I know that ice cubes that are larger melt slower because of their surface area. However, when you put more ice cubes in a cup, all the cubes melt slower than a cup with less cubes. I hope this makes sense.
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4$\begingroup$ Once the water is initially cooled down, the total melt rate is determined by heat transfer into the cup. $\endgroup$– Jon CusterCommented Dec 5, 2023 at 16:28
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3$\begingroup$ Is there water in the cup? If the cup contains only an ice-air mixture, is the cup on its side, so that the dense chilled air can flow out and be replaced by warm air? $\endgroup$– rob ♦Commented Dec 5, 2023 at 16:39
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2$\begingroup$ To answer this question, there is water in the cup and the cup is right side up. $\endgroup$– user386598Commented Dec 5, 2023 at 19:52
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1$\begingroup$ I think the space between cubes stacked is not useful surface area for absorbing heat, it's just one cold cube touching another cold cube, surrounded by cold air which @rob points out may not be able to circulate. So you essentially only have the the upper part of the cup to work with which is again a "surface" relative to a volume of ice $\endgroup$– Sidharth GhoshalCommented Dec 6, 2023 at 1:40
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$\begingroup$ When you spread it out sufficiently, the amount does not matter -> clue ;-). $\endgroup$– Peter - Reinstate MonicaCommented Dec 7, 2023 at 13:31
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3 Answers
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Water has heat of fusion about $ H_{fus} =80~\text{cal/g}$, this means that for melting 1 gram of ice you need about 80 calories of heat. Obviously having more ice (monolithic piece or fragmented pieces) needs more energy to melt it as $$ E_{melt} = H_{fus}\cdot m_{ice} \tag 1$$
If the environment supplies a constant rate of heat, then sure you need more time for melting that amount of ice. Unless, while adding more ice into a cup, at the same time you can increase ice contact surface area sharply, as heat transfers into the ice: $$\frac {dQ}{dt} \propto A \tag 2$$
This can be done for example by crushing the ice cubes into tiny lumps (or just put some fresh snow into the cup, but not recommended 😆🤣)
answered Dec 5, 2023 at 18:11
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10$\begingroup$ Also, the more ice you have, the less water you have (assuming we're filling the cup), so less energy is required to bring the water down to ice temperature. $\endgroup$ Commented Dec 5, 2023 at 20:48
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The reason is still basically a variation of the square cube law. A pile or cup of ice cubes have more volume to melt relative to the effective surface area to perform heat transfer compared to if you separate the cubes and lay them out. This is due to restricted airflow between cubes. Not unlike why you wear a sweater when you are cold.
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6$\begingroup$ You forgot an obvious pun about the ice cube law. $\endgroup$– pipeCommented Dec 6, 2023 at 11:52
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3$\begingroup$ @pipe rolls eyes. But then you would probably say I should have said "rolls ice". $\endgroup$– DKNguyenCommented Dec 6, 2023 at 15:18
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$\begingroup$ @pipe but ice cubes aren't always square, so it doesn't work. $\endgroup$ Commented Dec 6, 2023 at 18:31
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1$\begingroup$ @MarkRansom I would argue they are never square, seeing as how they are cubes not squares. $\endgroup$– DKNguyenCommented Dec 6, 2023 at 20:06
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Hey I think the statement: ice cubes that are larger melt slower because of their surface area is not truly correct. you know greater the surface (through which heat transfer occurs) lesser will be the time required for transferring heat. So if you increase the surface area of body (made of ice) keeping the volume constant the time required for it to melt will be lesser! It's actually not surface area but volume(mass) of ice cube which is responsible for it's slow melting. The aggregate of ice cubes is still a giant ice cube with small voids in it. It definitely contains more ice than aggregate of lesser ice cubes, hence takes more time.
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3$\begingroup$ Good catch. Yeah it's surface area relative to the volume. And surface area increases by the square (x^2)of the dimensions while volume increases by the cube (x^3) of the dimensions. So volume increases much faster than surface area as an object is scaled up equally in all three dimensions. $\endgroup$– DKNguyenCommented Dec 5, 2023 at 17:14