Energy and particles · GCSE Physics
Specific heat capacity
Teacher-written GCSE Physics revision on specific heat capacity: ΔE = mcΔθ, the energy to raise 1 kg by 1 °C, water’s high value, and the required practical with power, time and heat loss.
Specific heat capacity is the energy to raise 1 kg by 1 °C. ΔE = mcΔθ. Water’s large c means lots of energy for a small temperature change.
The important bits
What you need to know
- 1
Specific heat capacity c is the energy needed to raise the temperature of 1 kg of a substance by 1 °C (or 1 K). Units: J/kg°C.
- 2
ΔE = mcΔθ, where ΔE is energy in joules, m is mass in kilograms, c is specific heat capacity, and Δθ is the temperature change in °C.
- 3
Water has a high specific heat capacity (about 4200 J/kg°C), so it stores a lot of energy for a small temperature rise — useful in heating systems, slow to boil.
- 4
Metals have much smaller c values, so they heat up and cool down more quickly for the same energy input.
- 5
Rearrange to find c: c = ΔE / (mΔθ). In the electrical practical, ΔE is often taken as Pt, with P in watts and t in seconds.
- 6
Heat loss to the air and the beaker makes the experimental c larger than the data-book value, because some of the measured energy did not stay in the sample.
- 7
Temperature change Δθ is final minus initial. Do not use the actual temperature in the formula unless you are finding a difference.
- 8
The same energy into a smaller mass produces a larger Δθ, which is why a thin pan of water boils faster than a full kettle.
Quotations worth analysing
Short evidence. Real method.
“ΔE = mcΔθ”
Mass in kilograms, energy in joules, temperature change — not the thermometer reading on its own. Mixing grams with joules inflates c by 1000.
“Water has a high specific heat capacity.”
That single fact explains radiators, climate moderation by oceans, and why a hot-water bottle stays warm. Link the number to the job.
“Experimental values of c are often too high because of energy lost to the surroundings.”
The joulemeter or P × t counts all electrical energy in. Not all of it raises the temperature of the block or the water.
Go deeper
c is a material property; Δθ is the story of this sample
Two kilograms of water need twice the energy of one kilogram for the same temperature rise, but c itself does not change. That is what “specific” means: per kilogram. Copper’s c is much smaller than water’s, so a copper pan heats quickly while the soup inside lags. In calculations, list m, c and Δθ before you multiply. If a question gives a starting temperature of 18 °C and a finishing temperature of 34 °C, Δθ is 16 °C, not 34 °C. Students who put 34 into the formula invent a fantasy heating job. If the question gives power and time, energy in is Pt only if you assume no losses. Say that assumption, then calculate.
Go deeper
The required practical is really an energy-account problem
An aluminium block of known mass is heated with an immersion heater. You measure time, current and potential difference (or read a joulemeter), and temperature rise with a thermometer in the hole, using oil for thermal contact. Then c = Pt / (mΔθ) or E / (mΔθ). Insulation around the block reduces losses but cannot remove them. The measured energy is therefore larger than the energy that stayed in the aluminium, so calculated c comes out high. That evaluation sentence is worth as much as the arithmetic. Using a lower current for longer, or lagging the block, improves the result. Do not polish the table and forget to mention heat loss.
See the idea in action
A 1500 W kettle heats 0.50 kg of water from 20 °C to 100 °C. c for water = 4200 J/kg°C. Useful energy = mcΔθ = 0.50 × 4200 × 80 = 168 000 J. Time if all that energy stayed in the water: t = E/P = 168 000 / 1500 = 112 s. In the kitchen it takes 140 s, so total energy from the mains = 1500 × 140 = 210 000 J. The extra 42 000 J heated the kettle and the air. Experimental c using 210 000 J would be 210 000 / (0.50 × 80) = 5250 J/kg°C, higher than 4200 because of those losses.
Exam technique
Turn knowledge into marks
Write ΔE = mcΔθ, convert mass to kilograms, use the temperature change not the thermometer reading, and if energy is electrical use E = Pt with time in seconds. Mention heat loss when evaluating a practical.
Common mistakes
Do not give these marks away
- 01
Using mass in grams in ΔE = mcΔθ, or putting the final temperature in instead of Δθ.
- 02
Leaving time in minutes when calculating electrical energy as P × t.
- 03
Ignoring heat loss, then wondering why the experimental c is larger than the data-book value.
How much energy is needed to raise 0.20 kg of water by 15 °C? c = 4200 J/kg°C.
A1260 J
B12 600 J
C56 000 J
D840 J
Show the answer
12 600 J. ΔE = mcΔθ = 0.20 × 4200 × 15 = 12 600 J. 1260 J is a factor-of-ten slip; 56 000 J used the wrong temperature or mass.
Quick questions
If this is the bit you searched
What does specific heat capacity mean?
The energy required to raise the temperature of 1 kg of a substance by 1 °C. In symbols, ΔE = mcΔθ.
Why is water used in heating systems?
Its high specific heat capacity means it can carry a large amount of energy without an enormous temperature change, then release that energy in radiators.
Why is experimental c often too high?
Some of the measured electrical energy heats the surroundings and the container, not only the sample, so ΔE in the formula is too large.
Is 1 °C the same size as 1 K for these calculations?
Yes. A temperature change of 10 °C is a change of 10 K. You do not need to convert 20 °C into kelvin to use ΔE = mcΔθ.