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Weed_O_Whirler t1_jeah6ou wrote

The shaved ice would cool down the soup faster, because it would melt faster. But after both were completely melted, it would have cooled down the same.

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KRA2008 t1_jed09ku wrote

This is why I have a nerd rant in my back pocket all the time about those gigantic whiskey ice cubes. They're sold as cooling your whiskey without diluting it, and that's kind of true in that it dilutes it more slowly, but by diluting it more slowly it also cools it more slowly, meaning the whiskey is warmer while you're drinking it. On top of that many whiskey experts suggest a splash of water in your whiskey to begin with.

The important thing here is that the heat of enthalpy of water/ice is huge compared to the heat capacity of either liquid water or ice. Ice cools by melting!

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snakeskinrug t1_jeejlw3 wrote

I use those big cubes and like them precisely because they have less surface area than a handful of ice the same size and won't cool it as fast. You want the whiskey cooled, but not ice cold. You can always speed up the cooling if you want by swirling and increasing the convection rate.

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KRA2008 t1_jeer5tm wrote

Well that’s fine, that’s a preference thing, but it’s just false advertising to claim it works like conventional ice in every way but better.

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PyrrhoTheSkeptic t1_jeava9k wrote

Assuming that everything else is the same (ice is same temperature in both cases, not considering the effects of the surrounding area, the thermal properties of the bowl, etc.), they would cool it the same amount, though the shaved pieces would melt faster, cooling it faster. However, the soup would likely be cooler after the ice cube melted, than the soup after the shaved pieces melted, because more time would pass by then, so the soup would lose more heat to the surrounding area, to the air and through the bowl over that greater period of time. But that is including the effects of heat radiating into and out of the bowl and directly into the air over that greater period of time, and is not just a change in temperature from the ice cube.

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ICLab t1_jeauhkz wrote

This is slightly ambiguously worded, so I will address the wording in answering the question.

There are two ways to read what you're asking depending on how you interpret "cooling more". Normally, people would use cooling "more" to describe a lower equilibrium temperature (more change in temperature units regardless of time), and cooling "faster" to denote a greater rate of cooling (more change in temp per unit of time)

  1. "What are the relationship between *rates of cooling* of shaved and block ice with an identical mass and temperature?"
  2. "What are the equilibrium temperatures of the soup + shaved ice system and the soup + block ice system?"

The answer to question 1 is that the rate of cooling of the shaved ice will be greater than that of the block ice. This is due to the greater surface area of the shaved ice leading to more interactions per time. These interactions dictate the cooling effect on the soup, so more -> faster cooling rate.

To answer question 2, if the two samples of ice are identical mass and temperature (and composition), they will both have the same "heat capacity" and the total amount of cooling they both provide will be the same. Even though the block ice will take longer, when it finally finishes the soup will be the same temperature as that with the shaved ice added.

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Junder21 t1_jebkn4y wrote

Forgetting the naturally slower and longer time to cool with a cube in which room temperature would assist in its attempt to cool it down resulting in different temps.

At least I believe a cube sitting in the same temp room as the shaved ice would not be the same temp as the room temperature would take it’s time on heat loss too whilst it sits the prolonged period whereas after the shaved ice melts you take a bite compared to a ice cube and taking a bite after it melts. I dunno.

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