Changes of state and specific latent heat - AQA GCSE Physics Exam Questions

Practise 36 exam-style questions on Changes of state and specific latent heat for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

36 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Define the specific latent heat of a substance.

Model answer

The specific latent heat of a substance is the amount of energy required to change the state of one kilogram of the substance with no change in temperature.

Explanation

Two ideas are needed: the energy is for a change of state of 1 kg, and the temperature does not change while it happens. Do not confuse this with specific heat capacity, which is the energy needed to raise the temperature of 1 kg by 1 °C.

Mark scheme

1 mark
the energy needed to change the state of one kilogram of the substance
allow energy needed to melt / boil / freeze 1 kg; do not accept energy needed to raise the temperature of 1 kg
1 mark
with no change in temperature
allow at a constant temperature

Question 2

3 marks

A pan contains 250 g of water boiling at 100 °C.

Calculate the energy needed to turn all of the water into steam.
Give your answer in kilojoules.

specific latent heat of vaporisation of water = 2 260 000 J/kg

Use the equation:
thermal energy for a change of state = mass × specific latent heat

Model answer

m = 250 g = 0.250 kg
E = m × L
E = 0.250 × 2 260 000 = 565 000 J
E = 565 kJ

Explanation

Specific latent heat is given per kilogram, so the mass must be converted from grams to kilograms first (divide by 1000). Then divide the answer in joules by 1000 to give kilojoules.

Mark scheme

1 mark
mass = 0.250 kg
an answer of 565 (kJ) scores 3 marks
1 mark
E = 0.250 × 2 260 000 (= 565 000 J)
allow ecf from an incorrect mass conversion
1 mark
565 (kJ)
allow ecf; an answer of 565 000 with no conversion to kJ scores 2 marks

Question 3

3 marks

Two identical glasses of lemonade are at the same temperature. 50 g of ice at 0 °C is added to one glass and 50 g of water at 0 °C is added to the other.

Explain why the ice cools the lemonade more than the water does.

Model answer

Both the ice and the cold water absorb energy from the lemonade as they warm up. However, the ice must first melt, and melting needs extra energy (the latent heat of fusion) without any rise in temperature. This extra energy also comes from the lemonade, so more energy is removed from the drink with ice in it and its temperature falls further.

Explanation

Ice at 0 °C and water at 0 °C have the same temperature but not the same internal energy: the water has more, because energy was needed to melt it. That is why ice is much better at cooling drinks than the same mass of very cold water.

Mark scheme

1 mark
the ice has to melt before its temperature can rise
1 mark
melting needs energy (latent heat of fusion), which is transferred from the lemonade
allow 334 000 J/kg is absorbed from the drink
1 mark
so more energy is removed from the lemonade (melting and then warming, compared with warming only), so its temperature falls further

Question 4

4 marks

A freezer removes energy from 0.75 kg of water at 0 °C at a rate of 125 W (125 J every second).

Calculate the time it takes for all of the water to freeze.
Give your answer in minutes.

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

Model answer

E = m × L = 0.75 × 334 000 = 250 500 J
P = E ÷ t, so t = E ÷ P
t = 250 500 ÷ 125 = 2004 s
t = 2004 ÷ 60 = 33.4 minutes ≈ 33 minutes

Explanation

Freezing releases the same energy per kilogram as melting needs, so E = m L gives the energy the freezer must remove. Then rearrange power = energy ÷ time to find the time in seconds and divide by 60 to get minutes.

Mark scheme

1 mark
E = 0.75 × 334 000
an answer of 33 (minutes) scores 4 marks; allow 33.4 minutes
1 mark
E = 250 500 (J)
1 mark
t = 250 500 ÷ 125 = 2004 (s)
allow ecf
1 mark
33 (minutes)
allow 33.4 minutes; allow ecf; 2004 s (not converted) scores 3 marks

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