Track energy through a phase change

A flat thermometer reading can accompany energy transfer

At 80, 150 and 230 J/g, the heating investigation reports the same 60 °C temperature. Compare the solid and liquid masses at these three settings. A constant temperature does not imply that energy stopped entering: the phase fraction changes.

The fictional pure material begins solid at 20 °C. Its solid specific heat is 2 J/(g K), melting point is 60 °C, latent heat of fusion is 150 J/g and liquid specific heat is 3 J/(g K). Pressure is fixed and each displayed state is an equilibrium state. There is no boiling, decomposition or heat loss in the modeled range. These are assigned properties, not water data.

Divide the process into three energy accounts

First warm the solid. The energy per gram required to reach the melting point is 2 × (60 − 20) = 80 J/g. For 100 g this is 8000 J. At exactly 80 J/g the solid has reached melting onset; the liquid fraction is zero.

Next melt the material. For melted mass m, Qmelt = mLf, where Lf is latent heat of fusion in J/g. Melting all 100 g requires another 15000 J. This energy changes the material’s microscopic arrangement and interactions without raising its equilibrium temperature. The total supplied energy at melting completion is therefore 8000 + 15000 = 23000 J, or 230 J/g.

Finally warm the liquid. Above 230 J/g, only the excess energy warms the liquid in this model. At 260 J/g the excess is 30 J/g. Dividing by liquid specific heat 3 J/(g K) gives a 10 K rise, so temperature is 70 °C. For 100 g, the ledger is 8 kJ warming solid + 15 kJ melting + 3 kJ warming liquid = 26 kJ.

These steps are energy accounting, not three separate substances or three fixed time intervals. A different material, pressure or phase history could require a different model.

Calculate a partial melt

At 150 J/g, the solid warming has used 80 J/g, leaving 70 J/g for melting. The liquid mass fraction is 70/150 = 0.4667, approximately 46.67%. A 100 g sample therefore contains about 46.67 g liquid and 53.33 g solid, all at 60 °C.

In the melting interval, if e is total supplied energy per gram, the liquid fraction is f = (e − 80)/150, with 80 ≤ e ≤ 230 J/g. The liquid mass is f times total mass. Below onset f=0; above completion f=1. Do not extend the interval formula to claim negative liquid mass or more liquid than the sample contains.

At fixed J/g, doubling mass doubles every total energy and phase mass but leaves temperature and phase fraction unchanged. At fixed total energy, doubling mass reduces energy per gram and can change both phase and temperature. Explain the controlled quantity before comparing samples.

Read the axes honestly

The horizontal axis is supplied energy per gram, not time. A flat segment says temperature is constant as energy is added; it does not directly tell you a melting duration. The sloped segments have slope ΔT/Δe = 1/c, so a larger specific heat gives a gentler slope on these axes.

If an ideal heater supplied a constant net power P with no losses, elapsed time for an energy interval Q would be t = Q/P. For the 100 g sample’s 15000 J melting interval and a net 50 W supply, the duration would be 300 s. This is a conditional calculation, not a time trace produced by the widget. Real heater input and net sample heating can differ because the container and surroundings also receive energy.

Connect the plateau to measurements

In a physical investigation, record temperatures together with measured input energy or power and elapsed time. Identify sensor lag, imperfect mixing, container capacity and heat losses. A rounded or sloped experimental plateau does not automatically disprove phase change; neither does one flat reading prove melting. Independent observations of phase and repeatable energy accounts strengthen the explanation.

Compare this lesson’s ledger with the shared chemistry heating lesson only when you want a different subject context; your physics notebook remains separate.

Transfer to a different material

A 60 g solid begins 15 K below its melting point. Its solid specific heat is 1.5 J/(g K) and latent heat is 120 J/g. It receives 4950 J with no losses. Determine whether it is still warming, partially melted or fully liquid, and calculate the phase masses.

Check the latent-heat transfer

Warming to the melting point needs 60 × 1.5 × 15 = 1350 J. The remaining 3600 J melts 3600/120 = 30 g. The sample is half liquid and half solid at the melting point. Full melting from the original state would require 1350 + 60 × 120 = 8550 J. Applying Q = mcΔT to all 4950 J would incorrectly skip the phase-change energy.