Specific Latent Heat

Put a thermometer in a pan of melting ice and you will see something surprising: while the ice is melting, the temperature stays stubbornly at $0\,°\text{C}$0°C, even though the hob is pouring energy in. The temperature only starts to climb once all the ice has melted. Where is that energy going? The answer is latent heat — the "hidden" energy needed to change a substance from one state to another without changing its temperature. This lesson, part of Topic P1 (Matter) of OCR Gateway Combined Science A, defines specific latent heat, works through the equation $E = mL$E=mL, distinguishes the latent heat of fusion from that of vaporisation, explains why the temperature stays constant during a change of state, and reads the flat plateaus of full heating and cooling curves.

By the end of this lesson you should be able to define specific latent heat, use and rearrange $E = mL$E=mL, distinguish the specific latent heats of fusion and vaporisation, explain why temperature stays constant during a change of state, and interpret heating and cooling curves.

This lesson is AO1 for defining specific latent heat and distinguishing fusion from vaporisation, strongly AO2 for applying and rearranging $E = mL$E=mL and explaining constant-temperature changes with the particle model, and AO3 for interpreting heating and cooling curves and comparing the energies of successive stages.

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What Is Latent Heat?

Latent heat is the energy needed to change the state of a substance while its temperature stays constant. The word latent means "hidden", because this energy does not show up as a temperature rise — the thermometer reading does not move while the change of state is happening.

The specific latent heat ($L$L) of a substance is the energy needed to change the state of $1 \text{ kg}$1 kg of the substance with no change in temperature. Its unit is $\text{J/kg}$J/kg (joules per kilogram).

There are two specific latent heats, one for each change of state:

• the specific latent heat of fusion, $L_f$Lf​ — the energy to melt (or freeze) $1 \text{ kg}$1 kg at the melting point;
• the specific latent heat of vaporisation, $L_v$Lv​ — the energy to boil (or condense) $1 \text{ kg}$1 kg at the boiling point.

("Fusion" is the scientific word for melting, and "vaporisation" for boiling.)

Exam Tip: Specific latent heat is the energy to change the state of $1 \text{ kg}$1 kg with no temperature change. Compare this with specific heat capacity, which is the energy to raise the temperature of $1 \text{ kg}$1 kg by $1\,°\text{C}$1°C — latent heat changes the state, capacity changes the temperature.

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The Specific Latent Heat Equation

The energy needed to change the state of a substance is:

$E = mL$E=mL

where $E$E is the energy transferred (in $\text{J}$J), $m$m is the mass (in $\text{kg}$kg) and $L$L is the specific latent heat (in $\text{J/kg}$J/kg). You use $L_f$Lf​ for melting or freezing and $L_v$Lv​ for boiling or condensing.

The equation rearranges to:

$E = mL \qquad m = \dfrac{E}{L} \qquad L = \dfrac{E}{m}$E=mLm=LE​L=mE​

Notice there is no $\Delta\theta$Δθ in this equation — because the temperature does not change during a change of state, only the mass and the latent heat matter.

Substance / change	Specific latent heat / $\text{J/kg}$J/kg
Ice melting (fusion of water)	$3.34 \times 10^{5}$3.34×105
Water boiling (vaporisation of water)	$2.26 \times 10^{6}$2.26×106

Worked Example 1 — Energy to melt ice

How much energy is needed to melt $2 \text{ kg}$2 kg of ice at $0\,°\text{C}$0°C? (Specific latent heat of fusion of water $= 3.34 \times 10^{5} \text{ J/kg}$=3.34×105 J/kg.)

Step 1 — write the equation: $E = mL$E=mL.

Step 2 — substitute: $E = 2 \times 3.34 \times 10^{5}$E=2×3.34×105.

Step 3 — calculate: $E = 6.68 \times 10^{5} \text{ J}$E=6.68×105 J (= $668\,000 \text{ J}$668000 J).

Answer: $6.68 \times 10^{5} \text{ J}$6.68×105 J.

Worked Example 2 — Energy to boil water

Calculate the energy needed to boil away $0.5 \text{ kg}$0.5 kg of water already at $100\,°\text{C}$100°C. (Specific latent heat of vaporisation of water $= 2.26 \times 10^{6} \text{ J/kg}$=2.26×106 J/kg.)

Step 1 — write the equation: $E = mL$E=mL.

Step 2 — substitute: $E = 0.5 \times 2.26 \times 10^{6}$E=0.5×2.26×106.

Step 3 — calculate: $E = 1.13 \times 10^{6} \text{ J}$E=1.13×106 J.

Answer: $1.13 \times 10^{6} \text{ J}$1.13×106 J (= $1\,130\,000 \text{ J}$1130000 J). Notice how much more energy boiling needs than melting — vaporising completely separates the particles, which takes far more energy than just loosening them.

Worked Example 3 — Rearranging for mass

A freezer removes $1.002 \times 10^{6} \text{ J}$1.002×106 J of energy from water at $0\,°\text{C}$0°C to freeze it. What mass of ice is produced? (Specific latent heat of fusion $= 3.34 \times 10^{5} \text{ J/kg}$=3.34×105 J/kg.)

Step 1 — rearrange for $m$m: $m = \dfrac{E}{L}$m=LE​.

Step 2 — substitute: $m = \dfrac{1.002 \times 10^{6}}{3.34 \times 10^{5}}$m=3.34×1051.002×106​.

Step 3 — calculate: $m = 3 \text{ kg}$m=3 kg.

Answer: $3 \text{ kg}$3 kg of ice is produced. (Freezing releases the same latent heat that melting absorbs, so the same equation applies.)

Worked Example 4 — Rearranging for specific latent heat

It takes $84\,500 \text{ J}$84500 J to completely melt $0.25 \text{ kg}$0.25 kg of a substance at its melting point. Calculate its specific latent heat of fusion.

Step 1 — rearrange for $L$L: $L = \dfrac{E}{m}$L=mE​.

Step 2 — substitute: $L = \dfrac{84\,500}{0.25}$L=0.2584500​.

Step 3 — calculate: $L = 338\,000 \text{ J/kg} = 3.38 \times 10^{5} \text{ J/kg}$L=338000 J/kg=3.38×105 J/kg.

Answer: $3.38 \times 10^{5} \text{ J/kg}$3.38×105 J/kg — very close to the value for water, suggesting the substance could be ice.

Exam Tip: There is no temperature term in $E = mL$E=mL, so do not try to multiply by $\Delta\theta$Δθ. Keep mass in kilograms, and watch the powers of ten in the latent-heat values — write them in standard form to avoid dropping a zero.

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Why Vaporisation Needs So Much More Energy Than Fusion

One of the most striking facts in this topic is that boiling a kilogram of water takes almost seven times as much energy as melting a kilogram of ice. Understanding why deepens your grasp of the particle model and is exactly the sort of comparison a higher-mark question rewards.

When a solid melts, the particles do not move very far apart at all. They break free of their fixed positions in the lattice so that they can slide past one another, but they stay close together and touching — a liquid is only slightly less dense than the solid it came from. The forces of attraction between the particles are loosened, not fully broken, so only a modest amount of energy is needed. That is why the latent heat of fusion is comparatively small.

When a liquid boils, by contrast, the particles must be pulled completely apart, moving from touching one another to being far apart with large empty spaces between them. This means the forces of attraction between the particles have to be overcome entirely, not merely loosened — a much bigger job, requiring far more energy. On top of that, as the gas forms it expands enormously and has to push back the surrounding air to make room for itself, which takes still more energy. Both effects — fully separating the particles and pushing back the atmosphere — mean the latent heat of vaporisation is much larger than the latent heat of fusion. This is the particle-level reason the boiling plateau on a heating curve is so much longer than the melting plateau.

Everyday Consequences of Latent Heat

The large latent heat of vaporisation has consequences you meet every day. Steam scalds far more severely than boiling water at the same temperature. Both are at $100\,°\text{C}$100°C, so you might expect them to be equally dangerous — but when steam touches your skin it first condenses back to liquid water, and in doing so it releases its large latent heat of vaporisation directly into your skin, on top of the energy released as that water then cools. Boiling water only delivers the cooling energy, so it does far less damage. The hidden latent heat is what makes steam burns so serious.

Evaporative cooling is the flip side of the same idea. When sweat evaporates from your skin, it must absorb its latent heat of vaporisation from somewhere — and it takes that energy from your skin, cooling you down. This is your body's main way of losing heat on a hot day or during exercise, and it is why a breeze (which speeds evaporation) feels so refreshing. The same principle keeps a refrigerator cold: a fluid is deliberately evaporated inside the cold compartment, absorbing latent heat from the food, and is then condensed again outside the fridge to release that heat to the room.

Recognising that latent heat is absorbed when a substance changes to a more spread-out state (melting, boiling) and released when it changes to a more closely packed state (condensing, freezing) lets you explain all of these situations from a single idea.

Exam Tip: If asked why vaporisation needs more energy than fusion, say that melting only loosens the forces between particles (they stay touching) whereas boiling fully separates them (and the gas must push back the air). For steam burns, the key is that condensing steam releases its latent heat into the skin.

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