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. All experimental data are fictional. Data: T(K) = T(°C) + 273.15; He 4.003 g mol⁻¹, Xe 131.29 g mol⁻¹.
Question 1 (multiple choice · foundation)
A copper block at 60 °C is placed in contact with an identical copper block at 20 °C. Which statement about the particles is correct at the moment they first touch?
- (A) Every atom in the 60 °C block is moving faster than every atom in the 20 °C block.
- (B) The atoms in the 60 °C block have a greater average kinetic energy than the atoms in the 20 °C block.
- (C) The atoms in the 20 °C block are not moving.
- (D) The 60 °C block contains heat and the 20 °C block contains cold.
Answer and explanation
Answer: (B). Temperature measures the average kinetic energy of the particles, so the warmer block has the greater average.
- (A) ignores the spread of energies. Some atoms in the warm block are slower than some atoms in the cool block; only the averages differ.
- (C) is wrong: particles in any sample above absolute zero are moving (in a solid, vibrating). 20 °C is 293 K.
- (D) treats heat and cold as substances. A body has a temperature and energy; heat is energy in transit, and “cold” does not exist as a thing that moves.
Question 2 (multiple choice · core)
A hot stone is dropped into a bucket of cool water and left until the temperatures stop changing. Which statement describes the system once thermal equilibrium is reached?
- (A) The particles in the stone and the water have stopped colliding.
- (B) The stone and the water have the same total energy.
- (C) Energy still passes between the stone and the water in collisions, but the net transfer is zero.
- (D) There is still a net transfer of energy from the stone to the water, but it is slower.
Answer and explanation
Answer: (C). At thermal equilibrium the average kinetic energies are equal. Collisions continue and pass energy both ways, but in equal amounts, so there is no net transfer and the temperatures stay constant.
- (A) Particles never stop moving or colliding; equilibrium is dynamic.
- (B) Equal temperature means equal average kinetic energy per particle. The bucket of water has many more particles, so it has much more total energy.
- (D) describes the approach to equilibrium, not equilibrium itself. If net energy still moved from the stone, its temperature would still be falling.
Question 3 (multiple choice · core)
A sealed flask contains a mixture of helium and xenon at 300 K. Which statement correctly compares the atoms of the two gases?
- (A) Average kinetic energy: He = Xe. Average speed: He > Xe.
- (B) Average kinetic energy: He = Xe. Average speed: He = Xe.
- (C) Average kinetic energy: He > Xe. Average speed: He > Xe.
- (D) Average kinetic energy: He < Xe. Average speed: He = Xe.
Answer and explanation
Answer: (A). The gases are at the same temperature, so their atoms have the same average kinetic energy. Because KE = ½mv², the lighter helium atoms must move faster on average: about √(131.29 / 4.003) = 5.7 times faster.
- (B) confuses equal kinetic energy with equal speed; that would only be true for particles of equal mass.
- (C) is wrong because temperature, not particle mass, sets the average kinetic energy.
- (D) wrongly assumes heavier atoms have more kinetic energy at the same temperature, and then that speeds are equal, which contradicts KE = ½mv².
Question 4 (multiple choice · core)
A sample of oxygen is heated from 27 °C to 177 °C. By what factor does the average kinetic energy of the molecules increase?
- (A) 0.667
- (B) 1.50
- (C) 2.25
- (D) 6.56
Answer and explanation
Answer: (B). Average kinetic energy is proportional to the Kelvin temperature: 27 °C = 300.15 K and 177 °C = 450.15 K, so the factor is 450.15 / 300.15 = 1.50.
- (A) is the ratio upside down (300.15 / 450.15). The temperature rose, so the kinetic energy must increase.
- (C) squares the ratio, as if kinetic energy were proportional to T².
- (D) divides the Celsius values (177 / 27). Celsius ratios have no meaning for kinetic energy.
Question 5 (constructed response · core)
A sealed metal can of a fictional drink, at 12.0 °C, is placed in an insulated tank of water at 46.0 °C. The temperatures are recorded with thermometers that read to ±0.1 °C.
| Time (min) | 0 | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|---|
| Drink (°C) | 12.0 | 29.7 | 36.2 | 38.6 | 39.5 | 39.8 | 39.9 |
| Tank water (°C) | 46.0 | 42.2 | 40.8 | 40.3 | 40.1 | 40.0 | 40.0 |
(a) State the direction of the net energy transfer. Justify your answer with the data. (b) Explain, in terms of the particles, how the energy is transferred. (c) The drink warms by 17.7 °C in the first 2 minutes but by only 0.3 °C between 8 and 10 minutes. Explain why. (d) At 12 minutes the two readings differ by 0.1 °C. Has thermal equilibrium been reached? Justify your answer.
Worked solution and suggested Marlbridge rubric
(a) From the tank water to the drink. The water’s temperature falls (46.0 → 40.0 °C) while the drink’s rises (12.0 → 39.9 °C).
(b) The water molecules have a greater average kinetic energy than the particles of the can and the drink. Water molecules collide with the particles of the can wall and, on average, pass energy to them; the can’s particles pass energy on to the drink’s particles by further collisions. So the average kinetic energy of the drink’s particles rises and that of the water molecules falls.
(c) At the start the temperature gap is 34.0 °C, so the difference in average kinetic energy is large and much more energy passes into the drink than back out. By 8 minutes the gap is only 0.6 °C; collisions pass nearly equal amounts of energy each way, so the net transfer, and the temperature change, is small.
(d) Yes, within the precision of the data. A difference of 0.1 °C is no larger than the ±0.1 °C precision of the thermometers, so the temperatures cannot be shown to differ; both are about 40.0 °C. (Also accept: “practically yes; the readings are still converging very slowly”, if linked to the precision.)
| Point | What earns it |
|---|---|
| 1 | Water → drink, supported by the water’s fall and the drink’s rise in temperature |
| 1 | Water molecules have a greater average kinetic energy than the drink’s particles |
| 1 | Energy passed by collisions (water → can wall → drink), on average from faster to slower particles |
| 1 | (c) Rate of net transfer depends on the size of the temperature (average kinetic energy) gap, which shrinks over time |
| 1 | (d) Equilibrium reached (or effectively reached) because the 0.1 °C difference is within the thermometer precision |
Do not award the second point for “the water has more heat”; the answer must refer to average kinetic energy of particles.
Question 6 (constructed response · core)
Two students discuss a large block of iron and a small iron nail, both at 25 °C, placed in contact on an insulated tray.
- Student P says: “Energy will flow from the large block to the nail, because the block has much more energy.”
- Student Q says: “Once two objects reach thermal equilibrium, their particles stop colliding, so nothing moves between them.”
(a) Evaluate Student P’s claim. (b) Evaluate Student Q’s claim.
Worked solution and suggested Marlbridge rubric
(a) The claim is incorrect. The block does have more total energy, because it has many more atoms. But the direction of heat transfer depends on temperature, not total energy. Both objects are at 25 °C, so their atoms have the same average kinetic energy. In collisions at the contact surface, energy passes each way in equal amounts, so there is no net transfer.
(b) The claim is incorrect. At thermal equilibrium the atoms are still moving and still colliding with each other across the contact surface. Energy is still passed in individual collisions, but the energy passed from block to nail equals the energy passed from nail to block, so there is no net transfer. The temperatures stay constant because the transfers balance, not because they stop.
| Point | What earns it |
|---|---|
| 1 | (a) Incorrect: direction of transfer is set by temperature (average kinetic energy), not total energy |
| 1 | (a) Same temperature → same average kinetic energy → no net transfer |
| 1 | (b) Incorrect: particles keep moving and colliding at equilibrium |
| 1 | (b) Energy is exchanged in both directions in equal amounts, so the net transfer is zero |
Question 7 (constructed response · stretch)
An insulated box is divided into two halves by a thin metal wall. The left half holds gas A at 350 K; the right half holds gas B at 250 K. The gases cannot mix.
(a) Calculate the ratio of the average kinetic energy of the particles of A to that of B at the start. (b) Explain how energy moves from gas A to gas B even though the gas particles never meet. (c) Sketch the temperature of each gas against time. Describe the shape of each curve and where they end. (d) What is the ratio in (a) once thermal equilibrium is reached? Explain.
Worked solution and suggested Marlbridge rubric
(a) Average kinetic energy ∝ Kelvin temperature, so the ratio is 350 / 250 = 1.40.
(b) Gas A molecules collide with the atoms on the left face of the wall and, on average, pass energy to them, because A’s particles have the greater average kinetic energy. The wall’s atoms pass energy through the wall by colliding (vibrating against) their neighbours. On the right face, the wall’s atoms collide with gas B molecules and pass energy to them. The wall carries the energy across by collisions at every stage.
(c) Temperature on the vertical axis (K), time on the horizontal axis. Gas A’s curve starts at 350 K and falls, steeply at first and then more slowly. Gas B’s curve starts at 250 K and rises, steeply at first and then more slowly. Both level off at the same temperature, somewhere between 250 K and 350 K, and then stay flat.
(d) The ratio becomes 1.00. At thermal equilibrium the temperatures are equal, so the average kinetic energies of the two gases’ particles are equal.
| Point | What earns it |
|---|---|
| 1 | (a) 1.40, using Kelvin temperatures |
| 1 | (b) A’s particles pass energy to the wall’s particles by collisions, from higher to lower average kinetic energy |
| 1 | (b) Energy passed through the wall and on to B’s particles by further collisions |
| 1 | (c) A falls and B rises, each changing quickly at first and then more slowly |
| 1 | (c) Both curves end flat at the same temperature, between 250 K and 350 K |
| 1 | (d) 1.00, because equal temperatures mean equal average kinetic energies |
The exact final temperature cannot be found without knowing how much of each gas there is and what it is; any value strictly between 250 K and 350 K earns the fifth point.
How did you do?
- Q1 or Q3 wrong: re-read “What temperature means for particles” and “Same temperature, different particles” in the study guide.
- Q2 or Q6 wrong: go back to “Reaching thermal equilibrium”: collisions continue, only the net transfer stops.
- Q4 or Q7(a) wrong: work through Worked example 2 again and always convert to kelvin.
- Q5 or Q7(b)–(c) incomplete: compare your answer with Worked examples 1 and 3 and with Figure 2.
Then tick off the topic checklist.