Specific Heat Capacity

Specific heat capacity measures how much energy a material needs to warm up. It produces one of the most predictable calculations in IGCSE Physics 0625, which is exactly why examiners set it almost every session. Dropping its marks hurts.

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: $\text{energy} = \text{mass} \times \text{specific heat capacity} \times \text{temperature change}$. In symbols: $\Delta E = mc\Delta\theta$.

Quantity	Symbol	Unit
Energy transferred	$\Delta E$	joule (J)
Mass	$m$	kilogram (kg)
Specific heat capacity	$c$	J/(kg °C)
Temperature change	$\Delta\theta$	°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 Exam Question

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

Mark scheme:

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

Common Mistakes

• Substituting a single temperature instead of the change. $\Delta\theta$ 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 $\Delta\theta$ 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 $c$ is too high because some energy heats the surroundings, not the block”.

Exam Technique Tip

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. We drill the same five-step routine as our calculation guide (equation, substitute, rearrange, unit, significant figures) until it is automatic.

How This Is Examined

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 $\Delta E = mc\Delta\theta$ with electrical power ($E = Pt$) in one question. Core candidates focus on the qualitative link between energy and temperature rise; the full calculation with $c$ 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.