# Specific Heat Capacity

> A-Level Physics · CIE 9702
> Source: https://www.owlsprep.com/study/cie-9702-u20-specific-heat-capacity/

This sub-topic explains the relationship between thermal energy added to a substance and its resulting temperature change. You will learn key definitions, core calculations and experimental methods for measuring specific heat capacity, a common exam and practical topic.

**Prerequisites:** [Internal energy and temperature](https://www.owlsprep.com/study/cie-9702-u20-internal-energy-temperature/)

## Learning objectives

- Distinguish between heat capacity and specific heat capacity
- Calculate thermal energy changes using the relation $Q = mc\Delta T$
- Analyse experimental methods and evaluate sources of error

## Definitions and Core Relationship

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.

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

*Notation:* $c$

*Example:* Water has a specific heat capacity of $4200 \text{ J kg}^{-1} \text{K}^{-1}$

$$Q = mc\Delta T$$

Where $Q$ = thermal energy transferred, $m$ = mass of substance, $\Delta T$ = 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. $c_{\text{water}} = 4200 \text{ J kg}^{-1} \text{K}^{-1}$

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

## Measuring Specific Heat Capacity: Solids

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

> **Discrepancy Explanation**
>
> If heat is lost to the surroundings, the energy absorbed by the block is less than the total supplied. This means your calculated value of $c$ will be higher than the true value.

## Measuring Specific Heat Capacity: Liquids

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 ($c_{\text{copper}} = 400 \text{ J kg}^{-1} \text{K}^{-1}$). A 100 W heater runs for 2 minutes, and temperature rises by 10 °C. Calculate $c_{\text{oil}}$.

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. $$12000 = 400 + 2c_{oil} \implies c_{oil} = 5800 \text{ J kg}^{-1} \text{K}^{-1}$$

> **tip**
>
> Insulation is used in these experiments to reduce heat loss to the surroundings, which is the main source of error in this measurement.

## Common pitfalls

- **Wrong:** Using mass in grams instead of kilograms when $c$ is given in J kg⁻¹ K⁻¹
  - Why it fails: Units are inconsistent, leading to an answer 1000 times larger than the correct value
  - Correct: Always convert mass from grams to kilograms by dividing by 1000 before substitution
- **Wrong:** Forgetting to convert time from minutes to seconds when calculating electrical energy $E=Pt$
  - Why it fails: Power is measured in watts (joules per second), so time must be in seconds
  - Correct: Multiply time in minutes by 60 to convert to seconds before calculating energy
- **Wrong:** Converting ΔT from °C to K by adding 273, leading to an incorrect value
  - Why it fails: A change of 1 °C is equal to a change of 1 K, so only the difference matters
  - Correct: Use the difference in temperature in °C directly, it is numerically equal to ΔT in Kelvin
- **Wrong:** Claiming calculated c is lower than true value when heat is lost
  - Why it fails: Heat loss means less energy is absorbed by the substance than the energy we use in our calculation
  - Correct: State that the calculated value of c is higher than the true value when heat is lost to surroundings

## Cheatsheet

| Quantity | Symbol/Formula | Units | Notes |
| --- | --- | --- | --- |
| Specific heat capacity | $c = \frac{Q}{m\Delta T}$ | J kg⁻¹ K⁻¹ | Per 1 kg of substance |
| Heat capacity | $C = mc = \frac{Q}{\Delta T}$ | J K⁻¹ | For the whole object |
| Electrical energy | $Q = Pt = VIt$ | 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 |

## 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.

- [Specific Latent Heat](https://www.owlsprep.com/study/cie-9702-u20-specific-latent-heat/)
- [First Law of Thermodynamics](https://www.owlsprep.com/study/cie-9702-u20-first-law-of-thermodynamics/)

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