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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.
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
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.
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.
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.
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.
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.
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.
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?
Common mistakes to avoid
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
- OCR J249 official specification (opens in a new tab)
Check the assessed scope and practical or mathematical requirements.
- Read or explore this topic (opens in a new tab)
An external explanation or interactive activity to use alongside the lesson. It may include extension material.
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.
Keep practising
Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.