Different materials heat up by different amounts even when they receive the same energy. Specific heat capacity tells us how much energy a substance needs to increase its temperature, helping explain everything from hot metal pans to lakes warming slowly.
The equation also helps scientists measure energy transfers. A required practical uses an electric heater, a thermometer and a metal block to calculate specific heat capacity from experimental results.
On this page
- Key terms
- Understanding specific heat capacity
- Equation and calculations
- Required practical investigation
- Interpreting results and graphs
- Evaluating the investigation
- Key summary and quick questions
Key terms
- Thermal energy store: energy stored by a substance because of the energy of its particles.
- Specific heat capacity (c): energy needed to raise the temperature of 1 kg of a substance by 1°C.
- Temperature change (Δθ): final temperature minus initial temperature.
- Energy transferred (ΔE): energy supplied to or released by a substance, measured in joules.
- Gradient: the change in the vertical-axis value divided by the change in the horizontal-axis value.
Understanding specific heat capacity
A substance with a high specific heat capacity needs a lot of energy for a small temperature rise. Water has a specific heat capacity of approximately 4,200 J/kg°C, so 1 kg of water needs about 4,200 J to warm by 1°C.
A substance with a low specific heat capacity warms more for the same energy input, provided the masses are equal and little energy is lost. Copper is approximately 385 J/kg°C, so it takes much less energy than water to warm 1 kg by 1°C.
This is not the same as saying a material is a good conductor. Thermal conductivity describes how quickly energy passes through a material; specific heat capacity describes how much energy is needed for a temperature change.
Equation and calculations
Change in thermal energy = mass × specific heat capacity × temperature change
ΔE = m × c × Δθ
- ΔE = change in thermal energy, joules (J).
- m = mass, kilograms (kg).
- c = specific heat capacity, joules per kilogram per degree Celsius (J/kg°C).
- Δθ = final temperature − initial temperature, degrees Celsius (°C).
Rearrangements: c = ΔE ÷ (m × Δθ); Δθ = ΔE ÷ (m × c); m = ΔE ÷ (c × Δθ).
Worked example: calculating energy
How much energy is needed to warm 0.50 kg of aluminium (c = 900 J/kg°C) from 20°C to 30°C?
- Temperature change = 30 − 20 = 10°C.
- Write the equation: ΔE = m × c × Δθ.
- Substitute: ΔE = 0.50 × 900 × 10.
- Answer: ΔE = 4,500 J.
Worked example: calculating specific heat capacity
A 0.50 kg block gains 3,600 J of energy and its temperature rises by 7.4°C.
- Rearrange: c = ΔE ÷ (m × Δθ).
- Substitute: c = 3,600 ÷ (0.50 × 7.4).
- Calculate: 3,600 ÷ 3.7 = 973 J/kg°C (3 significant figures).
Watch out: mass must be in kilograms, not grams. Use the temperature change, not the final temperature. The equation describes a temperature change without a change of state.
Required practical: investigating specific heat capacity
Required Practical. The aim is to determine the specific heat capacity of a material by measuring electrical energy supplied and the corresponding temperature increase.
Apparatus and setup
- Metal block with holes for a low-voltage electric heater and thermometer or temperature probe.
- Balance, thermometer, stopwatch, insulation, power supply and connecting leads.
- Ammeter in series with the heater; voltmeter connected in parallel across the heater. Alternatively, use a joulemeter to measure electrical energy directly.
Variables
- Independent variable: electrical energy supplied to the block (increased by heating for longer).
- Dependent variable: the temperature rise of the block.
- Control variables: mass and material of the block, heater settings, thermometer position and insulation. When comparing materials, use blocks of equal mass and comparable conditions.
Method
- Use a balance to record the mass of the metal block in kg.
- Insert the heater and thermometer into their separate holes. Improve thermal contact using an approved small amount of thermal fluid, if instructed. Wrap the block in insulation.
- Connect the ammeter in series and the voltmeter in parallel across the heater. Have the circuit checked before switching on.
- Record the initial temperature. Switch on the low-voltage supply and start the stopwatch together.
- Record temperature, time, current and potential difference at fixed intervals, for example every 2 minutes. Check that current and potential difference remain approximately constant.
- Switch off the heater. Calculate the electrical energy supplied for each reading, and subtract the starting temperature to find Δθ.
- Plot a graph of temperature rise against electrical energy input, then calculate c using the equation or the gradient.
- For a repeat, allow the block to return close to its original temperature and use the same equipment and conditions.
Calculating electrical energy
If a joulemeter is not available, first calculate power: P = V × I. Then calculate energy: ΔE = P × t = V × I × t. Potential difference V is in volts, current I in amperes and time t in seconds.
Example: a heater uses 12 V and 2.5 A for 120 seconds. Power = 12 × 2.5 = 30 W. Energy supplied = 30 × 120 = 3,600 J.
Safety
Use a low-voltage supply and follow the teacher’s circuit instructions. The heater and metal block become hot, so avoid touching them until they cool. Keep liquids away from electrical connections; switch off the supply before changing the setup.
Interpreting results and graphs
The following illustrative results are for a 0.50 kg aluminium block heated at a steady 30 W from an initial 20.0°C. They are example data for practice, not measurements from a real experiment.
| Time (min) | Energy input (J) | Temperature (°C) | Rise Δθ (°C) |
|---|---|---|---|
| 0 | 0 | 20.0 | 0.0 |
| 2 | 3,600 | 27.4 | 7.4 |
| 4 | 7,200 | 34.1 | 14.1 |
| 6 | 10,800 | 40.1 | 20.1 |
| 8 | 14,400 | 45.5 | 25.5 |
The temperature increases as more electrical energy is supplied. However, the increases become slightly smaller over equal time intervals, suggesting that more energy is being lost to the surroundings as the block gets hotter.
Results graph: temperature rise (°C) against electrical energy supplied (J) for the illustrative dataset.
What the graph tells you
- Plot electrical energy (J) on the horizontal (x) axis and temperature rise (°C) on the vertical (y) axis.
- A steeper gradient means a greater temperature rise per joule. For equal masses, it indicates a lower specific heat capacity.
- For a straight section of the graph: gradient = Δθ ÷ ΔE = 1 ÷ (m × c). Therefore c = 1 ÷ (m × gradient).
- Choose two widely spaced points on the straight section of a best-fit line, not automatically two experimental points.
- If you plot energy (y) against temperature rise (x) instead, the gradient is m × c and c = gradient ÷ mass. Always check the axes.
Using example data
Using the first 4 minutes as an approximation, Δθ = 14.1°C and ΔE = 7,200 J. Gradient ≈ 14.1 ÷ 7,200 = 0.00196°C/J. Therefore c ≈ 1 ÷ (0.50 × 0.00196) = 1,020 J/kg°C (approximately).
The approximate reference value for aluminium is 900 J/kg°C, so this example result is higher. This does not prove the method is wrong: the heater also warms the surrounding air, insulation and parts of the apparatus. The simple calculation assumes that all electrical energy heats the block.
Evaluating the investigation
- Heat loss: some electrical energy warms the surroundings instead of the block. If this is ignored, the calculated specific heat capacity is often too high. Improve insulation and use a lid/cover where practical.
- Heating the apparatus: the heater and thermometer absorb some energy. Better thermal contact and accounting for the equipment reduce this error.
- Measurement uncertainty: a thermometer with coarse divisions gives uncertainty in Δθ. Use a temperature probe, take a sufficiently large temperature rise, and read consistently.
- Variable power: if V or I changes, assuming constant power gives incorrect energy values. Monitor both or use a joulemeter.
- Repeatability: repeat the experiment with the same method, compare c values and calculate a mean after checking unusual results. Repeating reduces the effect of random variation but does not remove systematic heat loss.
Common misconception: a large temperature rise does not mean a high specific heat capacity. For the same mass and energy input, it means low specific heat capacity.
Key summary
- Specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1°C.
- The energy-change equation is ΔE = m × c × Δθ, using joules, kilograms and °C.
- The practical heats an insulated metal block, measuring mass, energy supplied and temperature change.
- With temperature rise on the y-axis and energy on the x-axis, gradient = 1 ÷ (m × c).
- Energy lost to the surroundings makes the measured temperature rise smaller and can make calculated c too high.
Quick questions
1. What is specific heat capacity?
The energy needed to raise the temperature of 1 kg of a substance by 1°C.
2. Write the equation linking energy, mass, specific heat capacity and temperature change.
ΔE = m × c × Δθ.
3. A 0.50 kg metal block has c = 900 J/kg°C. How much energy raises its temperature by 10°C?
ΔE = 0.50 × 900 × 10 = 4,500 J.
4. A heater has a potential difference of 12 V and current of 2.5 A for 120 s. Calculate the electrical energy supplied.
ΔE = V × I × t = 12 × 2.5 × 120 = 3,600 J.
5. On a graph of temperature rise (y) against energy supplied (x), what does a steeper line mean for blocks of equal mass?
More temperature rise per joule, so a lower specific heat capacity.