babout 570
- Cabout 140
- Dabout 0.03
Two leaves on the same plant in bright sunshine are compared. Leaf B is coated with grease on both surfaces, which blocks the stomata; leaf A is untreated. The chart shows the air temperature and the temperature of each leaf after one hour. What is the best explanation for the difference between leaf A and leaf B?
- Awater evaporating from leaf A takes in its latent heat of vaporisation, cooling the leaf; leaf B cannot transpire
- Bleaf A contains more water, and its high specific heat capacity stops the leaf temperature rising
- Cwater in leaf A dissolves heat-absorbing ions, which are carried away from the leaf in the xylem
- Dhydrogen bonds form as water evaporates from leaf A, releasing heat into the air around the leaf
Quiz key
Oxygen attracts the shared electrons in each O-H bond more strongly than hydrogen, giving O a partial negative charge and each H a partial positive charge; because the molecule is bent, the δ− end and the δ+ end are on different sides, so the molecule as a whole is polar. Water is a covalent molecule, not a pair of ions. The electronegativity is the other way round: oxygen pulls the electrons. A linear molecule would have the dipoles cancelling (like carbon dioxide), and the charges are partial, not full.
A single hydrogen bond is much weaker than a covalent bond, but in liquid water vast numbers are constantly breaking and re-forming, and together they explain water's high specific heat capacity and latent heat of vaporisation. The strong covalent O-H bonds are inside each molecule; hydrogen bonds are between molecules. Hydrogen bonds exist in liquid water as well as in ice; ice just has a fixed, regular arrangement. Two δ+ hydrogens repel each other; a hydrogen bond is between a δ+ hydrogen and a lone pair on a δ− oxygen.
Glucose is a small polar molecule whose OH groups form hydrogen bonds with water, and potassium chloride is ionic, so the δ− oxygens surround K+ and the δ+ hydrogens surround Cl−; both dissolve. A triglyceride is non-polar and hydrophobic, so it does not dissolve. Cellulose has many OH groups, but they are hydrogen-bonded to neighbouring chains in microfibrils, so it is insoluble even though sodium chloride is soluble. Starch is insoluble (which is why it is a good store), and a triglyceride is insoluble too.
Mass = heat ÷ latent heat = 120 kJ ÷ 2.4 kJ g−1 = 50 g. 288 g multiplies instead of dividing. 28.6 g divides by 4.2, the specific heat capacity of water, which is about warming water, not evaporating it. 0.020 g divides the latent heat by the heat, the wrong way round.
Because a lot of energy is needed to raise the temperature of water (much of it goes into breaking hydrogen bonds), the temperature of cells changes slowly when heat is gained or lost, so enzymes work at a fairly steady rate. Cooling by evaporation is due to the latent heat of vaporisation and happens at body surfaces, not inside cells. A high specific heat capacity means heat is absorbed with little temperature change, not that it is lost quickly. The property is about temperature stability, not about how well enzymes dissolve.
In 5 min the water rises from 20 to 30 °C, so energy supplied = 100 × 4.2 × 10 = 4200 J. The oil receives the same 4200 J and rises from 20 to 41 °C, a change of 21 °C, so c = 4200 ÷ (100 × 21) = 2.0 J g−1 °C−1. 8.8 inverts the ratio of the temperature rises (4.2 × 21 ÷ 10), forgetting that a bigger rise means a lower specific heat capacity. 4.2 assumes all liquids behave like water. 0.48 is just the ratio 10 ÷ 21, without multiplying by the specific heat capacity of water.
Warming: 1.0 × 4.2 × (37 − 20) = 71.4 J. Evaporating: 2400 J. Ratio = 2400 ÷ 71.4 = about 34, which is why evaporation of a small amount of water is such an effective way to cool. About 570 divides 2400 by 4.2, forgetting the 17 °C temperature rise. About 140 divides 2400 by 17, forgetting the specific heat capacity. About 0.03 is the ratio the wrong way round.
Leaf A (28 °C) is cooler than the air (30 °C), while leaf B (34 °C) is warmer. Transpiration from leaf A evaporates water, and each gram needs a large amount of energy (the latent heat of vaporisation) to break hydrogen bonds, taken from the leaf. Leaf B cannot lose water vapour, so it heats up in the sun. Both leaves hold similar amounts of water, and a high specific heat capacity only slows warming; it cannot make a leaf cooler than the air. Dissolved ions do not carry heat away as a cooling mechanism, and xylem carries water into the leaf. Hydrogen bonds are broken, not formed, when water evaporates, so energy is absorbed, not released.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.