J249 · PAG5; P9Practical Activity GroupsCore

PAG5: Measuring Energy Changes

Specific heat capacity relates the energy transferred to a material to its mass and temperature change: E = mcΔT. Measure the mass of the material being investigated and record an initial and final temperature. Use the difference, not the final temperature alone. A temperature rise of 1 °C has the same size as a rise of 1 K; the two temperature scales do not have the same zero.

35 min Video by Science Shorts Practical Activity Groups
SHC - Science GCSE Physics Required Practical - Specific Heat CapacityCurated by Kepler, not produced by us. Watch it before the notes below.
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Before you start

  • Use your teacher’s tier checklist: this overview can include Higher-only material, and is not a tier-specific revision list.
  • OCR Gateway A separate GCSE Physics J249. P9 groups the practical techniques into PAG1–8; the activities here are suitable examples, and centres can choose alternatives covering the required apparatus and techniques.
  • Distinguish energy, power and temperature, and rearrange a three-factor equation.
  • Use the centre’s approved supervised method and equipment limits. These notes prepare you to explain and evaluate practical work; they do not replace carrying it out and recording the evidence.

Watch, pause, try

Video by Science Shorts. This practical overview develops electrical heating and heat-capacity reasoning. Use the board-specific written method: a metal block is not a substitute for Pearson’s required water investigation.

Viewing focus: Relate measured energy input to mass and temperature change

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: Identify one measurement needed for the calculation and one feature that improves the energy-transfer model. Explain the different jobs.

Show answer and reasoning

Mass, energy input and temperature change determine c. Insulation or improved thermal contact addresses an experimental limitation; it does not replace measuring those quantities.

After watching 2: Explain why energy lost from the apparatus can make the calculated c too high when the full electrical input is used.

Show answer and reasoning

The material receives less energy than the numerator assumes. Dividing that excessive input by its measured mΔT overestimates c.

What you'll be able to do

  • Relate measured energy input to mass and temperature change
  • Explain the purpose of insulation and thermal contact
  • Evaluate the direction of heat-loss and apparatus effects
1

Define the material and energy measurement

Specific heat capacity relates the energy transferred to a material to its mass and temperature change: E = mcΔT. Measure the mass of the material being investigated and record an initial and final temperature. Use the difference, not the final temperature alone. A temperature rise of 1 °C has the same size as a rise of 1 K; the two temperature scales do not have the same zero.

In an electrical-heating investigation, a suitable joulemeter measures transferred energy. Alternatively, steady measured potential difference and current give an input estimate E = VIt over the measured time. Identify where the energy meter is connected and what its reading includes. A heater’s nameplate power is not automatically its actual operating power, and the energy supplied to the apparatus is not all necessarily stored in the chosen material.

c=EmΔTc=\frac{E}{m\Delta T}
Use kg, J and K (or a temperature difference in °C) to obtain J kg⁻¹ K⁻¹.
2

Make temperature measurements representative

For a suitable metal block, a heater and temperature probe need good contact with the block, using the approved arrangement. Insulation reduces transfer to the surroundings. A poorly coupled probe can lag behind the material temperature. Keep the apparatus stationary and wait for an appropriate consistent reading rather than selecting a convenient peak without explaining the timing.

For water, measure water mass rather than including the vessel by accident. The approved heater must remain suitably immersed, and gentle mixing can reduce temperature differences within the liquid. A lid and insulation reduce unwanted transfers, but do not make the system perfectly isolated. State whether vessel and heater warming are neglected or accounted for in the model. Follow the centre’s electrical and hot-apparatus precautions.

3

Use the energy balance to explain a bias

If some measured input heats the surroundings or apparatus, the material receives less than the stated input. Dividing the full input by the material’s measured mΔT can therefore give an apparent heat capacity that is too high. Derive the direction from the equation instead of saying only “heat loss affects accuracy”. A small measured temperature rise also makes a fixed thermometer uncertainty a larger fraction of the result.

A series of energy and temperature readings can reveal an initial thermal lag or a changing heating rate. The gradient of input energy E against ΔT gives an apparent mc under the energy-balance approximation; the axes matter, and continuing losses can bias this slope even if their rate is steady. Compare repeated determinations after an appropriate reset and describe a specific improvement, such as reducing an exposed area or improving thermal contact. Repeats do not remove the same systematic loss in every run.

4

PAG5: energy-change and work contexts

Determining a metal’s specific heat capacity is the specification’s example for PAG5, which concerns suitable energy-change/transfer measurements and associated quantities such as work done. Use the measured mass, energy and temperature change to explain the chosen example and its limitations.

Other suitable energy activities can contribute across the course, such as measuring electrical energy used to lift a known load through a measured height. For that example, compare measured input with the useful increase mgh; define the system, measurement methods and transfer losses. The group is not restricted to one universally mandated metal-block recipe.

5

Worked applications

Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.

Example 1A 0.500 kg material receives a measured 4500 J input while its temperature rises from 21.0 °C to 31.0 °C. Estimate c, initially neglecting losses.
1ΔT = 31.0 − 21.0 = 10.0 K.
2c = 4500/(0.500 × 10.0).
Answer
900 J kg⁻¹ K⁻¹ under the stated model.
Example 2In a second illustrative run, 5000 J is supplied but only 4500 J heats a 0.500 kg material through 10.0 K. Compare using the input with using the energy received by the material.
1Using the full input gives 5000/(0.500 × 10.0) = 1000.
2Using energy received gives 4500/(0.500 × 10.0) = 900.
3The apparent estimate is 100/900 × 100 ≈ 11.1% higher.
Answer
Heat loss gives an excessive estimate in this particular calculation model.
Example 3A steady 12.0 V and 2.00 A are measured for 180 s. Why is the calculated energy not automatically equal to the increase in the water’s thermal store?
1Input estimate VIt = 12.0 × 2.00 × 180 = 4320 J.
2Some input can warm the vessel and heater or transfer to the surroundings.
3The water-only mcΔT model needs those neglected transfers evaluated.
Answer
4320 J electrical input; the water’s share requires an energy-balance judgement.

Common mistakes to avoid

Using the final temperature in place of its rise.
Calculate a difference from recorded readings.
Saying repeats remove heat loss.
Repeats can assess scatter; change the transfer conditions to reduce a systematic loss.

Key takeaways

  • You can relate measured energy input to mass and temperature change.
  • You can explain the purpose of insulation and thermal contact.
  • You can evaluate the direction of heat-loss and apparatus effects.

Apply it in an exam

Question 1: A steady 12.0 V and 2.00 A are measured for 180 s. Why is the calculated energy not automatically equal to the increase in the water’s thermal store?

Show answer and reasoning

4320 J electrical input; the water’s share requires an energy-balance judgement.

  • Input estimate VIt = 12.0 × 2.00 × 180 = 4320 J.
  • Some input can warm the vessel and heater or transfer to the surroundings.
  • The water-only mcΔT model needs those neglected transfers evaluated.

Explore further

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

  • What does insulation mainly help reduce?
  • If the full input is used although some heats the surroundings, which bias can result?

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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