Heating curves

Turning 135 g of ice at −15 °C into steam at 120 °C takes 416 kJ, and 305 kJ of that goes into boiling alone.

Put a block of ice on a stove that gives heat at a steady rate and record its temperature. The graph of temperature against heat added is a heating curve. It is a staircase. The sloping parts are where one state warms up. The flat parts are where the state changes. On those plateaus the burner keeps adding heat, but the thermometer does not move.

Two formulas

On a sloping part the heat raises the temperature, and the amount follows q = m c ΔT: mass times specific heat times temperature change. Specific heat is the heat needed to warm 1 g by 1 °C. For ice it is 2.09 J/g °C, for liquid water 4.18 J/g °C and for steam 1.86 J/g °C. Liquid water needs about twice as much heat per degree as ice or steam.

On a flat part the heat changes the state, and the amount follows q = n ΔH: moles times enthalpy of the change. To get moles, divide the mass by the molar mass, 18.02 g/mol for water. Melting uses ΔH fusion, 6.01 kJ/mol. Boiling uses ΔH vaporization at the boiling point, 40.67 kJ/mol.

Working through the example

The table follows OpenStax. The steps in joules are turned into kilojoules before adding. OpenStax rounds each step and reports 416 kJ; adding unrounded values gives 415.4 kJ, which the lab model above shows. Look at how lopsided the numbers are. Boiling takes almost three quarters of the total, more than five times the 56.4 kJ needed to take the water all the way from 0 °C to 100 °C. On the curve this is the long plateau at 100 °C.

Reading the slopes

The sloping parts are not equally steep. On a slope the temperature rises by q ÷ (m c), so the smaller the specific heat, the steeper the line. Ice (2.09 J/g °C) and steam (1.86 J/g °C) warm about twice as fast per kilojoule as liquid water (4.18 J/g °C). In the lab model the ice and steam sections are short and steep, and the water section is long and gentle. Change the mass and every step grows or shrinks in proportion, so the shape of the curve stays the same and only the heat axis stretches.

The same method answers the reverse question. OpenStax asks how much heat 94.0 g of water at 80.0 °C gives out as it cools, freezes and then cools to ice at −30.0 °C. Cooling the water gives 94.0 × 4.18 × 80.0 = 31.4 kJ, freezing gives 94.0 ÷ 18.02 × 6.01 = 31.4 kJ and cooling the ice gives 94.0 × 2.09 × 30.0 = 5.9 kJ, for 68.7 kJ in all.

What a cook can learn from it

Once a pot of water is boiling, turning the burner up does not make the water hotter. The temperature stays at the boiling point; the extra heat only makes the water boil away faster. Lowering the flame to a gentle boil cooks food at the same temperature and saves fuel. A cooling curve is the same staircase run backwards: steam cools, condenses at 100 °C while giving out heat, water cools, freezes at 0 °C and then the ice cools.