Will a block of ice that is 12"x12"x12" melt slower than a block sized 20"x20"x20"?
If Yes, then how much faster in the same climate (90f/29c weather outside)
If Yes, then how much faster in the same climate (90f/29c weather outside)
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I concur with billrussell42 and Steve4Physics. But I detect that there are two ways to take the term, "slower."
Yes, the larger cube will take longer to melt. In that sense, it melts more slowly. So if you place both blocks in identical environments at the same time (and assuming that both of them start out at the same temperature, because either one can be at any temperature at or below 0ºC!), then the smaller block will reach the "finish line" first -- it will melt completely away while there is still some of the larger block unmelted.
But in that setup, in equal times, the larger block will produce more melt water, and thereby diminish more in mass of ice, so in that sense, it melts more quickly!
It's sort of like placing two runners on a 400-m track, runner A at the 100-m start line, runner B, who runs at twice the speed of A, starting at the 400-m start-finish line. Runner B runs faster, but will finish later, than runner A.
Yes, the larger cube will take longer to melt. In that sense, it melts more slowly. So if you place both blocks in identical environments at the same time (and assuming that both of them start out at the same temperature, because either one can be at any temperature at or below 0ºC!), then the smaller block will reach the "finish line" first -- it will melt completely away while there is still some of the larger block unmelted.
But in that setup, in equal times, the larger block will produce more melt water, and thereby diminish more in mass of ice, so in that sense, it melts more quickly!
It's sort of like placing two runners on a 400-m track, runner A at the 100-m start line, runner B, who runs at twice the speed of A, starting at the 400-m start-finish line. Runner B runs faster, but will finish later, than runner A.
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Thx 4 the BA! Glad you found it helpful.
Be sure to check out oldprof's answer; it gives a sense of how to go about figuring the melting times. I was considering trying to put something like that together, but he has beat me to it.
And actually, I think what you need there is the rate of heat
Be sure to check out oldprof's answer; it gives a sense of how to go about figuring the melting times. I was considering trying to put something like that together, but he has beat me to it.
And actually, I think what you need there is the rate of heat
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.. transfer per unit area from air to ice (which will be proportional to the ∆T=29K, and so, equal for both blocks), along with the heat of fusion of water-ice (80kCal/kg). And the blocks must be suspended in air, not sitting on ground or pavement or in sunlight, which would complicate the problem.
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