Why is it that an ordinary chocolate bar (e.g. 4 x 12 pieces) breaks easily where it is supposed to when the chocolate bar is at room temperature or higher but breaks "randomly" when the chocolate is cold? By breaking randomly I mean that it cracks outside the crevasses created at the factory.

Below is a piece of a chocolate bar which has cracked "randomly" outside the crevasses. enter image description here

Below is a picture of a bar which breaks as it is supposed to at the crevasse (the bar is likely at or above room temp).

Picture: Normal temperature chocolate bar breaks correctly

  • 2
    $\begingroup$ Ductile to brittle transition in the mechanical properties of the (complex) cacao-fat-sugar mixture. Compare room temperature (or refrigerated) butter to frozen butter. $\endgroup$ – Jon Custer Mar 7 '18 at 18:19
  • $\begingroup$ I get that the properties change when the temperature changes but I'm curious to know why the crevasses lose their significance in the breaking process when temperature gets lower. $\endgroup$ – Crebit Mar 8 '18 at 14:52

Good question. Yes you would expect stress to concentrate, the material to yield at the minimum sections in the material. Our everyday experience with other materials tells us that. But chocolate seems to be different.

Although I don't know the answer, I'll take a stab at a hypothesis:

When you cool the bar below some temperature you create random crystalline dislocation boundaries that are not necessarily in the crevasses. So these actually become the 'weak links'. At higher temperatures the chocolate becomes semi-fluid and the dislocations disappear; the bar becomes monolithic.

And to test this hypothesis - sorry not sure how to go about designing an experiment. You'll have to ask a chemist or materials engineer to answer that question.

  • $\begingroup$ That is something I was guessing as well. It is interesting that the changes caused by the colder temperature start to dominate the strength provided by plain thickness and the crevasses become almost irrelevant in the breaking process. I'll wait to see if more detailed answers arise before accepting yours. Thanks! $\endgroup$ – Crebit Mar 8 '18 at 14:51

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