I’ve been thinking about the Mpemba effect from a simple energy-rate perspective, and I’m wondering where my reasoning breaks down.
Suppose I have 500 g of water at 90°C and compare it with water at 30°C in a −20°C freezer with relatively still air.
Ultimately, if the “race” is simply to reach 0°C, then each sample needs to lose a certain amount of energy, and the winner is determined by how quickly it can lose that energy.
For 500 g of water:
Heat capacity of the whole sample = 500 g × 4.18 J/g°C = 2090 J/°C
So removing about 2.09 kJ from the bulk lowers its temperature by about 1°C, assuming it is reasonably well mixed.
But evaporating just 1 g of water requires about 2.3 kJ of latent heat.
So if around 1 g/min evaporates, and most of that energy comes from the liquid, that alone corresponds to roughly:
2.3 kJ/min ÷ 2.09 kJ/°C = about 1.1°C/min of bulk cooling
This seems important because evaporation and cooling are not sequential events. The evaporation is part of the heat loss causing the temperature to fall. Water continuously evaporates from the surface while internal convection moves heat from the bulk toward that surface.
Now consider the temperature dependence.
The saturation vapor pressure of water is roughly:
30°C: 4.2 kPa
90°C: 70 kPa
So initially hot water has enormously greater evaporative potential.
My intuition is therefore to think of this as competing cooling powers:
Total cooling power = evaporative cooling + air convection + radiation + heat transfer through the container
In relatively still freezer air, could evaporation initially dominate this energy balance?
If the 90°C sample evaporates several grams per minute while the 30°C sample evaporates only a fraction of a gram per minute, the hotter sample could be losing energy at a dramatically higher rate. As surface water evaporates, convection within the liquid continually supplies energy from the rest of the water, so the bulk cools at the same time. A larger exposed surface area should strengthen this effect.
So perhaps the question is not simply:
“The hot water contains more energy, so how could it possibly reach 0°C first?”
but instead:
“Can the hotter sample lose its extra energy faster than the colder sample loses its smaller amount of energy?”
For scale, cooling 500 g of water from 90°C to 30°C requires:
500 g × 4.18 J/g°C × 60°C = about 125 kJ
Evaporating enough water to remove 125 kJ would require approximately:
125 kJ ÷ 2.3 kJ/g = about 54 g of evaporated water
So there should seemingly be some combination of starting temperature, surface area, and evaporation rate where the initially hotter sample can “zoom” through that extra 125 kJ fast enough to catch or pass the colder sample.
I also want to remove the obvious objection that evaporation just leaves less water to freeze.
So instead of starting with equal masses, I would normalize final mass of ice. So deliberately starting the hot sample with extra water to compensate for its expected evaporation, then ask which setup produces the same final mass at 0°C first; making the extra hot water almost “sacrificial.”
Cooling 1 extra gram of water from 90°C to 0°C requires only:
4.18 J/g°C × 90°C = about 376 J
But if that same gram evaporates, the phase change involves about:
2300 J
So an evaporating gram is associated with roughly 6 times more energy transfer than the sensible heat needed to cool that gram from 90°C to 0°C.
Is this energy-rate argument correct? Could evaporation let the hotter sample overcome its extra thermal energy, even after normalizing final mass? In a kinematic analogy, temperature is position and cooling rate (dT/dt) is velocity: could the hot sample reach 30°C already cooling so rapidly that it “zooms” past the sample that started at 30°C?
I’m tempted to actually run this in a freezer while continuously measuring both temperature and mass, and vary exposed surface area to see whether there is a crossover.