Physics 11–12 · Year 11

Rate of energy conversion in a circuit: heating water with a resistance coil

Module 4: Electricity and Magnetism (Electric Circuits)

Practical, model not builtMedium risk

School laboratory, not for home

In a school laboratory, with a teacher supervising, under the school's risk assessment. Not for home.

This site has no interactive model of its own. Where a step or a material names a Concept Studio model, simulation or tool, it has not been built; an external simulation a step names (for example PhET) is not part of this site.

The idea

Electrical power is the product of voltage and current, and in a resistor it appears as heat at the rate I squared R.

Safety card

Medium riskA teacher supervises

Setting: In a school laboratory, with a teacher supervising, under the school's risk assessment. Not for home.

Hazards

hot coil and water
coil run dry

Controls

switch on only when immersed
safety glasses

Note

Heat and hot liquids: record the activity in RiskAssess (https://www.riskassess.com.au/) and follow the Science ASSIST risk management information sheet (https://asta.edu.au/resource/ais-risk-management-and-risk-assessment/).

What you need

Nichrome heating coil of about 24 ohms in a calorimeter with 0.200 kg of water (the NSW Department of Education Module 3 guide further activity, used here for P = VI)
12 V supply, ammeter, voltmeter, thermometer or temperature probe reading to 0.1 degree, stopwatch, balance
Two 12 V lamps of different power for a brightness comparison

How to do it

Record the water mass and starting temperature; connect the coil with the meters.
Run at 12 V for 300 s, logging V and I every 30 s and the temperature every 60 s.
Compute electrical energy V I t and the water's energy gain m c delta T; compare them.
Run again at 6 V for 1200 s, four times as long, logging V and I as before: the meters show a quarter of the power, and the same electrical energy warms the water by close to the 12 V run's rise. (A 300 s run at 6 V warms the water by only 0.54 degrees, too little to compare with a thermometer reading to 0.1 degree.)
Compare the two lamps' brightness with their measured V I.

What you should see

At 12 V and 0.50 A the coil converts 6.0 W (R = 24 ohms, I^2 R = 6.0 W); over 300 s that is 1800 J, warming 0.200 kg of water by 2.2 degrees. Halving the voltage halves the current and quarters the power to 1.5 W, so a 6 V run of 1200 s delivers the same 1800 J and warms the water by close to the same 2.2 degrees, a little less because the longer run loses more energy to the surroundings. The thermal energy gained is less than the electrical energy supplied, the difference being loss to the calorimeter and the surroundings; the learner knows it worked when the two agree within about 10 per cent.

What changes

What you change: supply voltage (and hence current)
What you measure: rate of temperature rise (power)
What you keep the same: mass of water
same coil
insulation

Common misconceptions

Each of these ideas is wrong, and the activity is a chance to test it.

Voltage is the energy flowing; it is energy per unit charge, and power needs the current as well.
A thicker wire heats more because more current flows; at fixed current a thicker wire has less resistance and heats less.

Curriculum references

The NSW syllabus outcomes and Australian Curriculum v9 codes this activity supports. They are references, not a verified or complete curriculum alignment.

Physics Stage 6 Syllabus (2017), current: Year 11 until the end of 2026, Year 12 until Term 3 2027PH11-11 PH11/12-5
Physics 11-12 Syllabus (2025), not yet taught: Year 11 from Term 1 2027, Year 12 from Term 4 2027, first HSC examination 2028PY-11-03 PY-11WS-03
Australian Curriculum v9No Australian Curriculum v9 code is listed.

Sources

The pages the author read to write this activity.

All Concept Studio activities