☕ Kettles
Water needs a lot of energy to warm up, so a kettle needs a powerful heater. That same energy stays in the water, so a hot drink stays hot for a good while.
Change of State » Specific Heat Capacity
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 5.2P, 5.12P, 5.13P
- What specific heat capacity means and its unit, J/kg °C
- How to use change in thermal energy = mass × specific heat capacity × change in temperature
- How to rearrange the equation to find c, m or the temperature change
- Why the high value for water matters for kettles, radiators and the sea
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Unlock This CourseWhy some things heat up faster than others
This lesson is examined in Paper 2 only.
Put a metal spoon and a mug of water in the same sunshine. The spoon gets hot quickly. The water takes much longer. Different materials need different amounts of energy to warm up, even when the mass is the same. Specific heat capacity is the number that tells you how much.
Key terms:
- Specific heat capacity (c): the energy required to change the temperature of an object by one degree Celsius per kilogram of mass.
- Joule per kilogram degree Celsius (J/kg °C): the unit of specific heat capacity.
- Change in temperature (ΔT): the final temperature minus the starting temperature.
- Change in thermal energy (ΔQ): the energy transferred to or from the thermal store of an object.
The triangle symbol Δ is a Greek capital letter "delta". In physics it means "change in". So ΔT is "the change in temperature" and ΔQ is "the change in thermal energy". Each one is a single quantity. Never split ΔT into a separate "Δ" and "T".
What the number means
Water has a specific heat capacity of about 4200 J/kg °C. That means it takes about 4200 joules to warm 1 kg of water by 1 °C. Copper is about 390 J/kg °C. It takes only about 390 joules to warm 1 kg of copper by 1 °C.
| Material | Specific heat capacity (about) | Energy to warm 1 kg by 1 °C |
|---|---|---|
| Water | 4200 J/kg °C | 4200 J |
| Aluminium | 900 J/kg °C | 900 J |
| Copper | 390 J/kg °C | 390 J |
A big specific heat capacity means the material needs a lot of energy to warm up. It also gives out a lot of energy as it cools down. A small one means it warms up and cools down quickly.
The equation
On the formulae page
change in thermal energy = mass × specific heat capacity × change in temperature
ΔQ = m × c × ΔT
This equation is printed on the formulae page of the exam paper, so you do not have to learn it. You do have to know how to use it. Always use these units: ΔQ in joules (J), m in kilograms (kg), c in J/kg °C and ΔT in °C. If the mass is in grams, divide by 1000 first.
You can rearrange it with the usual triangle method:
| To find | Use |
|---|---|
| Energy | ΔQ = m × c × ΔT |
| Specific heat capacity | c = ΔQ ÷ (m × ΔT) |
| Mass | m = ΔQ ÷ (c × ΔT) |
| Change in temperature | ΔT = ΔQ ÷ (m × c) |
Worked examples
Worked example 1: finding the energy
How much energy is needed to warm 3.0 kg of water by 20 °C? (c of water = 4200 J/kg °C)
ΔQ = m × c × ΔT
ΔQ = 3.0 × 4200 × 20
ΔQ = 252 000 J
Worked example 2: working out the temperature change first
A copper block of mass 0.50 kg is heated from 20 °C to 100 °C. The specific heat capacity of copper is 390 J/kg °C. Calculate the energy transferred.
ΔT = 100 − 20 = 80 °C
ΔQ = 0.50 × 390 × 80
ΔQ = 15 600 J
Worked example 3: finding c
A 2.0 kg metal block needs 36 000 J to warm up by 25 °C. Calculate its specific heat capacity.
c = ΔQ ÷ (m × ΔT)
c = 36 000 ÷ (2.0 × 25)
c = 36 000 ÷ 50
c = 720 J/kg °C
Worked example 4: finding the mass
A heater transfers 84 000 J to some water. The temperature rises by 10 °C. Find the mass of the water. (c = 4200 J/kg °C)
m = ΔQ ÷ (c × ΔT)
m = 84 000 ÷ (4200 × 10)
m = 84 000 ÷ 42 000
m = 2.0 kg
Worked example 5: finding the temperature change
A 0.20 kg sample of cooking oil receives 12 000 J. Its specific heat capacity is 2000 J/kg °C. It starts at 18 °C. Find its final temperature.
ΔT = ΔQ ÷ (m × c)
ΔT = 12 000 ÷ (0.20 × 2000) = 12 000 ÷ 400 = 30 °C
Final temperature = 18 + 30 = 48 °C
Worked example 6: linking to electrical energy
A 12 V heater carries a current of 5.0 A for 840 s. All the energy goes into 0.50 kg of water. Find the temperature rise.
Energy = I × V × t = 5.0 × 12 × 840 = 50 400 J
ΔT = 50 400 ÷ (0.50 × 4200) = 50 400 ÷ 2100 = 24 °C
Why water's high value matters
🏡 Radiators
Hot water in a central heating system carries a lot of energy for each kilogram. As it cools in the radiator it gives that energy out to the room.
🌊 The sea
The sea takes in a lot of energy before its temperature rises much, and gives out a lot as it cools. So the sea warms up and cools down slowly, and places near it have milder temperatures.
Common mistakes
Common mistakes
- Using the final temperature as ΔT. You must subtract the start from the end.
- Leaving the mass in grams. Convert to kilograms first: 250 g = 0.25 kg.
- Writing the unit as J/kg. The unit is J/kg °C, because of the "per degree".
- Forgetting to add the temperature change back on to the starting temperature when asked for the final temperature.
Exam-style question
(a) State what is meant by the specific heat capacity of a material. [2 marks]
(b) An aluminium pan has a mass of 0.40 kg. Its specific heat capacity is 900 J/kg °C. Calculate the energy needed to warm the pan from 20 °C to 70 °C. [3 marks]
(c) Suggest why water is used in a central heating system. [2 marks]
Model answer
(a) The energy required to change the temperature (1) of an object by one degree Celsius per kilogram of mass (1).
(b) ΔT = 70 − 20 = 50 °C (1). ΔQ = m × c × ΔT = 0.40 × 900 × 50 (1). ΔQ = 18 000 J (1).
(c) Water has a high specific heat capacity (1), so each kilogram carries a lot of energy to the radiators and gives it out as it cools (1).
Exam tip
In (b) the starting and finishing temperatures are given, not ΔT. Write ΔT = 70 − 20 on its own line first. It earns a mark and stops you using 70 by mistake.