GCSE — physicsMedium4 min read

Internal Energy and Specific Latent Heat

Internal energy; Q = mL; specific latent heat of fusion and vaporisation

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# Internal Energy and Specific Latent Heat — GCSE Physics

When you heat a substance, its temperature rises — until it reaches a change of state. Then something interesting happens: you keep adding energy, but the temperature stays constant. Where does the energy go? The answer lies in understanding internal energy and latent heat.

1. Internal Energy

Internal energy is the total energy stored by the particles of a system: Internal energy=total kinetic energy+total potential energy of particles\text{Internal energy} = \text{total kinetic energy} + \text{total potential energy of particles}

  • Kinetic energy: due to particle motion (vibration, rotation, translation)
  • Potential energy: due to the forces (bonds) between particles

When You Heat a Substance

  • No change of state: Energy increases the KE of particles → temperature rises
  • During change of state: Energy increases the PE of particles (breaking/weakening bonds) → temperature stays constant

2. Specific Latent Heat

Specific latent heat (LL) is the energy needed to change the state of 1 kg of a substance without changing its temperature.

Q=m×L\boxed{Q = m \times L}

Where:

  • QQ = energy (J)
  • mm = mass (kg)
  • LL = specific latent heat (J/kg)

Two Types

Specific latent heat of fusion (LfL_f): Energy to change 1 kg from solid to liquid (or vice versa) at the melting point.

Specific latent heat of vaporisation (LvL_v): Energy to change 1 kg from liquid to gas (or vice versa) at the boiling point.

LvL_v is always much larger than LfL_f for the same substance because:

  • Boiling requires completely separating particles (breaking all bonds)
  • Melting only weakens the bonds (particles stay close together)

Values for Water

Property Value
LfL_f (fusion/melting) 334,000 J/kg
LvL_v (vaporisation/boiling) 2,260,000 J/kg

3. Heating Curves

A heating curve shows temperature vs. time (or energy) as a substance is heated from solid to gas.

  1. Solid phase: Temperature rises (particles gain KE, vibrate faster)
  2. Melting: Temperature constant at melting point (energy breaks bonds — solid → liquid). KE stays same, PE increases.
  3. Liquid phase: Temperature rises (particles gain KE, move faster)
  4. Boiling: Temperature constant at boiling point (energy breaks bonds — liquid → gas). KE stays same, PE increases.
  5. Gas phase: Temperature rises (particles gain KE)

The flat sections on the graph are where latent heat is being absorbed.

Worked Example: Example 1

Problem

Question: How much energy is needed to melt 3 kg of ice at 0°C? (Lf=334,000L_f = 334{,}000 J/kg)

Q=mL=3×334,000=1,002,000 J=1002 kJQ = mL = 3 \times 334{,}000 = 1{,}002{,}000 \text{ J} = 1002 \text{ kJ}

Solution

Worked Example: Example 2

Problem

Question: Calculate the energy needed to boil 0.5 kg of water already at 100°C. (Lv=2,260,000L_v = 2{,}260{,}000 J/kg)

Q=mL=0.5×2,260,000=1,130,000 J=1130 kJQ = mL = 0.5 \times 2{,}260{,}000 = 1{,}130{,}000 \text{ J} = 1130 \text{ kJ}

Solution

Worked Example: Complete Heating Problem

Problem

Question: Calculate the total energy to heat 2 kg of ice from −10°C to steam at 100°C. (cice=2100c_{\text{ice}} = 2100 J/kg°C, Lf=334,000L_f = 334{,}000 J/kg, cwater=4200c_{\text{water}} = 4200 J/kg°C, Lv=2,260,000L_v = 2{,}260{,}000 J/kg)

  1. Heat ice from −10°C to 0°C: E1=mcΔθ=2×2100×10=42,000E_1 = mc\Delta\theta = 2 \times 2100 \times 10 = 42{,}000 J
  2. Melt ice at 0°C: E2=mLf=2×334,000=668,000E_2 = mL_f = 2 \times 334{,}000 = 668{,}000 J
  3. Heat water from 0°C to 100°C: E3=mcΔθ=2×4200×100=840,000E_3 = mc\Delta\theta = 2 \times 4200 \times 100 = 840{,}000 J
  4. Boil water at 100°C: E4=mLv=2×2,260,000=4,520,000E_4 = mL_v = 2 \times 2{,}260{,}000 = 4{,}520{,}000 J

Total = 42,000+668,000+840,000+4,520,000=6,070,00042{,}000 + 668{,}000 + 840{,}000 + 4{,}520{,}000 = 6{,}070{,}000 J = 6070 kJ

Notice that the boiling step requires by far the most energy.

Solution

5. Practice Questions

    1. Define specific latent heat of fusion. (2 marks)
    1. Calculate the energy to melt 0.4 kg of ice. (Lf=334,000L_f = 334{,}000 J/kg) (2 marks)
    1. Explain why the temperature stays constant during boiling even though energy is being supplied. (3 marks)
    1. Why is LvL_v much greater than LfL_f for the same substance? (2 marks)
    1. A 2 kW heater is used to boil water. How long to convert 0.5 kg of water at 100°C to steam? (Lv=2,260,000L_v = 2{,}260{,}000 J/kg) (3 marks)

    Answers

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Summary

  • Internal energy = kinetic + potential energy of particles
  • Heating → higher temp (more KE) unless at change of state (then PE increases, temp constant)
  • Specific latent heat: Q=mLQ = mL
  • LfL_f = latent heat of fusion (melting/freezing)
  • LvL_v = latent heat of vaporisation (boiling/condensing)
  • Lv>>LfL_v >> L_f because vaporisation requires fully separating particles
Tags:
#gcse#physics#particle-model#internal-energy#latent-heat

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