AQA GCSE Physics Foundation (8463), Foundation tier · Particle model of matter › Internal energy and energy transfers
Practise Specific heat capacity. 9 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.
Specific heat capacity tells you how much energy is needed to raise the temperature of 1 kg of a material by 1 °C. You need to define it and use ΔE = m c Δθ, which is on the Physics equation sheet. Expect 2 to 4 mark calculations, often with a rearrangement, unit conversions or a link to power, plus explanations of why one material heats up faster than another.
Key facts
ΔE = m c Δθ (on the equation sheet)
ΔE in J, m in kg, c in J/kg °C, Δθ in °C
Specific heat capacity: energy to raise the temperature of 1 kg by 1 °C
Δθ = final temperature − starting temperature
c = ΔE ÷ (m × Δθ)
Energy from a heater: E = P t (P in W, t in s)
High c: more energy needed per °C, so it heats up and cools down more slowly
Notes
What specific heat capacity means
diagram
When you heat a substance without changing its state, its temperature rise depends on the mass heated, the type of material and the energy supplied.
The specific heat capacity (c) of a substance is the amount of energy needed to raise the temperature of 1 kg of the substance by 1 °C.
Its unit is J/kg °C. Water has a high specific heat capacity (about 4200 J/kg °C); metals such as copper and aluminium have much lower values.
A material with a high specific heat capacity needs more energy for each °C rise, so with the same heating it warms up more slowly. It also transfers more energy to the surroundings for each °C that it cools.
Same mass, same energy: the lower the specific heat capacity, the bigger the temperature rise.
The equation
change in thermal energy = mass × specific heat capacity × temperature change: ΔE = m c Δθ (on the equation sheet).
ΔE in joules (J), m in kilograms (kg), c in J/kg °C, Δθ in degrees Celsius (°C).
Δθ is a change: a rise from 15 °C to 40 °C is Δθ = 25 °C.
Rearranged: c = ΔE ÷ (m Δθ); m = ΔE ÷ (c Δθ); Δθ = ΔE ÷ (m c).
Convert first: grams to kg (÷ 1000) and kJ to J (× 1000).
The same equation gives the energy transferred away when an object cools down.
Finding c by experiment
The required practical to measure specific heat capacity is in the Energy topic: you heat a block of known mass with an electric heater and record its temperature rise.
Energy supplied: read it from a joulemeter, or use E = P t (P = power in W, t = time in s).
Then calculate c = E ÷ (m Δθ).
How to answer each type of question
Calculate the energy needed to heat something
2 to 3 marksGrade 5
Choose ΔE = m c Δθ from the equation sheet.
Work out Δθ = final − starting temperature, and make sure the mass is in kg.
Substitute, calculate and give the unit (J).
Example. An electric kettle heats 0.80 kg of water from 20 °C to 100 °C. specific heat capacity of water = 4200 J/kg °C Calculate the energy transferred to the water.
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Complete the sentence. The specific heat capacity of a substance is the amount of energy needed to raise the temperature of ................ of the substance by ................ .
1 kg; 1 °C
Calculate the energy needed to increase the temperature of a 0.40 kg glass beaker by 25 °C. specific heat capacity of glass = 840 J/kg °C Use the equation: change in thermal energy = mass × specific heat capacity × temperature change
8400 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) Complete the sentence. The specific heat capacity of a substance is the amount of energy needed to raise the temperature of ................ of the substance by ................ .[2]
(b) Calculate the energy needed to increase the temperature of 2.0 kg of water by 15 °C. specific heat capacity of water = 4200 J/kg °C Use the equation: change in thermal energy = mass × specific heat capacity × temperature change[2]
Show the answer and mark scheme
(a)Answer: 1 kg; 1 °C
one kilogram / 1 kg
one degree Celsius / 1 °C
(b)Answer: 126 000 J
ΔE = 2.0 × 4200 × 15
126 000 (J)
Question 2Medium6 marks
An electric kettle transfers 151 200 J of energy to 0.60 kg of water. The water starts at a temperature of 18 °C. specific heat capacity of water = 4200 J/kg °C
(a) Calculate the final temperature of the water. Assume that all of the energy is transferred to the water. Use the Physics Equations Sheet.[3]
(b) The actual final temperature of the water was lower than the value you calculated. Give one reason why.[1]
(c) On a summer day, the temperature of the sea changes much less between day and night than the temperature of the land. Explain why. Use ideas about specific heat capacity.[2]
Show the answer and mark scheme
(a)Answer: 78 °C
151 200 = 0.60 × 4200 × Δθ
Δθ = 60 (°C)
final temperature = 18 + 60 = 78 (°C)
(b)Answer: Some of the energy heated the kettle itself and the surrounding air.
some of the energy was transferred to the kettle / to the surroundings / to the air
(c)Answer: Water has a much higher specific heat capacity than rock or soil, so much more energy must be transferred to change its temperature by each degree.
water has a higher specific heat capacity than the land / rock / soil
so more energy has to be transferred to (or from) each kilogram of water for each 1 °C change in temperature