Revision NotesExam QuestionsFlashcards
IBPhysicsHLRevision NotesThe Particulate Nature of MatterThermal Energy TransfersSpecific Heat Capacity

Syllabus Edition

First teaching 2023

First exams 2025

|

Specific Heat Capacity (HL IB Physics)

Revision Note

Katie M

Author

Katie M

Last updated

Specific Heat Capacity

  • The amount of thermal energy needed to change the temperature of an object depends on:
    • The change in temperature required ΔT, i.e. the larger the change in temperature the more energy is needed
    • The mass of the object m, i.e. the greater the mass the more energy is needed
    • The specific heat capacity c of the given substance, i.e. the higher the specific heat capacity the more energy is needed

Substance Specific Heat Capacity (J kg−1 K−1)
Water 4200
Ice 2200
Aluminium 900
Copper 390
Gold 130
  • The equation for the thermal energy transferred, Q, is then given by

Q space equals space m c increment T

  • Where:
    • m = mass of the substance (kg)
    • ΔT = change in temperature (K or °C)
    • c = specific heat capacity of the substance (J kg–1 K–1)
  • The specific heat capacity of a substance is defined as:

The amount of energy required to change the temperature of 1 kg of a substance by 1 K (or 1°C)

  • This definition can be explained when the above equation is rearranged for c:

c space equals space fraction numerator Q over denominator m increment T end fraction

  • This means the higher the specific heat capacity of a substance, the longer it takes for the substance to warm up or cool down
    • Note that the specific heat capacity is measured in J kg–1 K–1

Specific heat capacity of water and copper

Specific Heat Capacity Comparison of Water and Copper, for IB Physics Revision Notes

The high specific heat capacity of water means it heats up and cools down much slower than metals, such as copper

Worked example

A piece of copper of mass 50 g is heated until it reaches a temperature of 120 °C. The copper is removed from the heat and immediately placed into 250 mL of water at 25 °C.

The temperature of the water and copper is measured until they reach thermal equilibrium.

Determine the final temperature of the copper and water, in degrees Celsius (°C), assuming no heat is lost to the surroundings.

  • The specific heat capacity of water is 4200 J kg–1 K–1
  • The specific heat capacity of copper is 390 J kg–1 K–1

Answer:

Step 1: Write down the known quantities 

  • Mass of copper, mc = 50 g = 0.05 kg
  • Mass of water, mw = 250 ml = 0.25 kg (since 1 litre = 1 kg)
  • Initial temperature of copper, Tc = 120 °C
  • Initial temperature of water, Tw = 25°C
  • Specific heat capacity of water, cw = 4200 J kg–1 K–1
  • Specific heat capacity of copper, cc = 390 J kg–1 K–1

Step 2: Write down the equation for thermal energy 

Q space equals space m c increment T

Step 3: Equate the equations for the energy transferred from the copper to the water

  • The copper is at a higher initial temperature than the water, hence thermal energy will be transferred from the copper to the water
  • The energy lost by the copper = the energy gained by the water

negative Q subscript c space equals space Q subscript w

negative m subscript c c subscript c increment T subscript c space equals space m subscript w c subscript w increment T subscript w

Step 4: Determine the final temperature Tf of the copper and water

  • Since the water and copper reach thermal equilibrium, their final temperature Tf will be the same

negative m subscript c c subscript c open parentheses T subscript f space minus space 120 close parentheses space equals space m subscript w c subscript w open parentheses T subscript f space minus space 25 close parentheses space

negative 0.05 space cross times space 390 space cross times space open parentheses T subscript f space minus space 120 close parentheses space equals space 0.25 space cross times space 4200 space cross times space open parentheses T subscript f space minus space 25 close parentheses space

2340 space minus space 19.5 T subscript f space equals space 1050 T subscript f space minus space 26 space 250

1069.5 space T subscript f space equals space 28 space 590

T subscript f space equals space fraction numerator 28 space 590 over denominator 1069.5 end fraction space equals space 26.7 space degree straight C

Examiner Tip

You should notice that changes in temperature ΔT can usually be written in degrees Celsius (although this is not the SI base unit for temperature) and do not need to be converted into kelvin (K). This is because differences in absolute temperatures always correspond to differences in Celsius temperature.

If the question asks to determine the initial or final temperature of a substance, make sure you always check the unit of measure (°C or K) in which you are required to give your final answer.

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

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

the (exam) results speak for themselves:

    Test yourselfFlashcards

    Did this page help you?

    Previous:Changes of StateNext:Specific Latent Heat
    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.