Internal Energy and Energy Transfers

Internal energy of a substance is the total energy stored in its particles — sum of:

1. Kinetic energy of every particle (their motion: vibration in solids; movement in liquids/gases).
2. Potential energy stored in the intermolecular bonds.

When you HEAT a substance, you transfer energy from a hotter source to its particles. This increased internal energy can show up as:

• Higher temperature (particles move faster) — if not at a state change.
• State change at constant temperature (particles overcome the bond energies) — if at a melting or boiling point.

Worked example. Heat a beaker of water from 20 °C to 100 °C: temperature rises; internal energy rises; particles move faster.

Continue heating at 100 °C: water boils; the temperature stays at 100 °C; internal energy continues to rise because bonds between molecules are being broken (the potential-energy part of internal energy increases).

When you heat a sample of ice at a constant power, a temperature-vs-time graph looks like:

• Sloped sections: temperature rising → kinetic part of internal energy rising → particles moving faster.
• Flat sections (at melting and boiling): all the added energy is breaking inter-particle bonds → potential part of internal energy rising; temperature unchanged.

Cooling curve is the same shape but downward — flat sections at freezing and condensation.

When energy is transferred to (or from) a substance and its temperature changes:

$\Delta E = m \, c \, \Delta\theta$

• $\Delta E$ in J.
• $m$ in kg.
• $c$ in J/kg°C — specific heat capacity of the substance.
• $\Delta\theta$ in °C — temperature change.

Typical SHC values:

Substance	c (J/kg°C)
Water	4200
Ice	2100
Air	1000
Aluminium	900
Iron	450
Copper	385
Lead	130

Water's value is unusually high — this is why oceans buffer climate, why water makes a good coolant, and why coastal UK has milder winters than central regions.

Worked example. Heat 0.6 kg of iron from 18 °C to 80 °C. $\Delta E = 0.6 \times 450 \times (80 - 18) = 270 \times 62 = 16{,}740\,\text{J}$

Worked example (find c). A 0.20 kg block warms from 20 °C to 70 °C when supplied with 9000 J. $c = \Delta E / (m \Delta\theta) = 9000 / (0.20 \times 50) = 900\,\text{J/kg°C}$ This matches aluminium.

Specific latent heat (L) is the energy required to change 1 kg of a substance from one state to another, at constant temperature:

$E = m \, L$

• $E$ in J.
• $m$ in kg.
• $L$ in J/kg.

Two types:

• Specific latent heat of fusion ($L_f$) — energy per kg to melt a solid (or released when freezing). Water: 334 kJ/kg.
• Specific latent heat of vaporisation ($L_v$) — energy per kg to boil a liquid (or released when condensing). Water: 2260 kJ/kg.

$L_v$ >> $L_f$ because going from liquid → gas requires breaking essentially ALL inter-molecular bonds, while melting only loosens them.

Worked example. How much energy melts 0.5 kg of ice? $E = 0.5 \times 334{,}000 = 167{,}000\,\text{J} = 167\,\text{kJ}$

Worked example. How much energy boils 0.2 kg of water (already at 100 °C)? $E = 0.2 \times 2{,}260{,}000 = 452{,}000\,\text{J} = 452\,\text{kJ}$

For problems that span both temperature changes AND state changes (e.g. heat ice to steam), split into stages:

1. Warm ice from below 0 °C to 0 °C: $\Delta E = m c_{\text{ice}} \Delta\theta$.
2. Melt the ice at 0 °C: $E = m L_f$.
3. Warm water from 0 °C to 100 °C: $\Delta E = m c_{\text{water}} \Delta\theta$.
4. Boil the water at 100 °C: $E = m L_v$.
5. Warm steam above 100 °C: $\Delta E = m c_{\text{steam}} \Delta\theta$ (rarely tested at GCSE).

Worked example. Convert 0.1 kg of ice at 0 °C to water at 100 °C.

• Step A (melt): $0.1 \times 334{,}000 = 33{,}400\,\text{J}$.
• Step B (heat liquid): $0.1 \times 4200 \times 100 = 42{,}000\,\text{J}$.
• Total: $75{,}400\,\text{J}$ ≈ 75 kJ.