AQA GCSE Physics (8463), Higher tier · Particle model of matter › Internal energy and energy transfers
Practise Specific latent heat. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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Calculate the energy released when 0.40 kg of liquid stearic acid solidifies (freezes) at its melting point. Use the equation: thermal energy for a change of state = mass × specific latent heat
79 600 J
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
(a) What is meant by the specific latent heat of a substance? Tick (✓) one box.[1]
The energy needed to change the state of 1 kg of the substance with no change in temperature
The energy needed to raise the temperature of 1 kg of the substance by 1 °C
The temperature at which the substance changes state
The total energy of all the particles in the substance
(b) What is the name of the specific latent heat for a change of state from liquid to vapour? Tick (✓) one box.[1]
Specific latent heat of condensation
Specific latent heat of fusion
Specific latent heat of vaporisation
Specific heat capacity
(c) Calculate the energy needed to melt 0.25 kg of ice at 0 °C. specific latent heat of fusion of ice = 334 000 J/kg Use the equation: thermal energy for a change of state = mass × specific latent heat[2]
Show the answer and mark scheme
(a)Answer: The energy needed to change the state of 1 kg of the substance with no change in temperature
(b)Answer: Specific latent heat of vaporisation
(c)Answer: 83 500 J
E = 0.25 × 334 000
83 500 (J)
Question 2Medium6 marks
Lead melts at 327 °C. 9200 J of energy is needed to melt a 0.40 kg block of lead that is already at 327 °C.
(a) Calculate the specific latent heat of fusion of lead. Use the Physics Equations Sheet.[2]
(b) Explain the difference between specific heat capacity and specific latent heat.[2]
(c) Before it melted, the block was heated from 20 °C to 327 °C. specific heat capacity of lead = 130 J/kg °C Calculate the energy needed to heat the block from 20 °C to 327 °C. Use the Physics Equations Sheet.[2]
Show the answer and mark scheme
(a)Answer: 23 000 J/kg
9200 = 0.40 × L
L = 23 000 (J/kg)
(b)Answer: Specific heat capacity is the energy needed to raise the temperature of 1 kg by 1 °C; specific latent heat is the energy needed to change the state of 1 kg without changing its temperature.
specific heat capacity is the energy needed to raise the temperature of 1 kg (of a substance) by 1 °C
specific latent heat is the energy needed to change the state of 1 kg (of a substance) with no change in temperature
(c)Answer: 16 000 J (15 964 J) J
ΔE = 0.40 × 130 × 307
15 964 (J) / 16 000 (J)
Question 3Hard8 marks
A student wants to determine the specific latent heat of fusion of ice. The student has:
two funnels, each above an empty beaker
crushed ice at 0 °C
an electric immersion heater connected to a joulemeter
a stopwatch
a top-pan balance
(a) Describe a method the student could use to determine the specific latent heat of fusion of ice. Your answer should include how the student would use the results.[6]
(b) Explain why the method uses a second funnel of ice that has no heater in it.[2]
Show the answer and mark scheme
(a)Answer: Two funnels of crushed ice, one with the heater. Collect meltwater from both for the same time while the heater runs; E from the joulemeter; m = heated mass − control mass; L = E ÷ m.
fill both funnels with crushed ice; put the immersion heater into the ice in one funnel, surrounded by ice; the other funnel has no heater (control)
wait until water drips from both funnels at a steady rate, then put empty (weighed) beakers under the funnels
switch on the heater, start the stopwatch and record the joulemeter reading
after a set time, e.g. 5 minutes, switch off the heater, remove the beakers and record the new joulemeter reading
energy transferred E = difference in joulemeter readings
measure the mass of water collected in each beaker using the balance
mass melted by the heater m = mass from the heated funnel − mass from the control funnel
calculate L = E ÷ m (in J/kg, with m in kg); repeat and calculate a mean
Marked with levels of response: the full level descriptors are in the app.
(b)Answer: Energy from the room melts some ice in both funnels; subtracting the control mass leaves just the mass melted by the heater.
some ice melts in both funnels because of energy transferred from the surroundings / the warm room
the mass melted in the funnel without the heater is subtracted, so that only the mass melted by the energy from the heater is used