Question bank › GCSE Physics › Particle model of matter
Internal energy and specific latent heat
Answer on paper first, then reveal the mark scheme one mark at a time.
1Grade 4–53 marks
(a) A 0.25 kg block of ice at 0 °C melts completely.
specific latent heat of fusion of ice = 334 000 J/kg
Calculate the energy needed to melt the ice. Use the equation: energy for a change of state $=$ mass $\times$ specific latent heat [2]
(b) State what is meant by the specific latent heat of fusion of a substance. [1]
(a) $E = 0.25 \times 334\,000$(1)
(a) 83 500 J allow 83.5 kJ / 8.35 × 10⁴ J; a correct answer with no working scores 2 marks(1)
(b) the energy needed to change 1 kg of the substance from solid to liquid without a change in temperature all three ideas needed: 1 kg (allow per kilogram / unit mass), solid to liquid (allow melt), no temperature change (allow at constant temperature); allow the energy released when 1 kg changes from liquid to solid with no change in temperature(1)
Answer: (a) 83 500 J (b) energy to melt 1 kg with no change in temperature
Examiner tip: The definition needs '1 kg', 'solid to liquid' and 'no change in temperature' — leaving any one out loses the mark.
2Grade 6–7Higher6 marks
A 0.10 kg sample of a solid is heated by a 50 W heater. The graph shows how its temperature changes with time.
(a) State the melting point of the substance. [1]
(b) Explain why the temperature does not change between 4 and 10 minutes, even though the heater is still on. [2]
(c) Assume all the energy from the heater is transferred to the substance.
Calculate the specific latent heat of fusion of the substance. [3]
(a) 80 °C(1)
(b) the energy is used to break bonds / overcome forces between particles (as it melts) allow energy is used to change the state(1)
(b) so the internal energy (potential energy of the particles) increases but the kinetic energy of the particles / temperature does not(1)
Examiner tip: Two stages, two equations: heating the water ($mc\Delta\theta$) then boiling it ($mL$) — then add.
4Grade 1–3Foundation4 marks
A pan of soup is heated on a hob.
(a) Which statement describes the internal energy of the soup?
Tick (✓) one box. [1]
☐ The total kinetic energy of all the particles in the soup. ☐ The total kinetic energy and potential energy of all the particles in the soup. ☐ The energy needed to raise the temperature of 1 kg of the soup by 1 °C. ☐ The total potential energy of all the particles in the soup.
(b) Heating the soup increases its internal energy.
Give two effects that this increase in internal energy can have on the soup. [2]
(c) After some time, the soup boils. Its temperature stays at 100 °C while it boils.
Which energy of the particles in the soup increases while it boils at 100 °C?
Tick (✓) one box. [1]
☐ chemical energy ☐ kinetic energy ☐ nuclear energy ☐ potential energy
(a) The total kinetic energy and potential energy of all the particles in the soup. more than one box ticked scores 0(1)
(b) the temperature (of the soup) increases allow it gets hotter / warms up; allow the particles move faster(1)
(b) the soup changes state allow (water in the soup) boils / evaporates / turns into steam or vapour; ignore 'melts' unless the soup is described as frozen(1)
(c) potential energy more than one box ticked scores 0(1)
Answer: (a) total kinetic and potential energy of all the particles (b) temperature rises; change of state (c) potential energy
Examiner tip: Internal energy is kinetic and potential energy of the particles — at a fixed boiling temperature it is the potential energy that is rising, not the kinetic.
5Grade 4–54 marks
A wallpaper steamer boils water to make steam. The steam is used to loosen old wallpaper.
(a) Name the change of state when water turns into steam. [1]
(b) The water in the steamer is at 100 °C. While the steamer is in use, 678 000 J of energy is transferred to the water at 100 °C.
specific latent heat of vaporisation of water = 2 260 000 J/kg
Calculate the mass of water turned into steam.
Use the Physics Equations Sheet. [3]
(a) boiling allow evaporation / vaporisation; do not accept condensation(1)
(b) $678\,000 = m \times 2\,260\,000$(1)
(b) $m = \dfrac{678\,000}{2\,260\,000}$(1)
(b) 0.30 kg allow 0.3 kg / 300 g; a correct answer with no working scores 3 marks(1)
Answer: (a) boiling (b) 0.30 kg
Examiner tip: Rearrange $E = mL$ to $m = E \div L$ — dividing the wrong way round gives 3.3, which should look far too big for a steamer.
6Grade 4–55 marks
A student heated a test tube of solid wax until it was a liquid at 85 °C. She then let the liquid wax cool in the room and recorded its temperature every 30 seconds.
The graph shows her results.
(a) Use the graph to give the freezing point of the wax. [1]
(b) Give the state (or states) of the wax at 6 minutes. [1]
(c) Between 3 minutes and 9 minutes, energy is being transferred from the wax to the surroundings, but the temperature of the wax does not change.
Explain why. [2]
(d) Between 9 minutes and 16 minutes the temperature of the wax decreases more and more slowly.
Suggest why. [1]
(a) 55 °C allow 54–56 °C(1)
(b) solid and liquid allow 'a mixture of solid and liquid' / 'it is freezing (changing from liquid to solid)'; 'liquid' alone or 'solid' alone scores 0(1)
(c) the wax is changing state / freezing / turning from liquid to solid allow bonds (between particles) are forming / particles are being pulled together by attractive forces(1)
(c) the internal energy decreases because the potential energy of the particles decreases, but the kinetic energy (and so the temperature) stays the same allow energy released by bonds forming replaces the energy transferred to the surroundings; allow latent heat is released (as it freezes)(1)
(d) the temperature difference between the wax and the room/surroundings gets smaller (so energy is transferred more slowly) allow the wax is getting closer to room temperature; ignore 'it has less energy'(1)
Answer: (a) 55 °C (b) solid and liquid (c) freezing: potential energy of particles falls as bonds form, kinetic energy (temperature) constant (d) smaller temperature difference with the room
Examiner tip: The flat section of a cooling curve is the substance freezing — the energy given out comes from particles' potential energy, so the temperature does not drop.
7Grade 6–7Higher7 marks
(a) Describe the difference between the specific heat capacity and the specific latent heat of a substance. [2]
(b) A burn from steam at 100 °C is usually much more serious than a burn from the same mass of water at 100 °C.
2.0 g of steam at 100 °C condenses on a person's skin, forming water at 100 °C.
specific latent heat of vaporisation of water = $2.26 \times 10^6$ J/kg
Calculate the energy transferred to the skin as the steam condenses. [3]
(c) When 2.0 g of water at 100 °C cools to 37 °C on the skin, it transfers 529 J of energy to the skin.
Use this information and your answer to part (b) to explain why a burn from steam at 100 °C is more serious than a burn from water at 100 °C. [2]
(a) specific heat capacity is the energy needed to raise the temperature of 1 kg of the substance by 1 °C(1)
(a) specific latent heat is the energy needed to change the state of 1 kg of the substance with no change in temperature if neither definition mentions 1 kg (unit mass), max 1 mark; allow 'specific heat capacity involves a change of temperature, specific latent heat involves a change of state without a change of temperature' for 1 mark only(1)
(c) the steam transfers the latent heat (4520 J) as it condenses and then transfers a further 529 J as the water cools allow total ≈ 5000 J / 5049 J from the steam; allow ecf from (b)(1)
(c) so the steam transfers about ten times as much energy to the skin as the water does (only 529 J) allow 'much more energy (to the skin)'; accept a correct ratio of 9–10(1)
Answer: (a) SHC: energy to raise 1 kg by 1 °C; SLH: energy to change state of 1 kg at constant temperature (b) 4520 J (c) steam gives 4520 J + 529 J ≈ 5050 J, about 10× the 529 J from the water
Examiner tip: Convert grams to kilograms before using $E = mL$ — forgetting gives an answer 1000 times too big.
8Grade 6–76 marks
A block of ice at −20 °C is heated by a heater that transfers energy at a constant rate.
The ice is heated until all of it has turned into steam at 100 °C.
specific latent heat of fusion of water = 334 000 J/kg specific latent heat of vaporisation of water = 2 260 000 J/kg
Describe and explain how the temperature and the internal energy change from the start to the end of the heating.
Your answer should refer to the particles and explain why boiling takes much longer than melting. [6]
from −20 °C to 0 °C the ice (solid) warms: the temperature rises because the kinetic energy of the particles increases (they vibrate faster)(1)
at 0 °C the ice melts: the temperature stays constant (at 0 °C) until all the ice has melted(1)
while melting, the energy is used to break bonds / overcome attractive forces between particles, increasing their potential energy (not their kinetic energy)(1)
from 0 °C to 100 °C the liquid water warms: kinetic energy of the particles increases so temperature rises(1)
at 100 °C the water boils: temperature stays constant (at 100 °C) until all the water has turned into steam(1)
while boiling, the particles are separated completely (move far apart), so their potential energy increases a lot(1)
the internal energy increases throughout the whole process (energy is supplied all the time)(1)
the specific latent heat of vaporisation is (about 6.8 times) greater than the specific latent heat of fusion, so more energy is needed to boil the water than to melt the same mass of ice(1)
the heater supplies energy at a constant rate, so boiling takes (about 6.8 times) longer than melting(1)
more energy is needed to boil because the particles must be completely separated, whereas melting only loosens the bonds / particles stay close together(1)
Level
Marks
What it looks like
Level 3
5–6
A detailed, logically ordered account of all four stages (warming solid, melting, warming liquid, boiling). Clearly links temperature changes to the kinetic energy of the particles and the flat stages to changes of state that increase potential energy, and uses the latent heat values to explain why boiling takes longer than melting.
Level 2
3–4
Describes most of the stages in order, with some correct explanation in terms of particles or energy, e.g. temperature stays constant while melting and boiling because the energy is used to change state. The comparison of the time for melting and boiling may be missing or not explained.
Level 1
1–2
Some relevant, isolated statements, e.g. 'the ice melts at 0 °C and boils at 100 °C' or 'the temperature goes up', with little or no explanation.
0
0
No relevant content.
Answer: Temperature rises (KE up) for solid, flat at 0 °C while melting (PE up), rises for liquid, flat at 100 °C while boiling (PE up a lot); internal energy rises throughout; boiling takes ≈6.8× as long as melting because Lv is ≈6.8× Lf and the energy rate is constant.
Examiner tip: For a top-band answer, say which energy of the particles changes in each stage — kinetic when the temperature rises, potential during the flat changes of state.
9Grade 8–9Higher6 marks
An ice-cube tray contains 0.36 kg of water at 15 °C. The tray is placed in a freezer.
The freezer removes energy from the water at a constant rate of 40 W.
specific heat capacity of water = 4200 J/kg °C specific latent heat of fusion of water = 334 000 J/kg
Calculate the minimum time taken for all the water to turn into ice at 0 °C.
Give your answer in minutes, to 2 significant figures. [6]
$\Delta E = 0.36 \times 4200 \times 15 = 22\,680$ (J) (energy removed cooling the water to 0 °C)(1)
total energy $= 22\,680 + 120\,240 = 142\,920$ (J) allow ecf from either value(1)
$t = \dfrac{E}{P} = \dfrac{142\,920}{40}$ allow ecf from total(1)
$= 3573$ (s) allow 3570 s(1)
$= 60$ minutes (2 s.f.) converted to minutes and given to 2 s.f.; 59.6 min or 59.55 min scores 5 marks; using only one stage (22 680 J → 9.5 min or 120 240 J → 50 min) scores max 3(1)
Answer: 60 minutes (3573 s)
Examiner tip: Two stages: cool the water to 0 °C ($mc\Delta\theta$) then freeze it ($mL$) — then use $P = E/t$ and convert seconds to minutes.
10Grade 8–9Higher6 marks
A student used the apparatus shown to determine the specific latent heat of fusion of ice.
Funnels A and B were filled with crushed ice at 0 °C. An electric heater was placed in the ice in funnel A. There was no heater in funnel B.
The heater in funnel A was switched on for 6 minutes. Melted ice (meltwater) from both funnels was collected in beakers for the same 6 minutes.
Funnel A
Funnel B
Energy transferred by heater in J
27 600
0
Mass of meltwater collected in g
98.5
22.9
(a) Explain why the student used funnel B. [2]
(b) Calculate the specific latent heat of fusion of ice from the student's results. [3]
(c) The accepted value for the specific latent heat of fusion of ice is 334 000 J/kg.
The student's value is higher than this.
Suggest one reason why. [1]
(a) (funnel B shows) the mass of ice melted by energy from the surroundings / the room (not by the heater) do not accept 'it is a control' on its own(1)
(a) this mass is subtracted from funnel A's mass so that only the mass melted by the heater is used (in the calculation) allow so the mass of ice melted by the heater alone can be found(1)
(b) mass melted by heater $= 98.5 - 22.9 = 75.6$ g $= 0.0756$ kg(1)
(b) 365 000 J/kg allow answers rounding to $3.65 \times 10^5$ or $3.7 \times 10^5$ J/kg; 280 000 J/kg (98.5 g used, B not subtracted) scores 2 marks; 365 J/kg (mass in g) scores 2 marks(1)
(c) any one from: some energy from the heater was transferred to the surroundings / funnel / air rather than to the ice (e.g. heater not completely covered by ice); some meltwater stayed on the ice / in the funnel and was not collected (so measured mass too small) ignore 'human error' / 'heat loss' with no qualification(1)
Answer: (a) to find the mass melted by the surroundings, to subtract from A (b) $3.65 \times 10^5$ J/kg (c) e.g. some heater energy went to the surroundings, not the ice
Examiner tip: Use only the mass melted by the heater (A minus B) — dividing by the full 98.5 g is the most common lost mark.
Still losing marks on this topic?
Book a free introductory session and we'll find exactly which step is costing you.