The other day I was dipping out Neapolitan(strawberry, vanilla, & chocolate) ice cream and noticed that the chocolate ice cream tended to melt faster than that of the vanilla and strawberry. I couldn't remember in which order I dipped the ice cream so I decided to do a controlled experiment with it the next time I dipped it out.

I took three bowls of the same type/shape and dipped out the strawberry first, then vanilla, and chocolate last. Even dipping the chocolate ice cream last showed that the chocolate ice cream would melt faster than the strawberry or vanilla.

Neither the strawberry nor chocolate use artificial coloring, though the vanilla states that it does. Is there a reason as to why chocolate ice cream melts faster? Is it because of the darker coloring that attracts heat or perhaps the low melting temperature of chocolate in general?

  • $\begingroup$ The darker color hypothesis sounds reasonable, how can you change your experiment to test this hypothesis? $\endgroup$
    – Leonardo
    Dec 3, 2012 at 7:20
  • 1
    $\begingroup$ what about different melting points? $\endgroup$
    – mdpc
    Dec 3, 2012 at 20:05
  • $\begingroup$ @mdpc Could you explain, in your answer, why you think they have different melting points? Please include some chemistry concepts in your answer.. (I have converted to comment for now) $\endgroup$ Dec 4, 2012 at 2:50

2 Answers 2


It's a complicated problem and I'm going to add a hand-waving physics answer.

For most materials, at a given wavelength the albedo and and emissivity sum to roughly unity. So if something is close to white with an albedo of 0.9, it's emissivity in the visible range is about 0.1, and if something is dark, those are reversed with albedo ~0.1 and emissivity ~0.9.

However, in the thermal infrared many non-metallic materials have an emissivity of very roughly 0.9 no matter what color they are in the visible. Even glass and water which are transparent in the visible have a high infrared emissivity.

In direct sunshine a surface perpendicular to the light will receive about 1100 W/m^2 in the visible, and so chocolate ice cream will be heated much faster than vanilla. Strawberry is pink because it absorbs shorter wavelengths but there's not a lot of power there from the Sun's black-body spectrum, so strawberry is like vanilla in the Sun.

However on an overcast day, or indoors, the the light is roughly 50 to 100 time dimmer, or about 11 to 22 W/m^2

(You can prove that by shooting a photo at f/16 with a shutter speed of 1/ISO or 1/ASA. I'm dating myself, but if had ASA 64 film you'd shoot f/16 at 1/60th second outdoors. You'd get nothing if you left it that way and shot indoors (Sunny 16 rule)

So now we're only down to say 11 or 22 W/m^2 in the visible, but that's not all the light! The Stefan-Boltzmann law tells us

$$P = \sigma \epsilon T^4.$$

If you plug in 273K and 298K (0° and 25° C) and $\epsilon=0.9$, you get about 280 and 400 W/m^2. That means that you, the walls ceiling or clouds are heating your ice cream at 400 W/m^2 while it is only radiating 280 W/m^2 back, it's a difference of 100 W/m^2 or way more than the heating at visible wavelengths.

So I would propose that if the phenomenon happens indoors that it's not related to the color, and could be related to the recipe.

Maybe chocolate ice cream has more salt, depressing the melting point, and so it starts melting sooner because it melts at a lower temperature.

So I've just asked Does chocolate ice cream tend to have more salt than strawberry or vanilla?


This appears to be physics more than chemistry. A dark color means it emits less visible light but rather absorbes it with temperature rise. Have your next ice cream in a darker place as to visible and ir emisions to test this.

  • 2
    $\begingroup$ This is almost certainly a factor, but anecdotally I have also observed that chocolate ice cream is softer, even straight from the freezer. This would likely be due to some change in the mixture of substances in chocolate ice cream versus other flavors. I wouldn't go so far as to say it's any one thing, but it's a different recipe so it will have different properties. $\endgroup$
    – KeithS
    Jul 23, 2013 at 22:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.