Skip to content
IGCSE Physics, Cambridge 0625, Malaysia
Core + Supplement sections

Specific Heat Capacity

Specific heat capacity for IGCSE Physics 0625: the ΔE = mcΔθ equation, a worked calculation with mark scheme, and the experiment for Paper 6.

Written by IGCSEPhysics Content Team · Physics subject adviser: K. S. Tan, 15+ years teaching IGCSE Physics · Checked against the Cambridge IGCSE Physics (0625) 2026 to 2028 syllabus

This lesson explains Specific Heat Capacity for Cambridge IGCSE Physics 0625. It separates the Core requirements from the additional Supplement work for Extended candidates. Pay close attention to equation choice, unit conversion and the final sense check. Work through the example before testing the same skill without notes.

What is specific heat capacity and what is the equation?

Specific heat capacity is the energy required per kilogram to raise the temperature of a substance by 1 °C. In words: energy=mass×specific heat capacity×temperature change\text{energy} = \text{mass} \times \text{specific heat capacity} \times \text{temperature change}. In symbols: ΔE=mcΔθ\Delta E = mc\Delta\theta.

QuantitySymbolUnit
Energy transferredΔE\Delta Ejoule (J)
Massmmkilogram (kg)
Specific heat capacityccJ/(kg °C)
Temperature changeΔθ\Delta\theta°C

Water’s value is famously high: 4200 J/(kg °C). Heating 1 kg of water by 1 °C takes 4200 J. Metals sit far lower (copper is about 385 J/(kg °C)), which is why a copper pan heats faster than the water inside it. Increasing an object’s temperature increases the internal energy of the object: the particles’ average kinetic energy rises.

Why does water’s high specific heat capacity matter?

It makes water an excellent coolant and a climate moderator. Car radiators use water because each kilogram absorbs a lot of energy for a small temperature rise. Coastal areas, including most of Peninsular Malaysia, have milder temperature swings because the sea warms and cools slowly. Either example earns application marks.

Worked example

A kettle transfers 126 000 J of energy to 0.50 kg of water. The water starts at 30 °C. The specific heat capacity of water is 4200 J/(kg °C). Calculate the final temperature of the water. Ignore energy losses. [4]

Worked solution:

  1. Equation: ΔE=mcΔθ\Delta E = mc\Delta\theta
  2. Rearrange: Δθ=ΔE÷(mc)\Delta\theta = \Delta E \div (mc)
  3. Substitute: Δθ=126000÷(0.50×4200)=126000÷2100\Delta\theta = 126\,000 \div (0.50 \times 4200) = 126\,000 \div 2100
  4. Δθ=60 °C\Delta\theta = 60\ \text{°C}, so final temperature =30+60=90 °C= 30 + 60 = 90\ \text{°C} (2 significant figures)

Original marking points:

  • M1: ΔE=mcΔθ\Delta E = mc\Delta\theta stated or correctly rearranged
  • M1: correct substitution: 126000÷(0.50×4200)126\,000 \div (0.50 \times 4200)
  • A1: Δθ=60 °C\Delta\theta = 60\ \text{°C}
  • A1: final temperature 90 °C with unit

These marking points belong to this original example. They are not an official Cambridge mark scheme.

Common errors and how to correct them

  • Substituting a single temperature instead of the change. Δθ\Delta\theta means final minus initial. A question giving 30 °C and 90 °C needs 60 in the equation, never 90.
  • Leaving mass in grams. 500 g must become 0.50 kg before substituting, or the answer is out by a factor of 1000.
  • Stopping at Δθ\Delta\theta when the question asks for the final temperature. Read the last line of the question again before moving on.
  • Quoting the unit as J/kg or J/°C. The full unit J/(kg °C) is required for definition marks.
  • In the experiment, ignoring energy lost to the surroundings. Mark schemes expect “measured cc is too high because some energy heats the surroundings, not the block”.

How to apply this in an exam

Write the standard equation first, substitute with units converted, then rearrange on paper, never in your head. Cambridge awards the method mark for a correct equation or substitution even when the arithmetic fails, so every visible line is insurance. The five-step routine (equation, substitute, rearrange, unit, significant figures) should become automatic through repeated practice.

Where this skill matters

All four written papers use this subtopic. Paper 2 tests the definition and one-step calculations. Paper 4 sets multi-step calculations like the worked example, often combining ΔE=mcΔθ\Delta E = mc\Delta\theta with electrical power (E=PtE = Pt) in one question. Core candidates focus on the qualitative link between energy and temperature rise; the full calculation with cc is Extended, so it stays off Papers 1 and 3. This is also a flagship practical: measuring c for a metal block with an immersion heater, thermometer, balance and joulemeter appears regularly on Papers 5 and 6, with questions on insulation, stirring and sources of error. Know the experiment as well as the equation.

Key concepts in Specific Heat Capacity

Work through each concept below. Every page explains the idea, the common exam mistakes and the calculation steps that earn marks.

Still unsure about Specific Heat Capacity?

A 0625 specialist can work through the student's current question and help identify which concept, calculation step or answer-writing skill needs attention.