6.3.2.2Temperature changes in a system and specific heat capacity
AQA GCSE Combined Science (8464), Higher tier · Physics › Particle model of matter › Internal energy and energy transfers
Practise Temperature changes in a system and specific heat capacity. 15 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.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Define specific heat capacityThe amount of energy needed to raise the temperature of 1 kg of a substance by 1 °C.
4
Substitute into ΔE = m c ΔθCalculate the change in thermal energy with m in kg, c in J/kg °C and Δθ in °C.
5
Work out the temperature change ΔθΔθ is the difference between the final and starting temperatures, not the final temperature.
6
Rearrange to find c, m or ΔθUse c = ΔE ÷ (m Δθ), m = ΔE ÷ (c Δθ) or Δθ = ΔE ÷ (m c).
6
Explain what affects a temperature riseThe rise depends on the mass heated, the material (its specific heat capacity) and the energy supplied.
7
Use heater power and time in calculationsFind the energy supplied with E = P t, then use it in ΔE = m c Δθ.
7
Explain why a measured c is too highEnergy is transferred to the surroundings, so the temperature rise is smaller and c = ΔE ÷ (m Δθ) comes out larger.
Notes
What specific heat capacity means
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.
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).
Your value is usually higher than the true value. Some energy is transferred to the surroundings instead of the block, so the temperature rise is smaller than it should be. Insulating the block reduces this. grade 7+
Cheatsheet
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
A measured c is usually too high: energy is transferred to the surroundings grade 7+
How to answer each type of question
Calculate the energy needed to heat something
2 to 3 marks5
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.
Substitute into ΔE = m c Δθ, then rearrange for the unknown.
Calculate and round as the question asks.
Example. A 450 g copper block is heated. 6000 J of energy is transferred to the block. specific heat capacity of copper = 385 J/kg °C Calculate the temperature rise of the block. Give your answer to 2 significant figures.
Find the energy supplied: E = P t, with the time in seconds.
Find Δθ from the starting and final temperatures.
Calculate c = E ÷ (m Δθ) and give the unit J/kg °C.
If asked about accuracy, explain the effect of energy transferred to the surroundings.
Example. A student heats a 1.0 kg aluminium block with a 50 W heater for 10 minutes. The temperature of the block rises from 20.0 °C to 52.0 °C. (a) Calculate the specific heat capacity of aluminium given by these results. Give your answer to 2 significant figures. (4 marks) (b) The accepted value is 900 J/kg °C. Explain why the student's value is higher than this. (2 marks)
Show the model answer
(a) E = 50 × 600 = 30 000 J (1) Δθ = 52.0 − 20.0 = 32.0 °C (1) c = 30 000 ÷ (1.0 × 32.0) (1) c = 937.5 = 940 J/kg °C (to 2 s.f.) (1) (b) Some of the energy from the heater is transferred to the surroundings, not to the block (1). So the temperature rise is smaller than it would otherwise be, which makes the calculated value of c too large (1).
Explain why one material heats up more than another
2 marks6
Say what is the same (the energy supplied and the mass).
Compare the specific heat capacities: the lower one needs less energy for each kg to rise by 1 °C.
Example. Equal masses of water and cooking oil are heated by identical heaters for the same time. The temperature of the oil rises more than the temperature of the water. Explain why.
Show the model answer
Both liquids have the same mass and receive the same amount of energy (1). The oil has a lower specific heat capacity, so less energy is needed to raise the temperature of each kilogram by 1 °C (1).
Shortcuts and memory tricks
Both Δs in ΔE = m c Δθ stand for 'change': a change in energy and a change in temperature.
Sense check with water: about 4200 J warms 1 kg by 1 °C, so heating 1 kg of water by 80 °C needs roughly 340 000 J.
Double the mass or double the temperature rise and you double the energy needed.
The unit J/kg °C reminds you of the rearrangement: c = joules ÷ (kg × °C).
Calculator tip: for c = ΔE ÷ (m Δθ), put brackets round m × Δθ, or divide twice: ΔE ÷ m ÷ Δθ.
Where marks are lost
Using the final temperature instead of the temperature change for Δθ.
Leaving the mass in grams, which makes the answer 1000 times out.
Leaving the time in minutes when using E = P t.
Using ΔE = m c Δθ during a change of state. The temperature does not change then: use E = m L instead.
Rearranging wrongly, e.g. c = ΔE × m ÷ Δθ. You must divide by both m and Δθ.
Exam technique
The equation is on the equation sheet, but you must choose it and know what each symbol means.
Always write the substitution line with numbers: it usually earns a mark even if the final answer goes wrong.
Give J/kg °C as the unit of specific heat capacity, and round to the significant figures asked for.
In practical questions, describe the error precisely: energy is transferred to the surroundings (so the measured temperature rise is too small).
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
Question 3Hard8 marks
A student heated a 0.20 kg copper block to 95 °C in a beaker of hot water. The student then quickly moved the block into 0.30 kg of water at 18 °C in an insulated cup. specific heat capacity of copper = 385 J/kg °C specific heat capacity of water = 4200 J/kg °C
(a) Explain why the temperature of the block decreases and the temperature of the water in the cup increases until they are the same.[2]
(b) Calculate the final temperature of the block and the water. Assume that no energy is transferred to the cup or to the surroundings. Give your answer to 3 significant figures. Use the Physics Equations Sheet.[4]
(c) The student measured the final temperature as 21.8 °C. Suggest two reasons why this was lower than the value calculated.[2]
Show the answer and mark scheme
(a)Answer: Energy is transferred from the hot block to the cooler water until both are at the same temperature, when there is no net energy transfer.
energy is transferred from the hotter block to the colder water (by heating)
until they are at the same temperature, when there is no longer a net transfer of energy
(b)Answer: 22.4 °C
energy transferred from the copper = energy transferred to the water
0.20 × 385 × (95 − T) = 0.30 × 4200 × (T − 18)
7315 − 77T = 1260T − 22 680 / 1337T = 29 995
T = 22.4 (°C)
(c)Answer: The block cooled while it was being moved, and some energy was transferred to the cup and to the surroundings.
the block cooled / transferred energy to the air while it was being moved to the cup
energy was transferred from the water to the cup / thermometer
energy was transferred from the water to the surroundings (the cup was not perfectly insulated / had no lid)