Every calculation in this module assumes all the energy supplied goes into the material being heated. That is a model, and real equipment does not obey it exactly. Questions ask you to spot where energy escapes, say how it changes a result, and suggest how to reduce the effect.
Here the ideas from specific heat capacity and latent energy meet real measurement.
Where does the energy go in a real heating experiment?
Imagine a metal block or a container of water heated by an electrical heater, with the temperature measured by a thermometer. Only part of the electrical energy ends up as a temperature rise in the sample.
Energy is also transferred to the container, to the heater and thermometer themselves, and to the air around by conduction, convection, radiation and evaporation. The hotter the sample gets above room temperature, the faster these losses occur.
- Name the loss: for example, energy transferred to the surroundings.
- Say the direction: less energy raises the temperature than the calculation assumes.
- State the effect on the result: the observed temperature rise is smaller than predicted, or a value worked out from the supplied energy is too high.
- Suggest a reduction: lagging, a lid, a shorter test, stirring.
Worked example
An electric kettle rated 2000 W heats 1.0 kg of water from 20 °C to 100 °C in 200 s. (Invented example data.) Use c = 4200 J/kg °C. Compare the ideal case with the measured case.
Step 1, useful energy: Q = 1.0 × 4200 × 80 = 336 000 J.
Step 2, ideal time with no loss: t = 336 000 ÷ 2000 = 168 s.
Step 3, energy actually supplied: E = 2000 × 200 = 400 000 J.
Step 4, energy lost: 400 000 − 336 000 = 64 000 J.
Step 5, efficiency: 336 000 ÷ 400 000 = 0.84, which is 84%.
Step 6, effect on c: a student who used the supplied energy would calculate c = 400 000 ÷ (1.0 × 80) = 5000 J/kg °C. That is about 19% above the accepted 4200, and the loss explains why.
The mistake to watch for
A common slip is to blame human error in general terms instead of naming the physics.
Mistaken answer: “The result is wrong because the student measured the temperature badly.”
This does not explain why the value is always too high, and it earns no marks for physics.
The correction is to name the energy transfer and its direction: some of the supplied energy heated the container and surroundings, so the water gained less energy than was supplied. Dividing the larger supplied energy by the smaller temperature rise gives a c that is too high.
Check yourself
1. A 0.50 kg copper block (c = 390 J/kg °C) is heated by a 500 W heater for 60 s. With no losses, what temperature rise would be expected?
Show answer
E = 500 × 60 = 30 000 J. m × c = 0.50 × 390 = 195 J/°C. Δθ = 30 000 ÷ 195 ≈ 154 °C (about 150 °C).
2. In question 1 the measured rise is 120 °C. What percentage of the supplied energy heated the block?
Show answer
Useful energy = 195 × 120 = 23 400 J. Fraction = 23 400 ÷ 30 000 = 0.78, so 78%. The remaining 6600 J went elsewhere.
3. A student divides the electrical energy supplied by mass × temperature rise to find c. Will the result be too high or too low, and why?
Show answer
Too high. Some supplied energy was lost, so the temperature rise is smaller than it would be with no loss. Dividing the full supplied energy by the smaller rise gives a larger c.
Where this leads next
With all five lessons done, test the whole module with the heat calculations practice set. If energy losses in other devices interest you, work, energy and efficiency covers the same idea from the efficiency side.
Turning a half-formed idea into a precise examiner-friendly sentence is something a teacher can do live in online one-to-one Physics tuition.