Specific Heat Capacity (Edexcel IGCSE Physics (Modular): Unit 1): Revision Note

Exam code: 4XPH1

Ashika

Written by: Ashika

Reviewed by: Caroline Carroll

Updated on

Specific heat capacity

  • The specific heat capacityc of a substance is defined as:

The amount of energy required to raise the temperature of 1 kg of the substance by 1 °C per kilogram of mass (J/kg °C)

  • Different substances have different specific heat capacities

    • If a substance has a low specific heat capacity, it heats up and cools down quickly (ie. it takes less energy to change its temperature)

    • If a substance has a high specific heat capacity, it heats up and cools down slowly (ie. it takes more energy to change its temperature)

Examples of specific heat capacity

Specific heat examples, downloadable AS & A Level Physics revision notes

Low vs high specific heat capacity

  • How much the temperature of a system increases depends on:

    • The mass of the substance heated

    • The type of material

    • The amount of energy put into the system in the form of thermal energy

Calculating specific heat capacity

  • The specific heat capacity of a substance can be calculated using the equatiion:

change in thermal energy  = mass × specific heat capacity  × change in temperature

ΔQ = mcΔT

  • Where:

    • ΔQ = change in thermal energy, in joules (J)

    • m = mass, in kilograms (kg)

    • c = specific heat capacity, in joules per kilogram per degree Celsius (J/kg °C)

    • ΔT = change in temperature, in degrees Celsius (°C)

Worked Example

Water of mass 0.48 kg is increased in temperature by 0.7 °C. The specific heat capacity of water is 4200 J / kg °C.

Calculate the amount of thermal energy transferred to the water.

 Answer:

Step 1: Write down the known quantities

  • Mass, m = 0.48 kg

  • Change in temperature, ΔT = 0.7 °C

  • Specific heat capacity, c = 4200 J/kg °C

Step 2: Write down the relevant equation 

ΔQ = mcΔT

Step 3: Calculate the thermal energy transferred by substituting in the values

ΔQ = (0.48) × (4200) × (0.7) = 1411.2

Step 4: Round the answer to 2 significant figures

ΔQ = 1400 J

Examiner Tips and Tricks

This equation will be given on your equation sheet, so don't worry if you cannot remember it, but it is important that you understand how to use it. You will always be given the specific heat capacity of a substance, so you do not need to memorise any values.

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Build on this topic

Ashika

Author: Ashika

Expertise: Physics Content Creator

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.