Two Flat Spots on the Heating Curve
A student places 100 g of crushed ice in a beaker on a 100 W hot plate and stirs while recording temperature every 30 seconds. The data show: a rise from −10 °C to 0 °C; a flat stretch at exactly 0 °C lasting about 6 minutes; a rise from 0 °C to 100 °C; and a long flat stretch at exactly 100 °C. The student expected steady warming and is confused that the temperature stops changing twice. Reference values for 100 g of this substance: melting requires about 3.3 × 10⁴ J, and boiling requires about 2.3 × 10⁵ J.
1. Explain why the temperature holds at 0 °C for six minutes even though the hot plate keeps delivering energy.
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During that stretch the beaker contains a mixture of ice and liquid water, and the temperature cannot rise until the last ice is gone.
Temperature measures the average kinetic energy of the particles. Energy arriving from the hot plate is not increasing particle kinetic energy; it is being used to break the attractions holding water molecules in the ice lattice and let them move past one another. That energy becomes potential energy of the arrangement, so the thermometer does not move.
So the flat region is not a pause in energy transfer. It is a period when all incoming energy is spent on a change of state rather than a change of temperature. The plateau ends only when the final bit of ice has melted; after that, additional energy raises the temperature again.
(The same reasoning applies at 100 °C, where the energy instead separates molecules completely during boiling.)
2. Check the 6-minute plateau against the rest of the data. Is it consistent with a 100 W hot plate melting 100 g of ice? Show your reasoning.
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A 100 W hot plate delivers 100 J every second.
$$E = Pt = (100\ \text{J/s})(6\ \text{min} \times 60\ \text{s/min}) = (100)(360) = 3.6 \times 10^{4}\ \text{J}$$
Melting 100 g of ice requires about 3.3 × 10⁴ J. The plate supplied about 3.6 × 10⁴ J in that time - slightly more than the minimum needed.
That small excess is expected and makes the data credible: some energy warms the beaker and stirrer, and some is lost to the surrounding air. A real apparatus would show a plateau slightly longer than the ideal calculation predicts, never shorter. A measured plateau of, say, 1 minute would signal a problem - too little energy for the mass of ice - while 6 minutes fits the equipment as described.
3. The plateau at 100 °C lasts much longer than the plateau at 0 °C. Explain why, using the reference values.
Read the worked answer
The hot plate delivers energy at the same rate in both cases, so the duration of a plateau is set by how much energy that change of state requires.
$$\frac{2.3 \times 10^{5}\ \text{J}}{3.3 \times 10^{4}\ \text{J}} \approx 7$$
Boiling 100 g of water requires roughly seven times the energy needed to melt it. At a fixed 100 W, that means the 100 °C plateau should last about seven times as long - on the order of 40 minutes rather than 6.
The physical reason is that melting only loosens molecules enough to let them slide past each other, while boiling must pull them entirely apart and send them off as gas. Separating molecules completely takes substantially more energy than loosening the lattice, and the data reflect that difference directly.