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HKDSE Physics Recap: Internal Energy, Heat Capacity, Latent Heat, Change of State and Evaporation | Tia (2026)

HKDSE Physics — Internal Energy, Heat Capacity, Latent Heat, Change of State and Evaporation

Subject: Physics

Tia (2026)

Issue date: Friday, 9 October 2026

Total: 70 marks  |  Suggested time: 90–100 minutes  |  Data: specific heat capacity of water = 4200 J kg−1 °C−1; specific latent heat of fusion of ice = 3.34 × 105 J kg−1; specific latent heat of vaporisation of water = 2.26 × 106 J kg−1. Energy exchange with the surroundings is to be ignored only when the question says so.

01

Section A — Heat and Internal Energy

Multiple Choice

Each question has one correct answer. Choose the best option. (1 mark each)

  1. 1 kg of steam at 100 °C and 1 kg of water at 100 °C are compared. Which of the following statements is correct?
    1. The molecules of the steam have a greater average kinetic energy and the same total potential energy as those of the water.
    2. The molecules of the steam have the same average kinetic energy as those of the water, but a greater total potential energy.
    3. The molecules of the steam have the same average kinetic energy and the same total potential energy as those of the water.
    4. The molecules of the steam have a greater average kinetic energy and a smaller total potential energy than those of the water.
  2. Two solid blocks P and Q are at the same initial temperature. The same amount of energy is supplied to each block and no energy is lost. The mass of P is one-third of the mass of Q, and the rise in temperature of P is twice that of Q. What is the value of specific heat capacity of Pspecific heat capacity of Q ?
    1. 2/3
    2. 6
    3. 1/6
    4. 3/2
  3. A metal block of mass 0.20 kg at 100 °C is dropped into 0.30 kg of water at 20 °C in a well-insulated container of negligible heat capacity. The final steady temperature is 25 °C. What is the specific heat capacity of the metal?
    1. 84 J kg−1 °C−1
    2. 394 J kg−1 °C−1
    3. 420 J kg−1 °C−1
    4. 1260 J kg−1 °C−1
  4. A heater supplies energy at a constant rate of 600 W to 2.0 kg of a liquid for 100 s. The temperature of the liquid rises by 10 °C. 20% of the energy supplied by the heater is lost to the surroundings. What is the specific heat capacity of the liquid?
    1. 1920 J kg−1 °C−1
    2. 2400 J kg−1 °C−1
    3. 3000 J kg−1 °C−1
    4. 3600 J kg−1 °C−1
  5. A 300 W heater melts 0.45 kg of a solid which is already at its melting point. The solid takes 5.0 minutes to melt completely, and 10% of the energy supplied by the heater is lost to the surroundings. What is the specific latent heat of fusion of the solid?
    1. 1.8 × 105 J kg−1
    2. 2.0 × 105 J kg−1
    3. 2.2 × 105 J kg−1
    4. 1.6 × 105 J kg−1
  6. 0.10 kg of ice at 0 °C is added to 0.50 kg of water at 30 °C in a well-insulated container of negligible heat capacity. What is the final temperature of the contents?
    1. 11.7 °C
    2. 15 °C
    3. 25 °C
    4. 0 °C
  7. Which of the following would increase the rate of evaporation of water from a wet towel?
    (1) spreading the towel out so that its exposed surface area is larger
    (2) hanging the towel in a draught of dry air
    (3) keeping the towel in a closed room of very high humidity, at the same temperature
    1. (1) and (3) only
    2. (2) and (3) only
    3. (1) only
    4. (1) and (2) only
  8. An air conditioner is running steadily. Let H1 be the heat absorbed per second by the refrigerant in the evaporator, E the energy supplied per second to the compressor, and H2 the heat released per second by the refrigerant in the condenser. Assuming that there is no other energy loss, which of the following is correct?
    1. H2 = H1
    2. H2 < H1
    3. H2 > H1
    4. H2 = 0
  9. Which of the following correctly describes the functions of the compressor and the expansion valve in an air conditioner?
    1. The compressor raises the pressure and temperature of the refrigerant vapour so that heat can be released to the outdoor air. The expansion valve lowers the pressure of the liquid refrigerant so that it can evaporate at a low temperature.
    2. The compressor lowers the pressure of the refrigerant vapour so that it condenses. The expansion valve raises the temperature of the liquid refrigerant so that it can release heat.
    3. The compressor condenses the refrigerant vapour into liquid. The expansion valve evaporates the liquid refrigerant.
    4. The compressor cools the refrigerant vapour by evaporation. The expansion valve heats the refrigerant by compression.
  10. A solid sample is heated by a heater of constant power. Energy losses are negligible. Fig. 1 shows how the temperature of the sample changes with time. What is the value of specific heat capacity of the sample in the liquid statespecific heat capacity of the sample in the solid state ?
    0 20 40 60 80 100 120 0 3 6 9 12 15 18 Time / min Temperature / °C
    Fig. 1 — Temperature–time graph for a sample heated at constant power
    1. 1/3
    2. 1.5
    3. 3
    4. 2
1
2
3
4
5
6
7
8
9
10
10 marks
02

Section B — Heat and Internal Energy

True or False

For each statement, state whether it is True or False. If it is false, rewrite the statement correctly. (1 mark each: the mark for a false statement is awarded only if the rewritten statement is also correct.)

  1. The internal energy of a body is the sum of the kinetic energy and the potential energy of all its molecules. Therefore, when a ball is thrown through the air, the internal energy of the ball increases as its speed increases.
  2. When a solid melts at its melting point, the energy absorbed increases the potential energy of the molecules, while the average random kinetic energy of the molecules remains unchanged.
  3. Two iron blocks, of masses 1 kg and 2 kg, are both at 50 °C. They have the same internal energy, because their molecules have the same average kinetic energy.
  4. The heat capacity of a body depends on both its mass and the material it is made of, whereas the specific heat capacity depends only on the material (in a given state).
  5. When a hot body is placed in contact with a cold body, the energy lost by the hot body is always equal to the energy gained by the cold body, whether or not energy is exchanged with the surroundings.
  6. The specific latent heat of vaporisation of water is much greater than its specific latent heat of fusion, because in vaporisation the molecules are completely separated from one another against the intermolecular forces, and the substance also does work in expanding against the atmosphere.
  7. Evaporation takes place only when the temperature of a liquid has reached its boiling point, whereas boiling takes place at the surface of the liquid at any temperature.
  8. During evaporation, the molecules that escape from the surface of a liquid are, on average, those with higher kinetic energy. Therefore, if no energy is supplied, the liquid left behind becomes cooler.
  9. In an air conditioner, the compressor raises the temperature of the refrigerant vapour above the temperature of the outdoor air, so that heat can flow from the refrigerant to the outdoor air as the vapour condenses.
  10. An air conditioner, with both its indoor and outdoor parts inside a closed and thermally insulated room, runs continuously. The temperature of the room falls steadily because heat is removed from the room air.
10 marks
03

Section C — Comprehensive Structured Question 1

Internal Energy: Molecular Kinetic Energy and Potential Energy

A student investigates how the internal energy of water can be increased by doing work on it. In Fig. 2, a paddle is driven by a mechanical drive and stirs 0.50 kg of water in a thermally insulated container. The drive does work on the water through the paddle at a constant rate of 75 W for 10 minutes. The heat capacity of the container and the paddle together is 150 J °C−1. The temperature of the water is measured by a thermometer.

drive (75 W) water (0.50 kg) paddle thermometer thermally insulated container (hatched = insulation)
Fig. 2 — Apparatus for increasing the internal energy of water by doing work (not drawn to scale)

(a) State what is meant by the internal energy of a substance. (1 mark)

(b) Assume that no energy is lost to the surroundings.

  1. Describe the energy transfer in the system. Explain, in terms of the motion of the water molecules, why the temperature of the water rises. (2 marks)
  2. Calculate the rise in temperature of the water after 10 minutes. (3 marks)

(c) In the actual experiment, the temperature of the water rises by only 16.0 °C after 10 minutes. Calculate the percentage of the energy supplied by the drive that is transferred to the surroundings. (2 marks)

(d) One beaker contains 0.20 kg of ice at 0 °C. Another beaker contains 0.20 kg of water at 0 °C. State which one has the greater internal energy. Explain your answer in terms of the kinetic energy and the potential energy of the molecules. (2 marks)

10 marks
04

Section C — Comprehensive Structured Question 2

Heat Capacity: Heating and Mixing

A metal block of mass 0.80 kg is fitted with a heater and a thermometer. The heater supplies energy to the block at a constant rate of 48 W. The block is wrapped in a thick layer of insulating material. The heater is switched on at t = 0. The room temperature is 20.0 °C. The heat capacity of the heater and the thermometer is negligible. Table 1 shows the temperature of the block at different times.

Table 1

Time t / s 0 60 120 180 240 300
Temperature of block / °C 20.0 23.7 27.1 30.2 33.0 35.6

(a) State what is meant by the specific heat capacity of a substance. (1 mark)

(b) Assume that energy loss to the surroundings is negligible during the first 60 s. Use the data in Table 1 to calculate the specific heat capacity of the metal. (2 marks)

(c) The temperature rise of the block in each successive 60 s interval becomes smaller, although the rate at which the heater supplies energy is constant. Explain why. (2 marks)

(d) Using the result of (b), estimate the energy lost to the surroundings between t = 240 s and t = 300 s. (2 marks)

(e) The heater is then switched off. The block, at a temperature of 90.0 °C, is quickly transferred into 0.250 kg of a liquid X at 20.0 °C in a well-insulated container of negligible heat capacity. The final steady temperature is 60.0 °C. Using the result of (b), calculate the specific heat capacity of X. Assume that no energy is lost to the surroundings. (3 marks)

10 marks
05

Section C — Comprehensive Structured Question 3

Specific Latent Heat: Ice, Water and Steam

In this question, assume that no energy is exchanged with the surroundings and that the heat capacity of each container is negligible unless stated otherwise. The specific heat capacity of ice is 2100 J kg−1 °C−1.

(a) State what is meant by the specific latent heat of vaporisation of a substance. (1 mark)

(b) Three ice cubes, each of mass 20 g and at a temperature of −5.0 °C, are added to 0.250 kg of a drink at 25.0 °C in an insulated container. The drink may be treated as water.

  1. By calculation, show that all the ice melts. (2 marks)
  2. Calculate the final temperature of the drink. (2 marks)

(c) Steam at 100 °C is passed into 0.50 kg of water at 20.0 °C contained in a vessel of heat capacity 200 J °C−1, until the temperature of the water and the vessel reaches 60.0 °C. Calculate the mass of steam that has condensed. (3 marks)

(d) A scald caused by steam at 100 °C is generally more serious than a scald caused by the same mass of water at 100 °C. Explain why. (2 marks)

10 marks
06

Section C — Comprehensive Structured Question 4

Change of State: A Heating Curve

A solid substance Z of mass 0.20 kg, initially at 20 °C, is heated by a heater of power 400 W which is fully immersed in it. Fig. 3 shows how the temperature of Z varies with time. The temperature of Z remains constant between B and C, and between D and E.

0 20 40 60 80 100 120 0 100 200 300 400 500 600 40 140 560 Time / s Temperature / °C A B C D E
Fig. 3 — Variation of the temperature of Z with time

(a) State the state(s) of matter in which Z exists during each of the stages AB, BC, CD and DE. (2 marks)

(b) Describe how the average kinetic energy and the total potential energy of the molecules of Z change during AB and during BC. Hence explain why the temperature of Z does not change during BC although energy is continuously supplied. (2 marks)

In parts (c) and (d), assume that all the energy supplied by the heater is absorbed by Z.

(c) Calculate the specific heat capacity of Z in the liquid state. (2 marks)

(d) Calculate the specific latent heat of vaporisation of Z. (2 marks)

(e) In fact, energy is lost to the surroundings at a constant rate of 20 W throughout the experiment. Calculate the actual specific latent heat of fusion of Z. (2 marks)

10 marks
07

Section C — Comprehensive Structured Question 5

Evaporation and the Simple Working of an Air Conditioner

(a) Explain, in terms of the motion of molecules, why evaporation causes the liquid left behind to become cooler. (2 marks)

(b) During a run, 25 g of sweat evaporates from the skin of a runner of mass 60 kg. The specific latent heat of vaporisation of sweat is 2.4 × 106 J kg−1. Calculate the energy removed from the runner’s body by the evaporation. If this were the only energy transfer, calculate the fall in the body temperature of the runner. The average specific heat capacity of the runner’s body is 3500 J kg−1 °C−1. (2 marks)

Fig. 4 shows the simplified working of an air conditioner. A refrigerant circulates around the loop in the direction shown by the arrows through four components A, B, C and D. The refrigerant in A is at a temperature of about 5 °C and the refrigerant vapour entering C is at a temperature of about 60 °C. H1 is the heat absorbed per second from the room air, H2 is the heat released per second to the outdoor air, and E is the energy supplied per second to B.

Indoors (room) Outdoors A B C D low-pressure vapour high-pressure, hot vapour high-pressure liquid cold, low-pressure liquid H1 E H2
Fig. 4 — Simplified working of an air conditioner (the thick vertical line represents the wall; not drawn to scale)

(c)

  1. Name the components A, B, C and D. (1 mark)
  2. State the change of state of the refrigerant in A and in C, and state whether heat is absorbed or released by the refrigerant in each. (2 marks)

(d) Explain why the refrigerant vapour has to be compressed by B before it enters C. (1 mark)

(e) The air conditioner absorbs heat from the room at a rate H1 = 3.0 kW. Energy is supplied to the compressor B at a rate of 0.80 kW. Assume that there is no other energy loss.

  1. Calculate the rate H2 at which heat is released to the outdoor air. (1 mark)
  2. The specific latent heat of vaporisation of the refrigerant is 2.0 × 105 J kg−1. Assuming that all the heat absorbed in A is used to vaporise the refrigerant, calculate the mass of refrigerant which evaporates in A per second. (1 mark)
10 marks

Question 1 — Multiple Choice

1. B
2. D
3. C
4. B
5. A
6. A
7. D
8. C
9. A
10. C
  1. B — same average kinetic energy, greater total potential energy. Both are at 100 °C, so the average random kinetic energy of the molecules is the same. The energy absorbed during boiling (the latent heat) goes into increasing the potential energy of the molecules, which are far apart in steam. So steam has the greater total potential energy and the greater internal energy.
  2. D — 3/2. Q = mcΔT is the same for both: (mQ/3) cP (2ΔTQ) = mQ cQ ΔTQ, so (2/3) cP = cQ and cP ÷ cQ = 3/2. (2/3 is the inverted ratio.)
  3. C — 420 J kg−1 °C−1. Energy gained by water = 0.30 × 4200 × (25 − 20) = 6300 J. Energy lost by metal = 0.20 × c × (100 − 25) = 15c. So 15c = 6300 and c = 420. (394 comes from using 100 − 20 = 80 °C for the metal. 84 comes from forgetting the mass of the metal. 1260 comes from using 5 °C for the metal.)
  4. B — 2400 J kg−1 °C−1. Energy supplied = 600 × 100 = 60 000 J. Energy absorbed by the liquid = 0.80 × 60 000 = 48 000 J. c = 48 000 ÷ (2.0 × 10) = 2400. (3000 ignores the loss. 3600 adds the loss instead of subtracting it. 1920 applies the 80% twice.)
  5. A — 1.8 × 105 J kg−1. Energy supplied = 300 × (5.0 × 60) = 90 000 J. Energy absorbed = 0.90 × 90 000 = 81 000 J. l = 81 000 ÷ 0.45 = 1.8 × 105 J kg−1. (2.0 × 105 ignores the loss. 2.2 × 105 adds the loss.)
  6. A — 11.7 °C. Energy released when the water cools to 0 °C = 0.50 × 4200 × 30 = 63 000 J. Energy needed to melt all the ice = 0.10 × 3.34 × 105 = 33 400 J. Since 63 000 J > 33 400 J, all the ice melts and the remaining 29 600 J warms 0.60 kg of water: ΔT = 29 600 ÷ (0.60 × 4200) = 11.7 °C. (25 °C ignores the latent heat. 0 °C wrongly assumes that some ice remains.)
  7. D — (1) and (2) only. A larger exposed surface area (1) and dry moving air (2), which carries away the vapour and keeps the humidity at the surface low, both increase the rate of evaporation. In very humid air (3), many vapour molecules return to the towel, so the net rate of evaporation decreases.
  8. C — H2 > H1. By conservation of energy, H2 = H1 + E. Since E > 0, the heat released per second at the condenser is greater than the heat absorbed per second at the evaporator.
  9. A. The compressor raises the pressure and the temperature of the vapour (above the temperature of the outdoor air) so that it can condense and release heat outdoors. The expansion valve lowers the pressure of the liquid, so the refrigerant can evaporate at a low temperature in the evaporator. B, C and D confuse the functions of the components.
  10. C — 3. Constant power, so energy is proportional to time. Solid: 3 min for a rise of 60 °C, so cs ∝ 3 ÷ 60 = 0.05. Liquid: 6 min (12 to 18) for a rise of 40 °C, so cl ∝ 6 ÷ 40 = 0.15. Ratio = 0.15 ÷ 0.05 = 3. (2 is the ratio of the times only. 1.5 is the ratio of the temperature rises, 60 ÷ 40, alone.)

Question 2 — True or False

1. F
2. T
3. F
4. T
5. F
6. T
7. F
8. T
9. T
10. F
  1. False. Internal energy involves the random kinetic energy and the potential energy of the molecules. It does not include the kinetic energy of the ball as a whole. The internal energy of the ball does not increase just because the ball moves faster.
  2. True. The temperature is constant, so the average random kinetic energy is unchanged. The energy absorbed increases the potential energy, as the molecules move farther apart against the attractive forces.
  3. False. Same temperature means the same average kinetic energy per molecule, but the 2 kg block has twice as many molecules, so it has the greater internal energy (about twice as much).
  4. True. Heat capacity C = mc depends on the mass and on the material. The specific heat capacity c depends on the material (and its state) only.
  5. False. The energy lost by the hot body equals the energy gained by the cold body only if no energy is exchanged with the surroundings (e.g. the system is well insulated).
  6. True. (2.26 × 106 J kg−1 for water, against 3.34 × 105 J kg−1 for ice.) A much larger increase in potential energy is needed to separate the molecules completely, and work is also done against the atmosphere as the volume increases greatly.
  7. False. Evaporation takes place at the surface at any temperature below the boiling point. Boiling takes place at the boiling point (at a fixed pressure), with bubbles of vapour forming throughout the liquid.
  8. True. The more energetic molecules escape, so the average kinetic energy of the remaining molecules is lower, and the temperature of the liquid falls.
  9. True. Heat flows only from a hotter to a cooler region. The vapour in the condenser must be hotter than the outdoor air (about 60 °C against about 35 °C, for instance), and compression raises its temperature.
  10. False. The room temperature rises. The heat H2 = H1 + E released by the condenser is also released inside the same closed room, and it is greater than the heat H1 removed from the room air. The net energy added to the room is E, so the internal energy of the room increases.

Question 3 — Internal Energy: Molecular Kinetic Energy and Potential Energy

(a) Internal energy

The internal energy of a substance is the sum of the (random) kinetic energy and the potential energy of all the molecules in the substance. (1)

(b) Energy transfer and temperature rise

(i) The drive does work on the water through the paddle, so energy is transferred to the water (and the container), increasing their internal energy. (1)
The moving paddle makes the water molecules move faster in random motion, so their average random kinetic energy increases and the temperature of the water rises. (1)
(ii) Time = 10 × 60 = 600 s. Energy supplied = 75 × 600 = 45 000 J (1)
Total heat capacity = 0.50 × 4200 + 150 = 2250 J °C−1 (1)
ΔT = 45 0002250 = 20.0 °C (1)

(c) Energy lost

Energy gained by the water and container = 2250 × 16.0 = 36 000 J (1)
Energy lost = 45 000 − 36 000 = 9000 J
Percentage lost = 900045 000 × 100% = 20% (1)

(d) Ice and water at 0 °C

The water has the greater internal energy.
Both are at 0 °C, so the average molecular kinetic energy is the same. (1)
Energy (the latent heat of fusion) must be supplied to melt the ice, which increases the potential energy of the molecules. So the molecules of the water have the greater total potential energy. (1)
Common errors: (1) using 10 instead of 600 s in (b)(ii); (2) forgetting the heat capacity of the container and the paddle in (b)(ii); (3) saying in (d) that the kinetic energy of the molecules in water is greater than in ice at the same temperature.

Question 4 — Heat Capacity: Heating and Mixing

(a) Specific heat capacity

The specific heat capacity of a substance is the energy required to raise the temperature of 1 kg of the substance by 1 °C (without a change of state). (1)

(b) Specific heat capacity of the metal

Energy supplied in 60 s = 48 × 60 = 2880 J (1)
c = 28800.80 × (23.7 − 20.0) = 28802.96 = 973 J kg−1 °C−1 (1)

(c) Smaller temperature rises

As the block becomes hotter than the surroundings, the rate of energy loss to the surroundings increases (the temperature difference increases). (1)
The heater supplies energy at a constant rate, so the net rate at which the block gains energy decreases, and the temperature rises more slowly. (1)

(d) Energy lost between 240 s and 300 s

Heat capacity of the block = 0.80 × 973 = 778 J °C−1
Energy gained by the block = 778.4 × (35.6 − 33.0) = 778.4 × 2.6 = 2024 J (1)
Energy lost = 2880 − 2024 = 856 J (accept 850 J to 860 J) (1)

(e) Specific heat capacity of liquid X

Energy lost by the block = 778.4 × (90.0 − 60.0) = 23 352 J (1)
Energy gained by X = 0.250 × cX × (60.0 − 20.0) = 10.0 cX (1)
cX = 23 35210.0 = 2.34 × 103 J kg−1 °C−1 (1)
Marking note: In (e), accept any answer which follows consistently from the student’s answer to (b) (error carried forward). Accept 2.3 × 103 J kg−1 °C−1.
Common errors: (1) using the final temperature of 35.6 °C, or 300 s, in part (b) (heat loss is then significant); (2) using 90.0 − 20.0 as the temperature change of the block in (e); (3) using 48 W × 300 s as the energy lost in (d), instead of the energy that is not absorbed by the block.

Question 5 — Specific Latent Heat: Ice, Water and Steam

(a) Specific latent heat of vaporisation

The specific latent heat of vaporisation is the energy absorbed when 1 kg of a substance changes from liquid to gas (vapour) at its boiling point, without a change in temperature. (1)

(b) Ice cubes in a drink

Mass of ice = 3 × 0.020 = 0.060 kg
(i) Energy to warm the ice to 0 °C = 0.060 × 2100 × 5.0 = 630 J
Energy to melt all the ice = 0.060 × 3.34 × 105 = 20 040 J
Total energy needed = 20 670 J (1)
Maximum energy available as the drink cools to 0 °C = 0.250 × 4200 × 25.0 = 26 250 J
Since 26 250 J > 20 670 J, all the ice melts. (1)
(ii) Let the final temperature be T:
0.250 × 4200 × (25.0 − T) = 630 + 20 040 + 0.060 × 4200 × T (1)
26 250 − 1050T = 20 670 + 252T, so 5580 = 1302T
T = 4.29 °C (1)
Alternative for (ii): The excess energy = 26 250 − 20 670 = 5580 J warms 0.310 kg of water: ΔT = 5580 ÷ (0.310 × 4200) = 4.29 °C.

(c) Steam passed into water

Let the mass of steam condensed be m.
Energy released by the steam = m(2.26 × 106) + m(4200)(100 − 60.0) = 2.428 × 106 m (1)
Energy gained by the water and vessel = (0.50 × 4200 + 200)(60.0 − 20.0) = 2300 × 40.0 = 92 000 J (1)
m = 92 0002.428 × 106 = 0.0379 kg (37.9 g) (1)

(d) Steam scald

When steam condenses on the skin, it releases its latent heat of vaporisation (2.26 × 106 J for every kg) at constant temperature, in addition to the energy released when the condensed water cools. (1)
Boiling water releases only the energy from cooling. For example, for 1 kg cooling to 40 °C, steam releases 2.26 × 106 + 4200 × 60 = 2.51 × 106 J, whereas water releases 4200 × 60 = 2.52 × 105 J. The skin therefore receives much more energy from steam. (1)
Common errors: (1) forgetting to include the energy needed to warm the ice from −5.0 °C to 0 °C; (2) forgetting that the condensed steam (as water at 100 °C) also cools to 60.0 °C in (c); (3) forgetting the heat capacity of the vessel in (c).

Question 6 — Change of State: A Heating Curve

(a) States of Z

AB: solid; BC: solid and liquid (melting); CD: liquid; DE: liquid and gas (boiling).
(2 marks for all four correct; 1 mark for two or three correct)

(b) Molecular energies

During AB, the energy absorbed mainly increases the average kinetic energy of the molecules (the temperature rises). During BC, the average kinetic energy is unchanged but the total potential energy increases. (1)
During BC the energy supplied is used to overcome the attraction between the molecules as the solid changes to liquid (increasing the potential energy), not to make the molecules move faster, so the temperature stays constant. (1)

(c) Specific heat capacity of the liquid

CD takes 200 − 140 = 60 s. Energy = 400 × 60 = 24 000 J (1)
c = 24 0000.20 × (100 − 60) = 3000 J kg−1 °C−1 (1)

(d) Specific latent heat of vaporisation

DE takes 560 − 200 = 360 s. Energy = 400 × 360 = 144 000 J (1)
lv = 144 0000.20 = 7.2 × 105 J kg−1 (1)

(e) Specific latent heat of fusion with energy loss

BC takes 140 − 40 = 100 s. Net rate of energy supply to Z = 400 − 20 = 380 W
Energy absorbed by Z = 380 × 100 = 38 000 J (1)
lf = 38 0000.20 = 1.9 × 105 J kg−1 (1)
Check: If the loss were ignored, lf would be 400 × 100 ÷ 0.20 = 2.0 × 105 J kg−1. The actual value is lower, since not all the energy supplied is absorbed by Z.
Common errors: (1) reading the times from the axis incorrectly (e.g. using 200 s or 560 s instead of the durations 60 s and 360 s); (2) using the temperature change 100 − 20 in (c); (3) saying that the temperature stays constant during BC because “the heat is not absorbed”.

Question 7 — Evaporation and the Simple Working of an Air Conditioner

(a) Cooling by evaporation

The molecules near the surface with higher kinetic energy can overcome the attraction of the other molecules and escape from the liquid. (1)
The molecules left behind have a lower average kinetic energy, so the temperature of the liquid falls. (1)

(b) Sweat

Energy removed = 0.025 × 2.4 × 106 = 6.0 × 104 J (1)
Fall in temperature = 6.0 × 10460 × 3500 = 0.29 °C (1)

(c) Components and changes of state

(i) A = evaporator; B = compressor; C = condenser; D = expansion valve (1, all four correct)
(ii) In A, the refrigerant evaporates (liquid → vapour) and absorbs heat (latent heat) from the room air. (1)
In C, the refrigerant condenses (vapour → liquid) and releases heat (latent heat) to the outdoor air. (1)

(d) Role of the compressor

The compression raises the pressure and the temperature of the vapour (about 60 °C) above that of the outdoor air, so that heat can flow from the refrigerant to the outdoor air and the vapour can condense in C. (1)

(e) Energy balance and flow of refrigerant

(i) H2 = H1 + E = 3.0 + 0.80 = 3.8 kW (1)
(ii) Mass evaporating per second = 3000 J s−12.0 × 105 J kg−1 = 0.015 kg per second (1.5 × 10−2 kg s−1) (1)
Common errors: (1) saying that the evaporator “produces cold” instead of absorbing heat from the room air; (2) writing H2 = H1 and ignoring the energy supplied to the compressor; (3) saying that evaporation cools the liquid because “heat is lost” without mentioning the escape of the faster molecules.
Mark weighting: Section A (10) + Section B (10) + Question 3 (10) + Question 4 (10) + Question 5 (10) + Question 6 (10) + Question 7 (10) = 70 marks.
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