These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: specific heat capacity of liquid water, c = 4.18 J g⁻¹ °C⁻¹. A calculator is assumed. Keep full calculator values until the last step, then round to the significant figures of the data. All data sets are fictional.
Question 1 (multiple choice · foundation)
Sodium chloride dissolves readily in water, but cooking oil does not. Which statement best explains this?
- (A) The δ⁻ oxygen ends of water molecules surround Na⁺ ions and the δ⁺ hydrogen ends surround Cl⁻ ions; nonpolar oil molecules cannot form such attractions with water.
- (B) Water breaks the covalent bonds inside sodium chloride, but it cannot break the covalent bonds inside oil.
- (C) The δ⁺ hydrogen ends of water molecules surround Na⁺ ions, which pulls the ions apart; oil has no ions.
- (D) Oil is denser than water, so it sinks and cannot mix.
Answer and explanation
Answer: (A). Opposite charges attract. The partly negative oxygen faces the positive sodium ion; the partly positive hydrogens face the negative chloride ion. These hydration shells keep the ions apart. Oil is nonpolar, so water molecules stay hydrogen-bonded to each other instead.
- (B) treats sodium chloride as covalent. It is ionic, and dissolving separates ions; it does not break covalent bonds.
- (C) gets the orientation backwards: like partial charges (δ⁺ and Na⁺) repel, so hydrogens do not face Na⁺.
- (D) confuses floating with dissolving, and is also factually wrong: oil floats on water because it is less dense.
Question 2 (multiple choice · core)
How much energy is needed to warm 150 g of water from 15.0 °C to 25.0 °C?
- (A) 41.8 J
- (B) 359 J
- (C) 6.27 kJ
- (D) 15.7 kJ
Answer and explanation
Answer: (C). ΔT = 25.0 − 15.0 = 10.0 °C. q = m × c × ΔT = 150 g × 4.18 J g⁻¹ °C⁻¹ × 10.0 °C = 6270 J = 6.27 kJ.
- (A) leaves out the mass: 4.18 × 10.0 = 41.8 J is the energy for 1 g only.
- (B) divides by c instead of multiplying: 150 × 10.0 ÷ 4.18 = 359. The units would not come out as joules.
- (D) uses the final temperature (25.0 °C) as ΔT: 150 × 4.18 × 25.0 = 15 675 J.
Question 3 (multiple choice · core)
Which statement best explains why ice floats on liquid water?
- (A) Ice contains trapped bubbles of air, which lower its overall density.
- (B) In ice, hydrogen bonds hold each molecule in an open lattice, so the molecules are farther apart on average than in the liquid.
- (C) Freezing lengthens the covalent O–H bonds inside each water molecule, so each molecule takes up more space.
- (D) Ice has fewer hydrogen bonds than liquid water, so its molecules spread out.
Answer and explanation
Answer: (B). In ice every molecule is hydrogen-bonded to four neighbours at fixed angles. This open arrangement takes up more space than the liquid, so ice is less dense and floats.
- (A) comes from everyday cloudy ice cubes. Pure, clear ice with no bubbles still floats.
- (C) confuses covalent bonds within a molecule with hydrogen bonds between molecules. The density change comes from how molecules are arranged.
- (D) reverses the facts: ice has more stable hydrogen bonds than the liquid, and that is what holds the lattice open.
Question 4 (graph · core)
A student places 100 g of water and 100 g of a liquid labelled Y in identical beakers under the same lamp. She records the temperature every 2 minutes.
| Time / min | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| Water / °C | 20.0 | 21.5 | 23.0 | 24.5 | 26.0 |
| Liquid Y / °C | 20.0 | 23.0 | 26.0 | 29.0 | 32.0 |
(a) Calculate the rate of temperature rise of each liquid, in °C min⁻¹. (b) Calculate the energy absorbed by the water in 8 minutes. (c) Assuming liquid Y absorbed the same energy, calculate its specific heat capacity. (d) State which liquid would better buffer an organism against temperature change, and state one assumption you made in (c).
Worked solution and suggested Marlbridge rubric
(a) Water: (26.0 − 20.0) ÷ 8 = 6.0 ÷ 8 = 0.75 °C min⁻¹. Liquid Y: (32.0 − 20.0) ÷ 8 = 12.0 ÷ 8 = 1.5 °C min⁻¹.
(b) q = 100 g × 4.18 J g⁻¹ °C⁻¹ × 6.0 °C = 2508 J ≈ 2.5 kJ.
(c) c(Y) = q ÷ (m × ΔT) = 2508 J ÷ (100 g × 12.0 °C) = 2.09 J g⁻¹ °C⁻¹ (about 2.1). Alternative method: same energy and mass, so c is inversely proportional to the gradient: 4.18 × 0.75 ÷ 1.5 = 2.09.
(d) Water: its higher specific heat means a smaller temperature change for the same energy. Assumptions (any one): both beakers absorbed equal energy from the lamp; no energy was lost to the air, the beaker or by evaporation; both liquids stayed liquid.
| Point | What earns it |
|---|---|
| 1 | Both gradients correct with units |
| 1 | 2.5 kJ (or 2508 J) for the water, using ΔT = 6.0 °C |
| 1 | c(Y) ≈ 2.1 J g⁻¹ °C⁻¹ by either method |
| 1 | Water chosen with a reason linked to specific heat, plus a valid assumption |
Accept an answer to (c) carried forward from an error in (b).
Question 5 (constructed response · core)
A student wants to test whether air humidity affects the rate at which a leafy shoot takes up water. She has a potometer, clear plastic bags, a water spray and leafy shoots of one species.
(a) State a testable hypothesis. (b) Identify the independent variable, the dependent variable and an appropriate control (baseline) condition. (c) Give two variables she must keep constant, and why. (d) Predict the result and justify the prediction using cohesion and evaporation.
Model answer and suggested Marlbridge rubric
(a) “Increasing the humidity of the air around a leafy shoot decreases its rate of water uptake.” (A hypothesis predicting an increase is acceptable as a statement, but the prediction in (d) must be justified.)
(b) Independent variable: humidity around the leaves (normal room air versus air inside a bag sprayed with water). Dependent variable: rate of water uptake, mL h⁻¹, measured with the potometer. Control: the same shoot or matched shoots in normal room air (no bag), or in a dry bag so that the bag itself is not the only difference.
(c) Any two with a reason: light intensity (affects how open the stomata are); temperature (affects evaporation rate); leaf area (more surface, more evaporation); species and age of shoot; air movement.
(d) Uptake should be lower in humid air. With more water vapour outside the leaf, the difference in water vapour concentration between the inside of the leaf and the air is smaller, so less water evaporates from the stomata. Less evaporation means less pull on the water column, which is linked by cohesion (hydrogen bonds between water molecules) all the way down the xylem. So less water is drawn in at the cut stem.
| Point | What earns it |
|---|---|
| 1 | Testable hypothesis naming humidity and water uptake |
| 1 | Independent and dependent variables correct and a sensible baseline condition |
| 1 | Two valid controlled variables, each with a reason |
| 1 | Prediction linked to less evaporation and to cohesion transmitting the pull |
Accept other ways of raising humidity, for example a humidifier, if the reasoning is complete. Do not award the last point for “cohesion is stronger in humid air”: humidity changes evaporation, not cohesion.
Question 6 (constructed response · stretch)
A fictional survey recorded the mean daily temperature range (the difference between the daily maximum and minimum) in six towns at different distances from the sea.
| Town | P | Q | R | S | T | U |
|---|---|---|---|---|---|---|
| Distance from sea / km | 1 | 5 | 12 | 25 | 40 | 60 |
| Mean daily range / °C | 6.1 | 6.8 | 8.0 | 9.4 | 10.9 | 12.2 |
A student claims: “This proves that the sea’s high specific heat causes smaller temperature swings.”
(a) Describe the trend, using numbers from the table. (b) Explain why the trend is consistent with water’s high specific heat. (c) Evaluate whether the data prove the claim. Name one possible confounding variable.
Model answer and suggested Marlbridge rubric
(a) As distance from the sea increases, the daily temperature range increases, from 6.1 °C at 1 km to 12.2 °C at 60 km, a range twice as large (12.2 ÷ 6.1 = 2.0).
(b) Water needs a lot of energy to change temperature, because energy goes into breaking hydrogen bonds. So the sea warms little by day and cools little by night, and it moderates the temperature of the air over nearby land.
(c) The data show a correlation, so they support the claim but do not prove it. Towns farther inland may differ in other ways that also affect the temperature range, such as altitude, cloud cover or air humidity (drier inland air lets the ground lose heat faster at night). Only six towns were measured. A stronger test would compare towns at the same altitude and latitude, or measure directly how much the sea’s temperature changes compared with the land’s.
| Point | What earns it |
|---|---|
| 1 | Trend described with at least two quoted values |
| 1 | Mechanism: high specific heat linked to hydrogen bonds and to smaller changes near the sea |
| 1 | States correlation is not proof of cause |
| 1 | Names a plausible confounding variable and explains how it could affect the range |
Question 7 (calculation · stretch)
During a long run, an athlete loses 500 g of sweat, all of which evaporates from the skin. Take the heat of vaporization of water at skin temperature as 2.4 kJ g⁻¹, the athlete’s mass as 70 kg and the average specific heat of the human body as 3.5 J g⁻¹ °C⁻¹.
(a) Calculate the energy removed from the body by evaporation. (b) If this energy had stayed in the body instead, by how much would body temperature have risen? (c) Explain at the molecular level why evaporation cools the skin.
Worked solution
(a) Energy = 500 g × 2.4 kJ g⁻¹ = 1200 kJ (1.2 × 10⁶ J).
(b) Convert: 70 kg = 70 000 g; 1200 kJ = 1 200 000 J. ΔT = q ÷ (m × c) = 1 200 000 J ÷ (70 000 g × 3.5 J g⁻¹ °C⁻¹) = 1 200 000 ÷ 245 000 = 4.90 °C, so about 4.9 °C.
(c) To escape as vapour, a water molecule must break all of its hydrogen bonds to its neighbours. Only the most energetic molecules can do this. When they leave, the average kinetic energy of the remaining molecules falls, so the sweat, and the skin under it, cools.
Suggested mark points (3): 1 for 1200 kJ; 1 for about 4.9 °C with consistent units (g with J); 1 for linking escape to breaking hydrogen bonds and loss of the most energetic molecules.
Common errors: forgetting to convert kg to g gives 4900 °C; using c = 4.18 (water) instead of the body value gives 4.1 °C, which uses the wrong data. A rise of nearly 5 °C would be dangerous, which shows how much sweating matters.
How did you do?
- Q1 or Q3 wrong: re-read “Hydrogen bonds” and “Water as a solvent” in the study guide, and look again at Figure 1.
- Q2, Q4 or Q7 calculations wrong: rework Worked example 1; check you used ΔT, not the final temperature, and grams throughout.
- Q4 graph reading wrong: re-read “Reading Figure 2” in the study guide: a smaller gradient for the same energy means a larger specific heat.
- Q5 or Q6 incomplete: re-read Worked example 2, especially the parts on controls and on correlation versus causation.
Then tick off the topic checklist.