Effect of Temperature on Change of State

The six changes of state. Matter moves between the solid, liquid and gas states in three matched reversible pairs:

• Melting (fusion): solid to liquid, on absorbing heat, at a fixed melting point.
• Freezing (solidification): liquid to solid, on releasing heat, at exactly the same temperature as the melting point.
• Boiling: liquid to gas throughout the bulk of the liquid, on absorbing heat, at a fixed boiling point.
• Condensation: gas to liquid, on releasing heat.
• Sublimation: solid to gas directly, skipping the liquid state entirely — camphor, naphthalene, ammonium chloride, dry ice and iodine all do this.
• Deposition: gas to solid directly, the reverse of sublimation.

Why melting and boiling points are fixed. For a given pure substance at a fixed (normal, atmospheric) pressure, the melting point and boiling point are characteristic constants — they do not drift with the size of the sample or how fast it is heated. This constancy is exactly why chemists use melting point and boiling point as a test of purity: an unexpectedly wide or shifted melting range is a sign of an impure sample.

Latent heat. During any change of state, heat is absorbed or released at constant temperature, and that heat does not raise kinetic energy — it changes the arrangement of particles, either overcoming the inter-particle force (melting, boiling, sublimation) or letting it pull particles back together (freezing, condensation, deposition). This heat, and only this heat, is called latent heat, and the quantity for a given mass follows a simple relation:

where: $Q$ is the heat absorbed or released (in joules), $m$ is the mass undergoing the change of state (in kilograms or grams), and $L$ is the specific latent heat of the substance for that particular change — fusion or vaporisation — expressed in joules per kilogram (or per gram), a fixed property of the substance at that pressure. There are two named specific latent heats: the latent heat of fusion, the heat needed to convert unit mass of solid at its melting point fully into liquid with no temperature change, and the latent heat of vaporisation, the heat needed to convert unit mass of liquid at its boiling point fully into gas with no temperature change. Class 9 chemistry does not require using this formula numerically — it is enough to state latent heat in words — but seeing the relation makes clear why more massive samples need proportionally more heat to melt or boil, even though the temperature at which that happens never changes. This is why a heating curve — temperature on the vertical axis, time on the horizontal axis, heat supplied at a steady rate — has a distinctive shape: it rises as a solid warms, goes perfectly flat at the melting point while the solid melts, rises again through the liquid range, goes flat a second time at the boiling point while the liquid boils, and rises once more once everything is gas. The two flat plateaus are the graphical signature of latent heat being absorbed while temperature holds still.

This section explains the mechanism behind the two flat plateaus, since the underlying physics — not a numerical calorimetry formula — is what the syllabus expects. A rise in temperature is nothing but an increase in the average kinetic energy with which particles vibrate, jostle or move. As long as heat supplied only speeds up this motion, the thermometer climbs, exactly as it does while ice warms from $-10\,^\circ\text{C}$ toward $0\,^\circ\text{C}$. At the melting point, something different starts happening. The particles in a solid are held in fixed positions by a strong inter-particle force. To let them slide past each other as a liquid, that force must be weakened — and doing so costs energy. Every joule of heat supplied at $0\,^\circ\text{C}$ goes into this rearrangement rather than into faster vibration, so the kinetic energy, and hence the temperature, does not move. Only once every last bit of solid has turned to liquid does additional heat resume raising the kinetic energy, and the thermometer starts climbing again. Exactly the same logic explains the plateau at the boiling point, except that boiling must overcome the inter-particle force almost completely, letting particles escape the liquid's grip and move freely as a gas — which is why the latent heat of vaporisation for a given substance is always far larger than its latent heat of fusion (for water, roughly seven times as much energy per gram). Reverse the process — cool a gas — and the same amount of heat is released at the same fixed temperatures as particles re-form the bonds they had lost, which is why freezing point equals melting point and condensation happens at the boiling point.

Question: A tray holds $2\ \text{kg}$ of ice already at its melting point, $0\,^\circ\text{C}$. Using the specific latent heat of fusion of ice, $L_f = 334{,}000\ \text{J/kg}$, find the heat needed to melt all of it completely into water still at $0\,^\circ\text{C}$. Then, given the specific latent heat of vaporisation of water, $L_v = 2{,}260{,}000\ \text{J/kg}$, find the heat needed to boil away $2\ \text{kg}$ of water already at $100\,^\circ\text{C}$ completely into steam.

Step 1. Identify which formula applies. Both stages happen at a fixed temperature — a change of state, not a temperature rise — so $Q = mL$ applies directly, with the mass in kilograms and $L$ in joules per kilogram for that specific change.

Step 2. Heat to melt the ice: $Q_{\text{fusion}} = m \times L_f = 2 \times 334{,}000 = 668{,}000\ \text{J} = 668\ \text{kJ}$.

Step 3. Heat to boil the water: $Q_{\text{vap}} = m \times L_v = 2 \times 2{,}260{,}000 = 4{,}520{,}000\ \text{J} = 4520\ \text{kJ}$.

Step 4. Compare the two. Boiling the same mass of water needs about $6.8$ times more heat than melting it — a direct consequence of vaporisation having to overcome the inter-particle force far more thoroughly than fusion does, exactly the reasoning given above. Answer: melting the $2\ \text{kg}$ of ice needs $668\ \text{kJ}$, and boiling the $2\ \text{kg}$ of water needs $4520\ \text{kJ}$ — none of this heat raises the temperature at all.

Competitive stretch

If the tray instead held $2\ \text{kg}$ of ice at $-10\,^\circ\text{C}$, the total heat to turn it fully into steam at $100\,^\circ\text{C}$ would be the sum of five stages: warming the ice, melting it ($668\ \text{kJ}$, found above), warming the water, boiling it ($4520\ \text{kJ}$, found above), and finally warming the steam. Class 9 chemistry does not require calculating the three warming stages (that needs $Q = mc\Delta T$, outside this chapter's scope), but recognising that the two flat-plateau stages already found are the two largest single contributions to the total is worth remembering for the picture as a whole.

• Calling any solid-to-gas change "evaporation." Evaporation is a liquid turning to gas at its surface, below the boiling point; solid straight to gas is sublimation, a different process with a different name.
• Assuming temperature rises steadily throughout heating. It stops rising completely during the two plateaus, even though heat keeps flowing in — that heat is latent heat, not temperature-raising heat.
• Treating melting and boiling points as vaguely variable for a pure substance. At fixed atmospheric pressure they are exact constants, which is precisely why they are used to test purity.
• Forgetting that latent heat of vaporisation is always much larger than latent heat of fusion for the same substance, since boiling must overcome the inter-particle force far more completely than melting does.

Ask the tutor: curious why melting ice always needs so much less heat than boiling the same mass of water — ask your AI tutor to compare the two latent heats step by step with real numbers.