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Science Lab

Take the readings. Answer on your own data.

A 3D school lab in your browser. Set up each GCSE required practical, take the readings yourself, then answer exam questions written round your numbers.

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The Lab during a rates run: the gas syringe on the bench, and the notebook with the volume of hydrogen recorded every 10 seconds and the points plotted below the table.
Real screens from the Lab: a rates of reaction run, with its readings and graph.
  • 28 AQA required practicals
  • Nine steps as in your class
  • Free no account needed

Your readings become the question

One run of specific heat capacity. The rows you measured are the rows the exam question uses, so a new run means new numbers.

Table 1: Heating the block. The row at 60 s is the one on the bench. Question 02.3 uses the rows at 120 s and 600 s. The temperature at 360 s is ringed as anomalous.
Time in sEnergy transferred in JTemperature in °C
0022.0
60164322.5
120328624.0
180492925.5
240657227.5
300821629.0
360985929.0
42011 50232.5
48013 14534.0
54014 78835.5
60016 43237.5

02.3Use your readings at 120 s and 600 s in Table 1 to calculate the specific heat capacity of aluminium.[4 marks]

  1. ΔE = 16 432 − 3286 = 13 146 JM1, mark awarded
  2. Δθ = 37.5 − 24.0 = 13.5 °CM1, mark awarded
  3. c = 13 146 ÷ (1.002 × 13.5)M1, mark awarded
  4. c = 972 J/kg °CA1, mark awarded

4 marks out of 4

Fig. 1 One run of GCSE Physics required practical 1, specific heat capacity, in the Lab (seed 20261004, run C29HO). The bench is shown at one minute: the readings on its stopwatch, joulemeter and thermometer are row 2 of the table. Question 02.3 is built on the rows at 120 s and 600 s. Your own run gives different numbers.

One answer, marked line by line

Marks come from written mark points, so the scheme says exactly where a mark is earned. Below, question 02.3 with its mark points shown one by one: a worked slip, the fix, and the same question on a new run.

  1. The slip

    A student used the reading at 0 s

    1. ΔE = 16 432 − 0 = 16 432 J
    2. Δθ = 37.5 − 22.0 = 15.5 °C
    3. c = 16 432 ÷ (1.002 × 15.5)
    4. c = 1058 J/kg °C
    1. M1ΔE = 16 432 − 3286 = 13 146 J, no mark
    2. M1Δθ = 37.5 − 24.0 = 13.5 °C, no mark
    3. M1c = 13 146 ÷ (1.002 × 13.5), mark awarded
    4. A1c = 972 J/kg °C, mark awarded

    2 marks out of 4The method marks are carried forward from the student's own numbers.

  2. The reason

    The reason, and the marks to win back

    Your method is right, but you used the reading at 0 s. Use the readings at 120 s and 600 s, as the question asks.

    Not 0 s · 0 J · 22.0 °Cbut 120 s · 3286 J · 24.0 °C

    The two mark points to win back

    • ΔE = 16 432 − 3286 = 13 146 J
    • Δθ = 37.5 − 24.0 = 13.5 °C

    The same answer always gets the same mark, and every mark points to the line of the scheme it came from.

  3. A new run

    A new run, new readings, the same question

    Seed 20261005, run C29HP: at 120 s, 5364 J and 22.0 °C; at 600 s, 26 822 J and 44.5 °C.

    1. ΔE = 26 822 − 5364 = 21 458 J
    2. Δθ = 44.5 − 22.0 = 22.5 °C
    3. c = 21 458 ÷ (1 × 22.5)
    4. c = 954 J/kg °C
    1. M1ΔE = 26 822 − 5364 = 21 458 J, mark awarded
    2. M1Δθ = 44.5 − 22.0 = 22.5 °C, mark awarded
    3. M1c = 21 458 ÷ (1 × 22.5), mark awarded
    4. A1c = 954 J/kg °C, mark awarded

    4 marks out of 4Every run has its own numbers, so a right answer shows the method, not a remembered value.

Try specific heat capacity yourself Read the full specimen run

Measure

Take every reading yourself, on the bench and up close

A reading starts the way it does in class: from wherever your eye happens to be. To read a measuring cylinder you move your eye, on the 3D bench or in the 2D close-up, until it is level with the bottom of the meniscus. Off level, the scale lines up with the wrong mark. Nothing is right by default.

How density is measured

Where it startsEye 6 cm below · seems to read 17 cm³

At eye levelThe reading: 19 cm³

2D close-up

The scale goes up in 1 cm³ divisions. From where your eye is, the bottom of the meniscus looks 6.6 small divisions above the 10 mark. Your eye is 6 cm below the bottom of the meniscus.100 cm³eye level1020cm³
The scale goes up in 1 cm³ divisions. From where your eye is, the bottom of the meniscus looks 8.7 small divisions above the 10 mark. Your eye is level with the bottom of the meniscus.100 cm³eye level20cm³
Fig. 2Density, the first volume of sugar solution on the website’s specimen run (seed 20261004). The app’s 2D close-up: the side view and what that eye sees.

The nine steps your class does

  1. Plan
  2. Collect and check
  3. Set up
  4. Draw the table first
  5. Take readings
  6. Spot anomalies
  7. Make safe
  8. Process the data
  9. Conclude and evaluate

Your course, your numbers

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