Multiply heater power by time in seconds to find the input energy in joules. If all of that energy heats the sample and there is no change of state, divide it by the sample’s mass and specific heat capacity to find its temperature rise.
The 2027 SEC G3 Physics K323 syllabus covers power and specific heat capacity. This worked guide combines the two relationships and shows what changes when the question specifies energy losses.
- Convert minutes to seconds and match mass units to the specific heat capacity.
- Calculate the heater’s input energy using E = Pt.
- Use only the energy transferred to the sample in E = mcΔT.
- Add the temperature rise to the starting temperature if asked for the final temperature.
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Keep the two energy quantities clear
A power of 60 W means 60 J of energy is transferred each second. A heater running for two minutes therefore transfers energy for 120 seconds.
For a sample with specific heat capacity c, the energy required for a temperature change ΔT is mcΔT. Use kilograms when c is given per kilogram. A temperature difference of 1 °C has the same numerical size as a difference of 1 K.
The two energy amounts are equal in the ideal calculation. If the container warms or energy escapes to the surroundings, the sample receives less than the heater supplies.
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Worked example | Heat a 0.50 kg sample
A 60 W heater runs for two minutes. It heats a 0.50 kg sample with specific heat capacity 900 J/(kg °C). Assume all the heater’s energy reaches the sample. Find the rise from a starting temperature of 22 °C.
| Quantity | Calculation | Result |
|---|---|---|
| Heater input | 60 × 120 | 7,200 J |
| Sample heat capacity | 0.50 × 900 | 450 J/°C |
| Temperature rise | 7,200 ÷ 450 | 16 °C |
| Final temperature | 22 + 16 | 38 °C |
The 16 °C is the change in temperature. The final reading is 38 °C. Multiplying 450 J/°C by the rise of 16 °C returns the original energy of 7,200 J.
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Practise with energy losses
Suppose the same heater runs for the same time, but only 75% of its input energy heats the sample. The sample receives 0.75 × 7,200 = 5,400 J. Its temperature rises by 5,400 ÷ 450 = 12 °C, reaching 34 °C.
You can check by taking 75% of the original 16 °C rise. For the same mass and specific heat capacity, the temperature rise is proportional to the energy absorbed.
If the question gives an observed rise of 12 °C and asks for the fraction of input energy absorbed, work in reverse. Calculate 450 × 12 = 5,400 J, then divide by 7,200 J.
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Frequently asked questions
Why do I use 120 seconds instead of 2?
A watt is a joule per second. Seconds give the input energy in joules.
What if the sample melts during heating?
Energy can also be transferred during a change of state. Apply the temperature-rise relationship to the portions of heating without a change of state.
Where does the other 25% go?
It can warm other parts of the setup or be transferred to the surroundings, depending on the experiment.
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