Track energy through warming and melting

Start with an energy destination

The fictional sample begins as a solid at 20 °C. Before changing the controls, predict whether each added joule must raise its temperature. At a fixed pressure, a pure substance can absorb energy while solid becomes liquid at its melting temperature. During that interval, the amounts of the two phases change instead of the equilibrium temperature.

This model assigns a melting point of 60 °C, solid specific heat of 2 J/(g K), fusion heat of 150 J/g and liquid specific heat of 3 J/(g K). These are teaching values, not measured properties of water. The calculation assumes equilibrium and no energy losses, boiling or decomposition.

Build the calculation in three parts

For warming within one phase, q = mcΔT: q is heat added in joules, m is mass in grams, c is specific heat and ΔT is the temperature increase. Warming one gram from 20 to 60 °C costs 1 × 2 × 40 = 80 J. Warming 100 g costs 8,000 J.

Next, q = mL describes complete melting, where L is fusion heat per gram. Melting all 100 g costs another 15,000 J. The sample becomes entirely liquid at a cumulative 23,000 J, or 230 J/g. Only additional energy then warms the liquid. At 300 J/g, 70 J/g remains after melting, so the temperature is 60 + 70/3 = 83.333 °C.

At the initial widget setting of 150 J/g, 80 J/g warmed the solid and 70 J/g has melted part of it. Liquid fraction is 70/150 = 0.4667. A 100 g sample therefore contains about 46.667 g liquid and 53.333 g solid. Its temperature remains 60 °C.

Read the curve and test the boundary cases

The horizontal axis is heat added per gram, not time. A flat portion does not mean heating stopped. Compare 80, 150 and 230 J/g: the endpoints mark melting onset and completion; the middle point has both phases. The energy bars partition total input into warming solid, melting and warming liquid. Do not add the total bar to its own components.

Change mass from 100 to 200 g while keeping 150 J/g. Temperature and liquid fraction stay fixed, but total input doubles from 15 to 30 kJ. Equal temperatures do not mean equal total energy. Conversely, delivering the same total heat to different masses generally produces different specific energy inputs.

Melting changes the organization and interactions of particles; it need not break the covalent bonds inside molecules. Do not interpret this physical change as a chemical reaction. Mixtures, varying pressure, heat losses or nonequilibrium heating can depart from this ideal plateau.

Transfer: at 180 J/g and 75 g, calculate temperature, liquid mass and total input before trying the controls.

OpenStax phase transitions explains the physical basis of heating curves.