A gram of water will always act like a gram of water in its given state. And it is the basis of the VERY simple initial question. In the real world the question is theoretical, just as the snowball proposition, and, to my mind serves the same purpose, to illustrate the myriad real-world variables that have to be taken into consideration.
I'm talking about water. If we want to talk about salt water, hard water or carbonated water, then that's another discussion. You quite rightly brought up the effects of friction on a snowball traveling at a high rate of speed. As to our freezing water, others have brought up issues such as solubility of gases and minerals in the water which will affect the outcome.
My teacher might have tossed in the volume of the snowball as another red herring to drag us through a calculation of its density. He could have called it an ice cube instead of a snowball. His aim was to pose the snowball as an isolated system to test a)our ability to state kinetic energy (joules) in terms of equivalent calories and b) our ability to distinguish between velocity and momentum, and how those factors apply to such a case. If kinetic energy in joules is equal to the one half the mass (in kg) times the square of the velocity (in m/s), it all becomes simple, and the mass is irrelevant to the required answer. It works with one gram (not much of a snowball!), 500 grams, 5 kilos (the head of a small snowman). They will all, in isolation, melt at the calculated velocity. The only difference will be the total number of joules/calories released to do the melting when it hits.
If I recall, those of us that got it doubled the mass of the snowball to 2kg for ease of calculation, set the required number of joules equivalent to 10,000 calories (to raise 2kg of water 5*C).
1/2*2*v-squared = 41,840 joules v-squared = 41,840 joules v = 41,840sq/rt v = 204.54 m/s
That works out to about 450 mph, which is a pretty peppy fastball, so I don't expect anybody I know will me meltin' a snowball on my forehead.
Those that didn't get the answer for the most part reduced that velocity to 51+m/s to get back down to the given 500g mass.
As someone else pointed out, water can exist in both liquid and solid states at the 0*C freezing point and there is an exchange of energy involved in the transition either way that didn't enter into our high school calculations, and there are other factors that get into it, but the point of the exercise was that mass, in isolation, is irrelevant, to the calculation of heat exchange.
Which gets to other points made. Freezing water in an ice cube tray is an entirely different proposition from freezing water in a pipe. It's not just a matter of the surface area which may hinder or enhance the rate of heat exchange, but also the effects of pressure within a pipe as opposed to water free to contract or expand.
I don't mean to be a pain in the neck. I'm kind of enjoying the whole thread. Theory vs. real world.
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