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    Specific heat capacity required practical: method, results and errors

    By Engr. Mehmood Kakar · Updated

    Quick answer

    Heat a metal block of known mass with an electric heater, measure the energy supplied and the temperature rise, then use c = ΔE ÷ (m × Δθ). Find the energy with a joulemeter, or from E = V × I × t using a voltmeter, ammeter and stopwatch. Your answer is usually higher than the true value because some energy heats the surroundings instead of the block.

    What you are finding

    Specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1 °C. In this practical you supply a measured amount of energy to a metal block, measure how much its temperature rises, and work out c = ΔE ÷ (m × Δθ).

    The same method works for water in an insulated beaker. The numbers in the examples below are sample values to show the method, not real measurements.

    Equipment

    • A metal block (often 1 kg of aluminium or copper) with two holes: one for the heater, one for the thermometer
    • An immersion heater, low-voltage power supply and leads
    • A joulemeter, or an ammeter, a voltmeter and a stopwatch
    • A thermometer (or temperature probe) and a pipette of water or oil for the thermometer hole
    • Insulation to wrap round the block, and a top-pan balance to check its mass

    Method

    1. Measure the mass of the block on a balance.
    2. Wrap the block in insulation. Put the heater in the large hole and the thermometer in the small hole, with a few drops of water or oil to give good thermal contact.
    3. Connect the heater to the power supply through the joulemeter (or with the ammeter in series and the voltmeter in parallel across the heater).
    4. Record the starting temperature.
    5. Switch on and start the stopwatch. Note the ammeter and voltmeter readings; they should stay steady.
    6. After 10 minutes, switch off. Keep watching the thermometer and record the highest temperature it reaches, because the block keeps warming for a short time after the heater is off.
    7. Work out the energy supplied and the temperature rise, then calculate c.

    Working out c from results

    A 1.00 kg aluminium block is heated for 10 minutes. The heater runs at 12.0 V and 4.00 A. The temperature rises from 20.0 °C to 50.0 °C. Calculate the specific heat capacity.

    1. Power: P = V × I = 12.0 × 4.00 = 48.0 W
    2. Time: t = 10 × 60 = 600 s
    3. Energy: ΔE = P × t = 48.0 × 600 = 28 800 J
    4. Temperature rise: Δθ = 50.0 − 20.0 = 30.0 °C
    5. c = ΔE ÷ (m × Δθ) = 28 800 ÷ (1.00 × 30.0)
    6. c = 960 J/kg °C

    Data books give about 900 J/kg °C for aluminium, so this result is a little high. The next two sections explain why, and how a graph helps.

    Watch the time unit. Power is in watts, which are joules per second. Using 10 (minutes) instead of 600 (seconds) makes the energy, and your answer, 60 times too small.

    The graph method

    Instead of one final reading, record the temperature every minute and work out the energy supplied up to each reading (from the joulemeter, or P × t). Plot temperature on the y-axis against energy supplied on the x-axis.

    Because Δθ = ΔE ÷ (m c), the straight part of the graph has a gradient of 1 ÷ (m × c). So:

    The straight part of a temperature–energy graph for a 1.00 kg block has a gradient of 0.00110 °C per joule. Calculate c.

    1. gradient = 1 ÷ (m × c), so c = 1 ÷ (m × gradient)
    2. c = 1 ÷ (1.00 × 0.00110)
    3. c = 909 J/kg °C (3 s.f.)

    The graph usually curves at the start, because the heater has to warm up before the block does. Using only the straight part leaves out that start-up delay, which is one reason the graph method gives a better value than a single before-and-after reading.

    Why your value is too high

    • Energy lost to the surroundings. Some energy warms the air, the bench and the heater. It is counted in ΔE but doesn’t raise the block’s temperature, so c comes out too big. This is the main reason.
    • Poor thermal contact. An air gap round the thermometer makes it read lower than the block’s true temperature, so Δθ is too small and c too big.
    • Missing the peak. If you read the temperature the moment you switch off, you miss the last part of the rise.

    Improving the practical

    • Wrap the block in insulation, including the top, to reduce energy transfer to the surroundings.
    • Use oil or water in the thermometer hole for good thermal contact.
    • Use a thermometer or temperature probe with a better resolution (for example 0.1 °C).
    • Take readings every minute and use the gradient of the straight part of the graph.
    • Repeat the experiment and calculate a mean value of c.

    Exam questions on this practical

    Expect to be asked for a method (often 6 marks), a calculation from given readings, how to find the energy without a joulemeter, or why the value is higher than the data-book value. For the method, cover what you measure, how you measure it, the calculation and one improvement. For the “too high” question, name the cause and the effect: energy is transferred to the surroundings, so not all of the energy supplied raises the block’s temperature.

    The independent, dependent and control variables come up too. Here the energy supplied (or time of heating) is the independent variable, the temperature rise is the dependent variable, and the mass and material of the block are kept the same.

    Practice

    1. A joulemeter reads 18 000 J after heating a 0.500 kg copper block. Its temperature rises by 92 °C. Calculate the specific heat capacity of copper.

    c = ΔE ÷ (m × Δθ) = 18 000 ÷ (0.500 × 92) → c = 391 J/kg °C (3 s.f.)

    2. A heater runs at 10.0 V and 3.00 A for 5 minutes. How much energy does it supply?

    P = V × I = 30.0 W, t = 300 s → E = P × t = 30.0 × 300 → E = 9000 J

    3. A student’s value for aluminium is 1100 J/kg °C. Give one reason it is higher than 900 J/kg °C and one way to improve the experiment.

    Energy was transferred to the surroundings, so not all of the energy supplied raised the block’s temperature. Improve it by insulating the block (or using oil in the thermometer hole, or using the gradient of a graph).

    Frequently asked questions

    What is the equation for specific heat capacity?

    Change in thermal energy = mass × specific heat capacity × temperature change, ΔE = m c Δθ. Rearranged for the practical, c = ΔE ÷ (m × Δθ). It is on the AQA equation sheet.

    Why is the measured specific heat capacity higher than the real value?

    Some of the energy from the heater warms the air, the bench and the heater itself instead of the block. You count that energy in ΔE, but it does not raise the block’s temperature, so ΔE ÷ (mΔθ) comes out too big.

    Why do you put insulation around the block?

    Insulation reduces the energy transferred from the block to the surroundings, so more of the energy supplied raises the block’s temperature and your value of c is closer to the true value.

    Why put a drop of water or oil in the thermometer hole?

    It fills the air gap between the thermometer and the metal. Air is a poor conductor, so the liquid gives better thermal contact and a more accurate temperature reading.

    How do you find the energy supplied without a joulemeter?

    Measure the potential difference with a voltmeter and the current with an ammeter to find the power, P = V × I. Then multiply by the time in seconds: E = P × t.

    Engr. Mehmood Kakar

    Written by Engr. Mehmood Kakar

    Physics educator and engineer (MSc Physics, B.Ed, M.Ed) with 8 years of hands-on teaching, from college students to teacher training. About the author · LinkedIn (opens in a new tab)

    Checked against the AQA GCSE Physics (8463) specification on 15 November 2026.