🌡️ Define specific heat capacity and explain what it means physically
🔢 Use the equation E = mcΔT to solve problems involving energy, mass and temperature change
⚡ Describe and carry out the required practical using an electric heater and thermometer
📊 Compare the specific heat capacities of different materials and explain practical consequences
📉 Interpret and sketch cooling curves for substances
🔬 Evaluate sources of error in the practical and suggest improvements
What is Specific Heat Capacity?
When you heat an object, you transfer energy to it and its temperature rises. But different materials need different amounts of energy to achieve the same temperature rise — this is captured by the idea of specific heat capacity.
Specific Heat Capacity (c) is the amount of energy required to raise the temperature of 1 kg of a substance by 1 °C (or 1 K).
A high specific heat capacity means a substance needs a lot of energy to warm up — and it also releases a lot of energy as it cools down. Water has an exceptionally high specific heat capacity, which is why it is used in central heating systems and why coastal areas have milder climates than inland regions.
A low specific heat capacity means a substance heats up and cools down quickly. Metals like copper and aluminium have relatively low specific heat capacities compared to water, which is why metal objects on a hot day feel much hotter than surrounding water or wet sand.
The higher the specific heat capacity, the more energy is needed to change the temperature — and the more energy is stored for a given temperature rise.
The Equation: E = mcΔT
The relationship between energy transferred, mass, specific heat capacity and temperature change is given by:
E = m × c × ΔT
Symbol
Quantity
SI Unit
E
Energy transferred (thermal energy)
Joules (J)
m
Mass of the substance
Kilograms (kg)
c
Specific heat capacity
J/kg°C or J/kg·K
ΔT
Temperature change (T_final − T_initial)
°C or K
The equation can be rearranged to find any unknown:
m = E ÷ (c × ΔT) | c = E ÷ (m × ΔT) | ΔT = E ÷ (m × c)
ΔT is the change in temperature, not the final temperature. Always subtract: ΔT = T_final − T_initial. A temperature change of 1°C is the same size as a change of 1 K.
Note: 1 J/kg°C = 1 J/kg·K. You may see both notations in textbooks and data tables — they mean exactly the same thing.
Specific Heat Capacities of Common Materials
Here are approximate values of specific heat capacity for materials you will encounter at GCSE:
Material
c (J/kg°C)
Notes
Water
4,200
Very high — used in heating systems
Concrete / brick
880
Used in thermal storage walls
Aluminium
900
Good for cooking pans
Copper
390
Heats up quickly
Iron / steel
450
Common structural metal
Glass
670
Moderate heat storage
Air
1,000
Low density limits heat storage
Water's specific heat capacity (~4,200 J/kg°C) is about 10 times higher than copper (~390 J/kg°C). This means 1 kg of water needs about 10 times more energy than 1 kg of copper to rise by 1°C. This is why water is the preferred coolant in car radiators and central heating systems — it can absorb enormous amounts of thermal energy without large temperature changes.
Real-world applications: storage heaters use concrete or brick (which release heat slowly), while central heating uses water pipes. Both rely on high specific heat capacity materials.
Using an Electric Heater — Energy Input
In the required practical, an electric heater is used to supply a known amount of energy to a block of metal or water. The energy supplied by an electric heater can be calculated from:
E = P × t
Where P is the power of the heater in watts (W) and t is the time in seconds (s). The energy E is in joules (J).
You can also use:
E = V × I × t
Where V is voltage (V), I is current (A), and t is time (s). This is useful when the power rating of the heater is not known and you measure V and I with a voltmeter and ammeter.
By measuring the temperature change (ΔT) of a known mass (m) of material when a known amount of energy (E) is supplied, you can calculate the specific heat capacity:
c = E ÷ (m × ΔT) = (P × t) ÷ (m × ΔT)
In practice, your calculated value of c will be higher than the accepted value because some energy is lost to the surroundings (heat loss), rather than going entirely into heating the block. This is a systematic error.
Cooling Curves
A cooling curve is a graph of temperature (y-axis) against time (x-axis) as a substance cools down. Understanding cooling curves helps you interpret energy changes during cooling.
Key features of a cooling curve for a pure substance:
The temperature decreases gradually as the substance loses thermal energy to the surroundings.
At the melting/freezing point, the temperature stays constant for a period — this is a plateau on the graph. During this plateau, energy is being released (latent heat) but temperature does not change.
After solidification, the temperature continues to fall.
For a substance that does not change state (e.g. a metal block cooling from 80°C to room temperature), the cooling curve is a smooth curve that flattens out as it approaches room temperature — because the rate of heat loss decreases as the temperature difference between the object and surroundings decreases (Newton's Law of Cooling).
The gradient of a cooling curve tells you the rate of temperature change. A steeper gradient means faster cooling. Materials with a low specific heat capacity cool faster for the same mass and surface area.
On a cooling curve: steeper slope = faster cooling = lower specific heat capacity (for same mass and conditions). A flat plateau indicates a change of state — latent heat is being released at constant temperature.
Cooling curves are used in industry to identify unknown materials, check purity (impurities lower the freezing point), and design thermal management systems.
Example 1 — Basic E = mcΔT calculation
A 2.0 kg block of aluminium is heated from 20°C to 70°C. The specific heat capacity of aluminium is 900 J/kg°C. Calculate the energy transferred to the block.
1Write down the formula: E = m × c × ΔT
2Identify values: m = 2.0 kg, c = 900 J/kg°C, ΔT = 70 − 20 = 50°C
Example 2 — Finding specific heat capacity from practical data
In an experiment, a 1.5 kg copper block is heated using a 48 W electric heater for 5 minutes. The temperature rises from 18°C to 46°C. Calculate the specific heat capacity of copper from this data and suggest why it differs from the accepted value of 390 J/kg°C.
1Calculate energy supplied: E = P × t = 48 × (5 × 60) = 48 × 300 = 14,400 J
2Calculate temperature change: ΔT = 46 − 18 = 28°C
3Rearrange for c: c = E ÷ (m × ΔT) = 14,400 ÷ (1.5 × 28)
4Calculate denominator: 1.5 × 28 = 42
5Calculate c: c = 14,400 ÷ 42 ≈ 343 J/kg°C
6Evaluate: The calculated value (343 J/kg°C) is lower than the accepted value (390 J/kg°C). This could be because some energy was lost to the surroundings (thermal energy escaped to the air) so less energy actually went into the block than was supplied by the heater. The thermometer may also not have reached thermal equilibrium with the block.
c ≈ 343 J/kg°C. Lower than accepted value because of heat losses to surroundings.
Example 3 — Rearranging to find mass
A central heating system transfers 630,000 J of energy to water. The water temperature rises by 15°C. The specific heat capacity of water is 4,200 J/kg°C. What mass of water is in the system?
1Write the formula: E = m × c × ΔT
2Rearrange for mass: m = E ÷ (c × ΔT)
3Substitute values: m = 630,000 ÷ (4,200 × 15)
4Calculate denominator: 4,200 × 15 = 63,000
5Calculate mass: m = 630,000 ÷ 63,000 = 10 kg
m = 10 kg
Example 4 — Comparing two materials
A 3.0 kg block of iron (c = 450 J/kg°C) and a 3.0 kg block of copper (c = 390 J/kg°C) are both supplied with 27,000 J of energy. Calculate the temperature rise of each block and state which heats up more.
4Compare: Copper has a lower specific heat capacity, so the same energy causes a greater temperature rise in copper than in iron.
Iron: ΔT = 20°C | Copper: ΔT ≈ 23°C. Copper heats up more because it has a lower specific heat capacity.
Question 1. What does the symbol ΔT represent in the equation E = mcΔT?
Question 2. Water has a specific heat capacity of 4,200 J/kg°C. What energy is needed to heat 0.5 kg of water by 10°C?
Question 3. A 4.0 kg block of concrete (c = 880 J/kg°C) is heated by 25°C. Calculate the energy stored. Give your answer in J.
Question 4. In the required practical, why is the calculated value of specific heat capacity often higher than the accepted value?
Question 5. An electric heater with power 60 W heats a 2.0 kg metal block for 4 minutes. The temperature rises by 36°C. Calculate the specific heat capacity of the metal. Give your answer in J/kg°C.
Challenge 1 — Multi-step calculation
A solar thermal panel heats 80 kg of water from 12°C to 55°C during the day. At night the water cools back to 12°C, releasing energy into the house. The specific heat capacity of water is 4,200 J/kg°C.
(a) Calculate the energy stored during the day.
(b) If the same energy were used to heat a 400 kg concrete block (c = 880 J/kg°C) starting from 12°C, what would the final temperature of the concrete be?
(a) E = 80 × 4,200 × (55−12) = 80 × 4,200 × 43 = 14,448,000 J ≈ 14.4 MJ
(b) ΔT = E ÷ (m × c) = 14,448,000 ÷ (400 × 880) = 14,448,000 ÷ 352,000 ≈ 41.0°C
Final temperature = 12 + 41 = 53°C
Challenge 2 — Evaluating practical data
A student measures the specific heat capacity of water using a 100 W heater, 0.5 kg of water, heating for 3 minutes. The thermometer reads an initial temperature of 20°C and a final temperature of 34°C.
(a) Calculate the specific heat capacity of water from the student's results.
(b) The accepted value is 4,200 J/kg°C. Calculate the percentage error.
(c) Explain two reasons why the student's result differs from the accepted value.
(c) Possible reasons:
1. Heat losses to surroundings — energy escaped from the water/container to the air, so less energy went into heating the water, causing a larger temperature rise than expected for the energy supplied. This makes c appear smaller.
Note: Actually here c is smaller than accepted, suggesting more temperature rise per joule — possibly the thermometer did not fully mix/reach equilibrium, or the heater was not fully submerged so some energy went into the container rather than the water.
2. Energy absorbed by the container — the beaker or calorimeter also absorbed energy, so not all 18,000 J went into the water alone, meaning the actual energy heating the water was less than calculated.
Challenge 3 — Cooling curve interpretation
A student records the cooling of two 1 kg metal blocks (Block A and Block B) starting from 90°C in the same room conditions. After 10 minutes, Block A is at 50°C and Block B is at 65°C.
(a) Which block has the higher specific heat capacity? Explain your answer.
(b) Both blocks lose 8,000 J of energy in the first 10 minutes. Calculate the specific heat capacity of each block.
(c) Sketch and describe the expected shape of the cooling curves for both blocks on the same axes.
(a) Block B has the higher specific heat capacity. Despite losing the same amount of energy, Block B only cooled by 25°C (90→65°C) compared to Block A which cooled by 40°C (90→50°C). A higher specific heat capacity means more energy must be removed to produce the same temperature drop.
(c) Both curves start at 90°C and decrease, levelling off towards room temperature. Block A has a steeper initial gradient (cools faster). Block B has a shallower gradient throughout (cools more slowly). Both curves are not straight lines — they curve and flatten as the temperature approaches room temperature (smaller temperature difference = slower rate of heat loss).
Challenge 4 — Extended writing: Why is water used in central heating?
Using your knowledge of specific heat capacity, explain why water is used as the fluid in domestic central heating systems rather than a metal like copper or oil. Your answer should refer to the equation E = mcΔT and include a comparison of specific heat capacity values. [6 marks]
Model answer points:
• Water has a very high specific heat capacity of approximately 4,200 J/kg°C, much higher than copper (~390 J/kg°C) or typical oils (~1,700 J/kg°C).
• From E = mcΔT, a large c means a large amount of energy E can be stored for a given mass m and temperature change ΔT.
• This means water can absorb a large amount of thermal energy from the boiler without its temperature rising excessively, and can carry (transport) that energy around the system to radiators efficiently.
• At the radiator, the water releases a large amount of energy (E = mcΔT) for a relatively small temperature drop, heating the room effectively.
• Using copper would require a much larger mass or volume to store the same amount of energy, making the system heavier and more expensive.
• Water is also inexpensive, readily available, safe at typical heating temperatures, and does not corrode metal pipes significantly.
• Conclusion: Water's high specific heat capacity makes it the most practical and efficient fluid for transferring and storing thermal energy in central heating.
🔬 AQA Required Practical — Specific Heat Capacity
Aim
To measure the specific heat capacity of a solid material (e.g. aluminium or copper block) using an electric heater and thermometer, and compare the result with the accepted value.
Equipment
Metal block (aluminium or copper) with two holes drilled — one for the heater, one for the thermometer
Electric immersion heater (e.g. 25 W or 50 W)
Thermometer (or digital temperature probe)
Joulemeter (or ammeter + voltmeter + stopwatch)
Power supply (variable DC)
Electronic balance (to measure mass of block)
Insulating material (cotton wool or polystyrene) to wrap the block
Connecting wires
Method
Measure and record the mass (m) of the metal block using a balance.
Insert the electric heater into one hole and the thermometer into the other hole in the block. Add a drop of oil to each hole to improve thermal contact.
Wrap the block in insulating material (cotton wool) to reduce heat loss to the surroundings.
Connect the heater to the power supply and a joulemeter (or set up a voltmeter and ammeter circuit). Record the initial temperature T₁.
Switch on the power supply and simultaneously start the stopwatch (if not using a joulemeter).
Heat the block for a fixed time (e.g. 10 minutes = 600 s). Record the energy supplied (E from joulemeter, or E = V × I × t).
Record the final temperature T₂ when the heater is switched off.
Calculate ΔT = T₂ − T₁.
Calculate the specific heat capacity: c = E ÷ (m × ΔT).
Repeat with a different material and compare results.
Safety
⚠️ The metal block will become very hot — do not touch it during or immediately after the experiment. Use heat-resistant gloves if handling.
⚠️ Ensure all electrical connections are secure before switching on the power supply.
⚠️ Do not exceed the voltage rating of the heater.
⚠️ Keep water away from electrical connections.
Results Table
Material
Mass m (kg)
Initial Temp T₁ (°C)
Final Temp T₂ (°C)
ΔT (°C)
Energy E (J)
c = E ÷ (m × ΔT) (J/kg°C)
Accepted c (J/kg°C)
Aluminium
900
Copper
390
Analysis Questions
1. Your measured value of c is likely to be higher than the accepted value. Explain why.
Some of the electrical energy supplied by the heater is lost to the surroundings (the air around the block, the insulation, and the thermometer) rather than going entirely into the block. This means ΔT is smaller than it would be if all the energy went into the block, so the calculated c = E ÷ (m × ΔT) is larger than the true value.
2. Suggest two improvements to make the experiment more accurate.
1. Better insulation — wrap the block in more layers of insulating material or use a polystyrene enclosure to reduce heat loss to the surroundings.
2. Add oil/thermal paste to holes — ensure good thermal contact between the heater/thermometer and the block to reduce the amount of energy stored in air gaps.
3. Use a data logger — record temperature continuously and use the temperature at the moment the heater is switched off more precisely.
4. Repeat and average — repeat the experiment and take a mean value of c to reduce random errors.
3. Why is it important to measure the mass of the block rather than estimating it?
The mass m appears in the denominator of c = E ÷ (m × ΔT). A small error in m leads to a proportional error in c. Estimating mass (e.g. from the block's labelled size) may not account for the holes drilled into the block, which reduce the actual mass. An electronic balance gives a precise measurement and reduces this source of systematic error.
4. A student uses a 50 W heater for 8 minutes on a 1.0 kg aluminium block. The temperature rises from 21°C to 42°C. Calculate c and find the percentage error from the accepted value of 900 J/kg°C.
E = 50 × 480 = 24,000 J
ΔT = 42 − 21 = 21°C
c = 24,000 ÷ (1.0 × 21) ≈ 1,143 J/kg°C
% error = (1143 − 900) ÷ 900 × 100 = 243 ÷ 900 × 100 ≈ 27%
The calculated value is 27% higher than the accepted value, indicating significant heat losses to the surroundings.