Study Vault
All PostsFlashcardsResourcesAI ChatBlog
  1. Home
  2. /↳All Posts
  3. /↳Physics
  4. /↳Specific Heat Capacity and Latent Heat
Study VaultStudy Vault

Free, comprehensive study notes for CSEC students.

matthewlloydw@gmail.com

Navigate

  • Home
  • All Posts
  • Flashcards
  • Resources
  • AI Chat

Community

  • Contributors
  • Changelog
  • Suggest a Feature
  • My Suggestions
  • Bookmarks

Mathematics and Science

  • Mathematics
  • Additional Mathematics
  • Biology
  • Chemistry
  • Physics

Arts and Humanities

  • Caribbean History
  • Geography

Business and Human Development Studies

  • Principles of Accounts
  • Principles of Business
  • Economics

Modern Languages

  • English A (Language)
  • English B (Literature)
  • French
  • Spanish

Technical Studies and Creative Arts

  • Electrical & Electronic Technology
  • Information Technology

© 2026 Matthew Williams. Made with other contributors for all.

Physics

Specific Heat Capacity and Latent Heat

PDF
Matthew Williams
|May 20, 2026|7 min read
CSEC PhysicsLatent HeatPaper 01Paper 02Phase ChangeSection BSpecific Heat CapacityThermal Physics

Heat capacity, specific heat capacity (E = mcΔT), specific latent heat of fusion and vaporisation (E = mL), heating and cooling curves, and the distinction between evaporation and boiling.

Heat Capacity and Specific Heat Capacity

Different materials need different amounts of heat to reach the same temperature rise. This is described by two related quantities.

Heat capacity (CCC) of an object is the energy required to raise its temperature by 1 K (or 1 °C):

C=EHΔTC = \frac{E_H}{\Delta T}C=ΔTEH​​

Unit: J K⁻¹.

Specific heat capacity (ccc) of a substance is the energy required to raise the temperature of 1 kg of that substance by 1 K (or 1 °C):

c=EHmΔTc = \frac{E_H}{m \Delta T}c=mΔTEH​​

The working formula is:

EH=mcΔTE_H = mc\Delta TEH​=mcΔT

where EHE_HEH​ is heat energy in joules (J), mmm is mass in kilograms (kg), and ΔT\Delta TΔT is the temperature change in kelvin or degrees Celsius.

Unit of ccc: J kg⁻¹ K⁻¹.

The two quantities are related by: C=mcC = mcC=mc.

Selected Specific Heat Capacities

Substanceccc / J kg⁻¹ K⁻¹
Water4 200
Aluminium900
Copper400
Iron500
Air1 000

Water has the highest specific heat capacity of common liquids. A large mass of water (ocean, lake) can absorb or release enormous amounts of thermal energy with a small temperature change, this moderates coastal climates.

Example/Specific heat capacity from electrical heating (2023 Paper 02, Q1)

A 300 g liquid sample is heated by an electric heater rated at 105 W. Temperature readings are taken every 60 s. The temperature rises from 25.3 °C to 56.8 °C over 360 s.

Gradient of temperature-time graph:

G=ΔθΔt=56.8−25.3360−0=31.5360≈0.0875°C s−1G = \frac{\Delta\theta}{\Delta t} = \frac{56.8 - 25.3}{360 - 0} = \frac{31.5}{360} \approx 0.0875 \text{°C s}^{-1}G=ΔtΔθ​=360−056.8−25.3​=36031.5​≈0.0875°C s−1

Specific heat capacity (using G=P/mcG = P / mcG=P/mc):

c=PmG=1050.300×0.0875=1050.02625≈4 000J kg−1°C−1c = \frac{P}{mG} = \frac{105}{0.300 \times 0.0875} = \frac{105}{0.02625} \approx 4\,000 \text{J kg}^{-1} \text{°C}^{-1}c=mGP​=0.300×0.0875105​=0.02625105​≈4000J kg−1°C−1

Comparing with the table of liquids (cooking oil: 1700, paraffin: 2100, milk: 3900), the liquid is most likely milk.

Specific Latent Heat

When a substance changes state (melts, freezes, boils, or condenses), it absorbs or releases heat energy without any change in temperature. This hidden energy is the latent heat ("latent" means "hidden").

Specific latent heat (LLL) is the energy absorbed or released when 1 kg of a substance changes state without a change in temperature:

EH=mLE_H = mLEH​=mL

where EHE_HEH​ is in joules, mmm is mass in kilograms, and LLL is the specific latent heat in J kg⁻¹.

Two specific latent heats exist for each substance:

  • Specific latent heat of fusion (LfL_fLf​): energy to melt (or freeze) 1 kg of the substance at its melting point.
  • Specific latent heat of vaporisation (LvL_vLv​): energy to vaporise (or condense) 1 kg of the substance at its boiling point.

For water: Lf=3.4×105L_f = 3.4 \times 10^5Lf​=3.4×105 J kg⁻¹, Lv=2.26×106L_v = 2.26 \times 10^6Lv​=2.26×106 J kg⁻¹.

The latent heat of vaporisation of water is about 7 times its latent heat of fusion, it takes far more energy to convert water to steam than to melt ice.

Heating and Cooling Curves

When a pure substance is heated at a constant rate, a temperature-time graph shows flat regions during each phase change.

Heating curve for water: flat sections at 0 °C (melting) and 100 °C (boiling) represent latent heat being absorbed at constant temperature.

The slope of each rising section equals P/mcP / mcP/mc (rate of temperature rise at constant heating power). The flat sections have slope zero because the energy input goes into breaking intermolecular bonds rather than increasing kinetic energy.

Evaporation vs Boiling

Both processes convert liquid to vapour, but they differ in several ways:

PropertyEvaporationBoiling
LocationSurface of the liquid onlyThroughout the bulk of the liquid
TemperatureOccurs at all temperaturesOccurs at the boiling point only
RateSlow; depends on surface area, air movement, humidityRapid; requires continuous heat input
Temperature of remaining liquidFalls (faster molecules escape)Stays at the boiling point
BubblesNo bubbles formBubbles of vapour form inside the liquid

Evaporation cools the liquid because the fastest-moving molecules escape from the surface, lowering the average kinetic energy of those remaining. This is why sweating cools the body.

Example/Latent heat of vaporisation (2017 Paper 02, Q5)

An immersion heater rated at 150 W boils water for 5 minutes. The mass of water decreases from 0.28 kg to 0.26 kg.

Energy supplied by heater:

E=P×t=150×(5×60)=150×300=45 000JE = P \times t = 150 \times (5 \times 60) = 150 \times 300 = 45\,000 \text{J}E=P×t=150×(5×60)=150×300=45000J

Mass of water vaporised:

m=0.28−0.26=0.02kgm = 0.28 - 0.26 = 0.02 \text{kg}m=0.28−0.26=0.02kg

Specific latent heat of vaporisation:

Lv=EHm=45 0000.02=2 250 000J kg−1=2.25×106J kg−1L_v = \frac{E_H}{m} = \frac{45\,000}{0.02} = 2\,250\,000 \text{J kg}^{-1} = 2.25 \times 10^6 \text{J kg}^{-1}Lv​=mEH​​=0.0245000​=2250000J kg−1=2.25×106J kg−1

This is close to the accepted value of 2.26×1062.26 \times 10^62.26×106 J kg⁻¹ for water.

Exam Tip

In latent heat calculations, identify the mass that actually changes state, it is often the mass difference between the start and end of an experiment, not the total mass present.

On a heating curve, the flat region at 0 °C is melting (latent heat of fusion). The flat region at 100 °C is boiling (latent heat of vaporisation). Rising sections represent temperature change governed by specific heat capacity.

Previous in syllabus order
Gas Laws
Next in syllabus order
Heat Transfer