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: H 1.008, O 16.00 g mol⁻¹; for water, ΔH_fus = 6.01 kJ mol⁻¹ at 0 °C, ΔH_vap = 40.7 kJ mol⁻¹ at 100 °C and c(liquid) = 4.18 J g⁻¹ °C⁻¹. Substances with single-letter names (J, K, L, G, X) are fictional. Assume constant pressure throughout.
Question 1 (multiple choice · foundation)
A pure solid is heated steadily. At its melting point the temperature stays constant for several minutes, even though heating continues. Which statement explains this?
- (A) The heater switches off automatically at the melting point.
- (B) The energy supplied increases the average kinetic energy of the particles, but some energy is lost to the air.
- (C) The energy supplied is used to overcome attractions between particles, so their average kinetic energy does not change.
- (D) The energy of the system stays constant until all of the solid has melted.
Answer and explanation
Answer: (C). While the solid melts, the energy goes into separating particles from their neighbours, which raises their potential energy. Temperature reflects average kinetic energy, which stays the same until melting is complete.
- (A) The heater is still supplying energy; the question says heating continues.
- (B) If the average kinetic energy increased, the temperature would rise. Heat loss alone does not explain a perfectly flat section.
- (D) The energy of the system increases throughout melting; only the temperature is constant.
Question 2 (multiple choice · core)
What is the energy change when 45.0 g of steam condenses to liquid water at 100 °C?
- (A) 15.0 kJ released
- (B) 102 kJ released
- (C) 102 kJ absorbed
- (D) 1.83 × 10³ kJ released
Answer and explanation
Answer: (B). n = 45.0 g ÷ 18.016 g mol⁻¹ = 2.498 mol. Condensing is the reverse of vaporizing, so q = 2.498 mol × (−40.7 kJ mol⁻¹) = −101.7 kJ: 102 kJ is released.
- (A) uses ΔH_fus (2.498 × 6.01 = 15.0 kJ). Condensing is a gas-to-liquid change, so ΔH_vap applies.
- (C) has the right size but the wrong direction. Condensing forms attractions and is exothermic.
- (D) multiplies the mass in grams by ΔH (45.0 × 40.7), skipping the conversion to moles.
Question 3 (multiple choice · core)
The table gives data for four fictional liquids at their boiling points. Which liquid needs the most energy to vaporize a 10.0 g sample?
| Liquid | Molar mass (g mol⁻¹) | ΔH_vap (kJ mol⁻¹) |
|---|---|---|
| J | 30.0 | 24.0 |
| K | 60.0 | 42.0 |
| L | 18.0 | 18.9 |
| G | 120.0 | 54.0 |
- (A) J
- (B) K
- (C) L
- (D) G
Answer and explanation
Answer: (C). The samples have equal mass, so find the moles first, then q = n × ΔH_vap.
| Liquid | n in 10.0 g (mol) | q (kJ) |
|---|---|---|
| J | 0.333 | 8.0 |
| K | 0.167 | 7.0 |
| L | 0.556 | 10.5 |
| G | 0.0833 | 4.5 |
- (D) has the largest ΔH_vap per mole, which is the trap. Because its molar mass is so large, 10.0 g is only a small amount in moles.
- (A) and (B) need less energy than L because their energy per gram (0.80 and 0.70 kJ g⁻¹) is lower than L’s (1.05 kJ g⁻¹).
Question 4 (multiple choice · core)
A 0.250 mol sample of a pure solid is heated at a steady 100 J s⁻¹. Its temperature stays at its melting point for 60.0 s. What is its molar enthalpy of fusion?
- (A) 1.50 kJ mol⁻¹
- (B) 6.00 kJ mol⁻¹
- (C) 24.0 kJ mol⁻¹
- (D) 2.40 × 10⁴ kJ mol⁻¹
Answer and explanation
Answer: (C). Energy supplied during the flat section: q = 100 J s⁻¹ × 60.0 s = 6000 J = 6.00 kJ. This melts 0.250 mol, so ΔH_fus = 6.00 kJ ÷ 0.250 mol = 24.0 kJ mol⁻¹.
- (A) multiplies by the moles (6.00 × 0.250) instead of dividing.
- (B) is the energy for the whole sample, not per mole.
- (D) is the value in J mol⁻¹ (24 000) with the unit wrongly written as kJ mol⁻¹.
Question 5 (calculation · core)
A student’s hand is splashed with 20.0 g of steam at 100.0 °C. The steam condenses and the water cools to 45.0 °C on the skin.
(a) Calculate the total energy released by the steam. (b) Calculate the energy released if the hand were instead splashed with 20.0 g of liquid water at 100.0 °C that cooled to 45.0 °C. (c) Use your answers to explain why steam causes worse burns than boiling water.
Worked solution and suggested Marlbridge rubric
(a) Two steps.
- Condensing at 100.0 °C: n = 20.0 ÷ 18.016 = 1.110 mol; q = 1.110 mol × (−40.7 kJ mol⁻¹) = −45.18 kJ.
- Cooling the liquid from 100.0 °C to 45.0 °C: q = 20.0 g × 4.18 J g⁻¹ °C⁻¹ × (45.0 − 100.0) °C = −4598 J = −4.598 kJ.
Total: −45.18 + (−4.598) = −49.8 kJ, so 49.8 kJ is released.
(b) Only step 2 happens: 4.60 kJ released (q = −4.60 kJ).
(c) The steam releases about 11 times as much energy (49.8 ÷ 4.60 ≈ 10.8), almost all of it from condensing, which happens at 100 °C on the skin. More energy transferred to the skin causes more damage.
| Point | What earns it |
|---|---|
| 1 | Moles of water and the condensation energy (45.2 kJ) with a negative sign or “released” |
| 1 | Cooling step 4.60 kJ (4598 J) using q = mcΔT |
| 1 | Total 49.8 kJ, with J converted to kJ before adding |
| 1 | Part (c): links the extra energy to condensation releasing energy on the skin |
Accept 49.7–49.8 kJ depending on rounding. Do not award the third point if 45.18 and 4598 are added directly.
Question 6 (constructed response · core)
A 0.400 mol sample of fictional substance X is heated at constant pressure. The table shows its temperature after different amounts of energy have been added. Each phase change starts and finishes exactly at one of the readings shown.
| Energy added (kJ) | 0 | 0.6 | 1.6 | 2.6 | 4.35 | 6.1 | 12.1 | 18.1 | 18.5 |
|---|---|---|---|---|---|---|---|---|---|
| Temperature (°C) | −20.0 | −5.0 | −5.0 | −5.0 | 30.0 | 65.0 | 65.0 | 65.0 | 75.0 |
(a) State the melting point and the boiling point of X. (b) Calculate ΔH_fus and ΔH_vap for X. (c) Explain, in terms of the particles, why ΔH_vap is larger than ΔH_fus. (d) Give the energy change when all 0.400 mol of gaseous X condenses at 65.0 °C.
Worked solution and suggested Marlbridge rubric
(a) The temperature is constant at −5.0 °C (melting point) and at 65.0 °C (boiling point).
(b) Melting lasts from 0.6 kJ to 2.6 kJ: 2.0 kJ for 0.400 mol, so ΔH_fus = 2.0 ÷ 0.400 = 5.0 kJ mol⁻¹. Boiling lasts from 6.1 kJ to 18.1 kJ: 12.0 kJ for 0.400 mol, so ΔH_vap = 12.0 ÷ 0.400 = 30.0 kJ mol⁻¹.
(c) When X melts, the particles move apart only slightly and are still in contact, attracting their neighbours. When X boils, the particles separate completely, so nearly all of the attractions between them must be overcome. More attractions overcome per mole means more energy per mole.
(d) Condensing is the reverse of boiling: q = −12.0 kJ (12.0 kJ released; ΔH_cond = −30.0 kJ mol⁻¹).
| Point | What earns it |
|---|---|
| 1 | Both temperatures correct, identified from the constant readings |
| 1 | ΔH_fus = 5.0 kJ mol⁻¹, dividing the energy of the flat section by 0.400 mol |
| 1 | ΔH_vap = 30.0 kJ mol⁻¹, with the same method |
| 1 | Part (c): particles still attract in the liquid but are fully separated in the gas |
| 1 | Part (d): −12.0 kJ, with a negative sign or “released” |
Do not award the second or third point if the energy of the flat section is given as the molar enthalpy without dividing by the moles.
Question 7 (constructed response · stretch)
A student adds 50.0 g of ice at 0.0 °C to 250. g of a drink at 25.0 °C in an insulated cup. In a second cup, the student adds 50.0 g of liquid water at 0.0 °C to another 250. g of the drink at 25.0 °C. Treat the drink as water, and assume no energy is exchanged with the cup or the air.
(a) Show that the drink can supply enough energy to melt all of the ice. (b) Calculate the final temperature in each cup. (c) Explain why the ice cools the drink much more than the same mass of water at the same temperature.
Model answer and suggested Marlbridge rubric
(a) Energy to melt the ice: n = 50.0 ÷ 18.016 = 2.775 mol; q = 2.775 × 6.01 = 16.68 kJ. Energy the drink could give out by cooling all the way to 0.0 °C: 250. × 4.18 × 25.0 = 26 125 J = 26.1 kJ. Since 26.1 kJ > 16.7 kJ, all of the ice melts and the final temperature is above 0 °C.
(b) Cup with ice. Energy lost by the drink = energy to melt the ice + energy to warm the melted ice from 0.0 °C to T: 250. × 4.18 × (25.0 − T) = 16 680 + 50.0 × 4.18 × T 26 125 − 1045 T = 16 680 + 209 T, so T = 9445 ÷ 1254 = 7.5 °C.
Cup with cold water. 250. × 4.18 × (25.0 − T) = 50.0 × 4.18 × T, so T = (250. × 25.0) ÷ 300. = 20.8 °C.
(c) In both cups the cold water must be warmed from 0 °C. The ice must also melt first, and melting absorbs 6.01 kJ for every mole without any rise in its temperature. That extra energy comes from the drink, so the drink cools much further.
| Point | What earns it |
|---|---|
| 1 | Part (a): compares 16.7 kJ needed with 26.1 kJ available |
| 1 | Sets up an energy balance that includes the melting term and warming the melted ice |
| 1 | Final temperature with ice, about 7.5 °C |
| 1 | Final temperature with water, about 20.8 °C |
| 1 | Part (c): the extra energy absorbed by melting at constant temperature comes from the drink |
Accept 7.5–7.6 °C depending on rounding of the melting energy.
How did you do?
- Q1 or Q6(c) wrong: re-read “Why the temperature stays constant” and “Why ΔH_vap is much larger than ΔH_fus” in the study guide.
- Q2, Q3 or Q5 wrong: check you converted grams to moles before using ΔH, and J to kJ before adding.
- Q4 or Q6(b) wrong: practise Worked example 3; remember to divide the flat-section energy by the moles.
- Q7 incomplete: set out every step as in Worked example 2, then write one energy balance.
Then tick off the topic checklist.