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)
- "What are the relationship between *rates of cooling* of shaved and block ice with an identical mass and temperature?"
- "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.
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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