Specific heat capacity and specific latent heat

Specific heat capacity

ΔE = mcΔθ

• c is the specific heat capacity: the energy needed to raise the temperature of 1 kg of a substance by 1 K (or 1 °C), in J kg^-1 K^-1.
• A temperature change of 1 K is the same as a change of 1 °C.
• Water has a high value, about 4180 J kg^-1 K^-1, which is why it is used in heating systems.

Specific latent heat

ΔE = LΔm

• L is the specific latent heat: the energy needed to change the state of 1 kg of a substance without changing its temperature, in J/kg.
• Latent heat of fusion: solid to liquid. Latent heat of vaporisation: liquid to gas (much larger, because the molecules must be separated completely).

Heating curves

On a graph of temperature against time for a substance heated at a steady rate:

• sloping sections: the temperature rises (energy increases the molecules' kinetic energy)
• flat sections: the state changes at constant temperature (energy breaks bonds and increases the molecules' potential energy)

A steeper slope means a smaller heat capacity for that state.

Worked example

How much energy is needed to heat 0.50 kg of water from 20 °C to 100 °C and then boil it all away? (c = 4180 J kg^-1 K^-1, L = 2.26 x 10⁶ J/kg)

1. Heating: ΔE = 0.50 x 4180 x 80 = 167200 J
2. Boiling: ΔE = 0.50 x 2.26 x 10⁶ = 1130000 J
3. Total = 1297200 J

Answer: About 1.3 x 10⁶ J (1.3 MJ)

Check yourself

1. How much energy is needed to raise the temperature of 2.0 kg of water by 15 K? (c = 4180 J kg^-1 K^-1) Answer in kJ, to 3 s.f.
   Show answer

   125 kJ. ΔE = mcΔθ = 2.0 x 4180 x 15 = 125400 J, which is 125 kJ to 3 s.f.

2. Supplying 5000 J to a 0.40 kg metal block raises its temperature by 12 K. Find its specific heat capacity in J kg^-1 K^-1, to 3 s.f.
   Show answer

   1040 J kg^-1 K^-1. c = ΔE / (mΔθ) = 5000 / 4.8 = 1041.7, which is 1040 J kg^-1 K^-1 to 3 s.f.

3. How much energy is needed to melt 0.25 kg of ice at 0 °C? (L = 3.34 x 10⁵ J/kg) Answer in kJ.
   Show answer

   83.5 kJ. ΔE = LΔm = 3.34 x 10⁵ x 0.25 = 83500 J = 83.5 kJ.

4. 12 kJ of energy is supplied to 0.80 kg of aluminium (c = 900 J kg^-1 K^-1). Find the temperature rise in K, to 3 s.f.
   Show answer

   16.7 K. Δθ = 12000 / (0.80 x 900) = 12000 / 720 = 16.7 K.

5. A 2.0 kW kettle boils away 0.10 kg of water at 100 °C. How long does this take, in s? (L = 2.26 x 10⁶ J/kg)
   Show answer

   113 s. Energy = 0.10 x 2.26 x 10⁶ = 226000 J; t = E / P = 226000 / 2000 = 113 s.

6. Why does the temperature of ice stay at 0 °C while it melts, even though energy is being supplied?
   • The energy is lost to the surroundings
   • Ice cannot absorb energy
   • The energy breaks bonds, increasing the molecules' potential energy rather than their kinetic energy
   • The molecules' kinetic energy increases but their potential energy falls
   Show answer

   The energy breaks bonds, increasing the molecules' potential energy rather than their kinetic energy. Temperature depends on mean kinetic energy, which does not change during melting.

7. What is the unit of specific latent heat?
   • J/kg
   • J kg^-1 K^-1
   • J/K
   • W/kg
   Show answer

   J/kg. L = ΔE / Δm, so its unit is J/kg; no temperature change is involved.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson Edexcel IAL Physics Unit 5 (WPH15) question paper, June 2023: list of data, formulae and relationships

Last updated 7 November 2026. Spotted a mistake? Tell us.