Specific Heat Capacity

A-Level Physics· 15 min read

1. Definitions and Core Relationship★★☆☆☆⏱ 10 min

Different substances require different amounts of thermal energy to raise their temperature by a fixed amount, depending on their mass and material type. This relationship is quantified by the concept of specific heat capacity.

📘 Definition

Specific heat capacity

$c$

The amount of thermal energy required to raise the temperature of 1 kilogram of a substance by 1 Kelvin (or 1 °C). Units: $J kg^{- 1} K^{- 1}$

Example:

Water has a specific heat capacity of $4200 J kg^{- 1} K^{- 1}$

Q = mc\Delta T

Where $Q$ = thermal energy transferred, $m$ = mass of substance, $Δ T$ = change in temperature.

📐 Worked Example

Calculate the thermal energy required to raise the temperature of 0.5 kg of water from 20 °C to 100 °C. $c_{water} = 4200 J kg^{- 1} K^{- 1}$

1
List known values: $m = 0.5 kg$ , $c = 4200 J kg^{- 1} K^{- 1}$
2
Calculate temperature change:
3
$\Delta T = 100 - 20 = 80 \text{ K}$
4
Substitute into $Q = m c Δ T$ :
5
$Q = 0.5 \times 4200 \times 80 = 168000 \text{ J} = 168 \text{ kJ}$

Exam tip:

Always check that your mass units match the units of specific heat capacity. Most values given in exams use kg, so convert grams to kg before calculating.

2. Measuring Specific Heat Capacity: Solids★★★☆☆⏱ 15 min

A standard CIE practical uses an electrical heater to heat a solid block of known mass. We assume all electrical energy supplied by the heater is converted to thermal energy absorbed by the block (ignoring heat loss to surroundings for a basic calculation).

📐 Worked Example

A 1 kg aluminium block is heated by a 50 W heater for 10 minutes. Its temperature rises from 18 °C to 38 °C. Calculate the specific heat capacity of aluminium, assuming no heat loss.

1
Calculate total electrical energy supplied, $E = P t$ (power × time, time in seconds):
2
$E = 50 \times (10 \times 60) = 30000 \text{ J}$
3
Find temperature change: $Δ T = 38 - 18 = 20 K$
4
Rearrange $Q = m c Δ T$ to solve for $c$ :
5
$c = \frac{Q}{m\Delta T} = \frac{30000}{1 \times 20} = 1500 \text{ J kg}^{-1} \text{K}^{-1}$

3. Measuring Specific Heat Capacity: Liquids★★★☆☆⏱ 15 min

For liquids, we use an insulated calorimeter to hold the liquid, with the heater immersed directly in the liquid. If the question mentions the calorimeter, we must account for thermal energy absorbed by the calorimeter itself.

📐 Worked Example

0.2 kg of oil is placed in a 0.1 kg copper calorimeter ( $c_{copper} = 400 J kg^{- 1} K^{- 1}$ ). A 100 W heater runs for 2 minutes, and temperature rises by 10 °C. Calculate $c_{oil}$ .

1
Calculate total energy supplied:
2
$E = Pt = 100 \times (2 \times 60) = 12000 \text{ J}$
3
Total energy absorbed = energy to heat calorimeter + energy to heat oil:
4
$E = m_{cu}c_{cu}\Delta T + m_{oil}c_{oil}\Delta T$
5
Substitute values and solve:
6
$12000 = (0.1 \times 400 \times 10) + (0.2 \times c_{oil} \times 10)$
7
$12000 = 400 + 2c_{oil} \implies c_{oil} = 5800 \text{ J kg}^{-1} \text{K}^{-1}$

4. Common Pitfalls

Wrong move:

Using mass in grams instead of kilograms when $c$ is given in J kg⁻¹ K⁻¹

Why:

Units are inconsistent, leading to an answer 1000 times larger than the correct value

Correct move:

Always convert mass from grams to kilograms by dividing by 1000 before substitution

Wrong move:

Forgetting to convert time from minutes to seconds when calculating electrical energy $E = P t$

Why:

Power is measured in watts (joules per second), so time must be in seconds

Correct move:

Multiply time in minutes by 60 to convert to seconds before calculating energy

Wrong move:

Converting ΔT from °C to K by adding 273, leading to an incorrect value

Why:

A change of 1 °C is equal to a change of 1 K, so only the difference matters

Correct move:

Use the difference in temperature in °C directly, it is numerically equal to ΔT in Kelvin

Wrong move:

Claiming calculated c is lower than true value when heat is lost

Why:

Heat loss means less energy is absorbed by the substance than the energy we use in our calculation

Correct move:

State that the calculated value of c is higher than the true value when heat is lost to surroundings

5. Quick Reference Cheatsheet

Quantity	Symbol/Formula	Units	Notes
Specific heat capacity	$c = \frac{Q}{m Δ T}$	J kg⁻¹ K⁻¹	Per 1 kg of substance
Heat capacity	$C = m c = \frac{Q}{Δ T}$	J K⁻¹	For the whole object
Electrical energy	$Q = P t = V I t$	J	For heater experiments
Water	$c = 4200$	J kg⁻¹ K⁻¹	Common standard value
Copper	$c = 400$	J kg⁻¹ K⁻¹	Common calorimeter material

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Multiple choice calculation
2023 · 2
Experimental error analysis
2024 · 3
Practical measurement

What's Next

Understanding specific heat capacity is fundamental for further topics in thermodynamics, including latent heat of fusion and vaporization, and analysis of thermal energy transfer in closed and open systems. It is also a core practical skill regularly assessed in CIE A-Level Physics practical papers, so mastering experimental methods and sources of error here will help you with all other practical assessment questions. The energy-temperature relationship you learn here forms the basis for calculating heat exchange between substances at different temperatures, a common structured question in paper 2.